Application Of Integrals - Class 12 Commerce Maths - Extra Questions

The region bounded by the curves $$y={ x }^{ 2 },y=-{ x }^{ 2 }$$ and $${ y }^{ 2 }=4x-3$$ meets at $$(1,1)$$
Area at curve $$\displaystyle(OABCO)=2\left[ \int _{ 0 }^{ 1 }{ { x }^{ 2 } } dx-\int _{ \frac { 3 }{ 4 }  }^{ 1 }{ \left( \sqrt { 4x-3 }  \right)  } dx \right] $$
$$\displaystyle=2\left[ { \left( \frac { { x }^{ 3 } }{ 3 }  \right)  }_{ 0 }^{ 1 }-{ \left( \frac { { \left( 4x-3 \right)  }^{ \frac { 3 }{ 2 }  } }{ 3.\frac { 4 }{ 2 }  }  \right)  }_{ \frac { 1 }{ 4 }  }^{ 1 } \right] $$
$$\displaystyle=2\left( \frac { 1 }{ 3 } -\frac { 1 }{ 6 }  \right) =\frac { 1 }{ 3 } $$units

Consider the given lines.
$$2y+x=8, x=2, x=4$$

Area of the region bounded by these lines is
$$=2\times2+\dfrac{1}{2}\times 1\times 2$$
$$=4+1$$
$$=5$$

Hence, this is the answer.

Given that,

$${ x }^{ 2 }+{ y }^{ 2 }=1$$ is a unit circle while $${ (x-1) }^{ 2 }+{ y }^{ 2 }=1$$ is a circle with centre $$(!,0)$$ and radius $$=1$$ unit.
By solving $${ x }^{ 2 }+{ y }^{ 2 }=1$$ and $${ (x-1) }^{ 2 }+{ y }^{ 2 }=1$$
We have $${ (x-1) }^{ 2 }+{ x }^{ 2 }=1$$
$$\therefore$$ $$(x-1+x)(x-1-x)=0$$

$$y={x}^{2}+2$$ intersects Y-axis at $$(0,2)$$
Area of a region bounded by a Curve
$$y={x}^{2}+2$$, line $$y=x,x=3$$ and $$x=0$$ is $$A=\left| I \right| $$
where $$I=\int _{ 0 }^{ 3 }{ \left( { y }_{ 1 }-{ y }_{ 2 } \right)  } dx=\int _{ 0 }^{ 3 }{ \left( { x }^{ 2 }+2-x \right)  } dx={ \left[ \cfrac { { x }^{ 3 } }{ 3 } -\cfrac { { x }^{ 2 } }{ 2 } +2x \right]  }_{ 0 }^{ 3 }=9+6-\cfrac { 9 }{ 2 } $$

the area of the region is $$\int_{}{}tanx$$ from 0 to $$\pi/4$$

Put  $$y = {{{x^2}} \over 7}$$

In $${y^2} = 7x$$

Then $${{{x^4}} \over {49}} = 7x$$

$${x^4} - 343x = 0$$

$$x({x^3} - 343) = 0$$

$$x = 0,x = 7$$

$$x = 0,y = 0$$

$$x = 7,y = 7$$

$${\text{Now}}\;{\text{area}}\;\Delta {\text{ABC}}\;{\text{ = }}\;{\text{ar}}\left( {OABO} \right) - ar\left( {OCBO} \right)$$

$$\int_0^7 {\sqrt {7x} dx}  - \int_0^7 {{{{x^2}} \over 7}} dx$$

$${2 \over 3}\sqrt 7 \left( {{x^{3/2}}} \right)_0^7 - {1 \over 7}\left( {{{{x^3}} \over 3}} \right)_0^7$$

$${2 \over 3}\sqrt 7 \left( {{7^{3/2}}} \right) - {1 \over {21}}\left( {{7^3}} \right)$$

$${2 \over 3}\left( {49} \right) - {1 \over {21}} \times 343$$

$${{686 - 343} \over {21}} = {{343} \over {21}} = {{49} \over 3}$$

Because  $$ x^2 + y^2 = 9 ;  y^2= 8x$$

So $$x^2 + 8x = 9 $$

$$ x^2 + 8x – 9 =0 $$

$$ \implies (x+9)(x-1) =0$$

$$ \implies x=1,  x= -9 $$

Therefore $$y = \pm 2 \sqrt{2}$$

Point of intersections are $$p(1, 2\sqrt{2} , 1, -2 \sqrt{2} )$$

$$y_1 = \sqrt{8x} = 2 \sqrt{2} x^{1/2}$$

$$y_2 = \sqrt{9-x^2}$$

Area = Area OPAQO =  2 Area OPAMO

Area = 2 (Area OPMO + Area APMA)

$$ = 2 [ \int_{0}^{1} y_1 dx + \int_{1}^{3} y_2 dx]$$

$$ =2 [ \int_{0}^{1} 2\sqrt{2} x^{1/2} dx + \int_{1}^{3} \sqrt{ 3^2 – x^2} dx ]$$

$$ =2[ \dfrac{2 \sqrt{2}}{3/2} ( x^{3/2})_{0}^{1}  + ( \dfrac{x}{2} \sqrt{3^2 – x^2} + \dfrac{9}{2} \ sin^{-1} \frac{x}{3} )_{1}^{3}]$$

$$ = 2[ \dfrac{4 \sqrt{2}}{2} .1 + \dfrac{9}{2} \ sin^{-1} 1 - \dfrac{1}{2} \times \sqrt{8} - \dfrac{9}{2} \ sin^{-1} \dfrac{1}{3} ]$$

$$ = 2 [ \dfrac{\sqrt{2}}{3} + \dfrac{ 9 \pi}{4} - \dfrac{9}{2} \ sin^{-1} \dfrac{1}{3}] $$ sq. Unit 

Area of shaded region

Let AD represent the line $$y=3$$

$$y^2=4-x\implies y=0 $$ when $$x=4$$

Area of verandah$$=$$area of hall including verandah$$-$$area of hall

Area of AFGI = Area of CDHI 

Both the curves are defined for $$x>0$$
Both are positive when $$x>1$$ and negative when $$0<x<1$$
We know $$\displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ \left( \log { x }  \right) \rightarrow -\infty  } $$
Hence, $$\displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ \left( \log { x }  \right) \rightarrow -\infty  } $$ .Thus $$y$$-axis is asymptote of second curve.
And $$\displaystyle\lim _{ x\rightarrow { 0 }^{ + } }{ ex\log { x }  } ........ \left[ \left( 0 \right) \times \infty form \right] $$
$$\displaystyle=\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { e\log { x }  }{ \frac { 1 }{ x }  }  } \left( -\frac { \infty  }{ \infty  } form \right) $$$$\displaystyle =\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { e\left( \frac { 1 }{ x }  \right)  }{ \left( -\frac { 1 }{ { x }^{ 2 } }  \right)  }  } =0$$ (using L'hopital rule)
Thus, the first curve starts from $$(0,0)$$ but does not include $$(0,0)$$
Now, the given curves intersect.

Intersection points

From figure, the two curves represent  parabolas with vertices at $$ (0,0)$$ and $$(3,0)$$. They intersect at $$(1,1)$$ and $$(1,-1),$$ so the required area is 
here,

The area of the region bounded by the curve, $$y^2=x$$, the lines, $$x=1$$ and $$x=4$$, and the $$x$$-axis is the area $$ABCD$$.

The area of the region bounded by the curve, $$y^2=9x, x=2$$ and $$x=4$$, and the $$x$$-axis is the area $$ABCD$$.
Area of $$ABCD$$ $$=\displaystyle \int_2^4 y dx$$
$$=\displaystyle \int_2^43\sqrt x dx$$
$$=3\left [\frac {x^{\frac {3}{2}}}{\frac {3}{2}}\right ]_2^4$$
$$=2[x^{\frac {3}{2}}]_2^4$$
$$=2[(4)^{\frac {3}{2}}-(2)^{\frac {3}{2}}]$$
$$=2[8-2\sqrt 2]$$
$$=(16-4\sqrt 2)units$$

On solving, we get $$x = 0, \displaystyle \frac{a}{a^2 + 1}$$
Required area $$= \displaystyle \int_0^{\lambda/1 + \lambda^2} \left ( x - ax^2 - \frac{x^2}{a} \right ) dx$$
$$\displaystyle \Delta' = \frac{a (1 - a) (1 + a)}{(1 + a^2)^3}$$
$$\Rightarrow a = 1$$ gives maximum.

The line, $$x=a$$, divides the area bounded by the parabola and $$x=4$$ into two equal parts.
$$\therefore \mbox{Area}\ OAD=\text{Area} \ ABCD$$

The area of the smaller part of the circle, $$x^2+y^2=a^2$$, cut off by the line, $$x=\dfrac {a}{\sqrt 2}$$, is the are ABCDA.
It can be observed that the area ABCD is symmetrical about x-axis.
$$\therefore \text{Area} \ ABCD =2\times \mbox{Area} \ ABC$$

The area of the region bounded by the circle, 

The area of the region bounded by the curve $$x^2=4y, y=2$$ and $$y=4$$, and the $$y$$-axis is the area $$ABCD$$.
Area of $$ABCD$$ $$=\displaystyle \int_2^4x dy$$
$$=\displaystyle \int_2^4 2\sqrt y dy$$
$$=2\displaystyle \int_2^4\sqrt ydy$$
$$=2\left [\dfrac {y^{\frac {3}{2}}}{\frac {3}{2}}\right ]_2^4$$
$$=\dfrac {4}{3}[(4)^{\frac {3}{2}}-(2)^{\frac {3}{2}}]$$
$$=\dfrac {4}{3}[8-2\sqrt 2]$$
$$=\left (\dfrac {32-8\sqrt 2}{3}\right )units$$

$$BL$$ and $$CM$$ are drawn perpendicular to x-axis.
It can be observed in the following figure that,
$$Area (\Delta ACB)=Area (ALBA)+Area (BLMCB)-Area (AMCA)$$ ......... (1)
Equating of line segment $$AB$$ is
$$y-0=\dfrac {3-0}{1+1}(x+1)$$
$$y=\dfrac {3}{2}(x+1)$$

The given equations of lines are
$$2x+y=4 ......... (1)$$
$$3x-2y=6 ......... (2)$$
And, $$x-3y+5=0 .......... (3)$$
The area of the region bounded by the lines is the area of $$\Delta ABC$$. $$AL$$ and $$CM$$ are the perpendicular on $$x$$-axis.

The required area is represented by the shaded area $$OBCDO$$.
Solving the given equation of circle, $$4x^2+4y^2=9$$, and parabola, $$x^2=4y$$, we obtain the point of intersection as $$B\left (\sqrt 2, \dfrac {1}{2}\right )$$ and $$D\left (-\sqrt 2, \dfrac {1}{2}\right )$$.
It can be observed that the required area is symmetrical about $$y$$-axis.

The graph of $$y=\sin x$$ can be drawn as shown in the diagram.

$${\textbf{Step-1: Draw a suitable diagram with shaded area and find points.}}$$

                       

                     $$x = 0$$ $${\text{is line with points (0, 1) (0, 0).... etc}}$$

                     $$y = 1$$ $${\text{is line with points (0, 1) (1, 1).... etc}}$$

                     $$y = 4$$ $${\text{is line with points (0, 4) (1, 4).... etc}}$$

                     $${\text{Parabola meets at points (0, 0) with line}}$$ $$x = 0.$$

                     $$(\dfrac{1}{2}, 1)$$ $${\text{with line}}$$ $$y=1 (1, 4)$$ $${\text{with line}}$$ $$y = 4.$$ 

$${\textbf{Step-2: Find required area under parabola.}}$$                                                                                                                                      $$\text{At first calculate the area with respect to y axis}$$

                 $${\text{Required area (shaded one) = Area under parabola from}}$$ $$y = 1$$ $${\text{to}}$$ $$y =  4.$$

                 $${\text{Required Area(Shaded one)}}$$ $$ = \int_1^4 {xdy} $$

                                                                     $$ = \int_1^4 {\sqrt {\dfrac{y}{4}} dy} $$

                                                                     $$ = \dfrac{1}{2}\int_1^4 {{\kern 1pt} \sqrt y dy} $$

                                                                     $$ = \dfrac{1}{2}{\left[ {\dfrac{{{y^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}}} \right]_1}^4$$

                                                                     $$ = {\left[ {\dfrac{1}{2} \times \dfrac{2}{3}{y^{\dfrac{3}{2}}}} \right]_1}^4$$

                                                                     $$ = \dfrac{1}{3}\left( {{4^{\dfrac{3}{2}}} - {1^{\dfrac{3}{2}}}} \right)$$

                                                                     $$ = \dfrac{1}{3}\left( {8 - 1} \right)$$

                                                                     $$ = \dfrac{7}{3}sq.units$$

$${\textbf{Hence,Area under simple curve lying in the first quadrant is}}$$   $$ \mathbf{\dfrac{7}{3}sq.units.}$$

The area bounded by the curves, $$(x-1)^2+y^2=1$$ and $$x^2+y^2=1$$, is represented by the shaded area as
On solving the equations, $$(x-1)^2+y^2=1$$ and $$x^2+y^2=1$$, we obtain the point of intersection as $$A$$ $$\left (\dfrac {1}{2}, \dfrac {\sqrt 3}{2}\right )$$ and $$B\left (\dfrac {1}{2}, \dfrac {\sqrt 3}{2}\right )$$.
It can be observed that the required area is symmetrical about $$x$$-axis.
$$\therefore$$ Area $$OBCAO$$ $$=2\times Area \ OCAO$$
We join $$AB$$, which intersects $$OC$$ at $$M$$, such that $$AM$$ is perpendicular to $$OC$$.
The coordinates of $$M$$ are $$\left (\dfrac {1}{2}, 0\right )$$.
$$\Rightarrow Area\ OCAO= Area \ OMAO + Area\ MCAM$$
$$=\left [\displaystyle \int_0^{\frac {1}{2}}\sqrt {1-(x-1)^2}dx+\int_{\frac {1}{2}}^1\sqrt {1-x^2}dx\right ]$$
$$=\left [\displaystyle \frac {x-1}{2}\sqrt {1-(x-1)^2}+\frac {1}{2}\sin^{-1}(x-1)\right ]_0^{\frac {1}{2}}+\left [\dfrac {x}{2}\sqrt {1-x^2}+\dfrac {1}{2}\sin^{-1}x\right ]_{\dfrac {1}{2}}^1$$
$$=\left [-\displaystyle \frac {1}{4}\sqrt {1-\left (-\frac {1}{2}\right )^2}+\frac {1}{2}\sin^{-1}\left (\frac {1}{2}-1\right )-\frac {1}{2}\sin^{-1}(-1)\right ]+\left [\displaystyle \frac {1}{2}\sin^{-1}(1)-\frac {1}{4}\sqrt {1-\left (\frac {1}{2}\right )^2}-\frac {1}{2}\sin^{-1}\left (\frac {1}{2}\right )\right ]$$
$$=\left [-\displaystyle \frac {\sqrt 3}{8}+\frac {1}{2}\left (-\frac {\pi}{6}\right )-\frac {1}{2}\left (-\frac {\pi}{2}\right )\right ]+\left [\displaystyle \frac {1}{2}\left (\frac {\pi}{2}\right )-\frac {\sqrt 3}{8}-\frac {1}{2}\left (\frac {\pi}{6}\right )\right ]$$
$$=\left [-\dfrac {\sqrt 3}{4}-\dfrac {\pi}{12}+\dfrac {\pi}{4}+\dfrac {\pi}{4}-\dfrac {\pi}{12}\right ]$$
$$=\left [-\dfrac {\sqrt 3}{4}-\dfrac {\pi}{6}+\dfrac {\pi}{2}\right ]$$
$$=\left [\dfrac {2\pi}{6}-\dfrac {\sqrt 3}{4}\right ]$$
Therefore, required area $$OBCAO=2\times \left (\dfrac {2\pi}{6}-\dfrac {\sqrt 3}{4}\right )=\left (\dfrac {2\pi}{3}-\dfrac {\sqrt 3}{2}\right )$$ units

The vertices of $$\Delta ABC$$ are $$A(2,0), B(4,5)$$, and $$C(6,3)$$.

The equations of sides of the triangles are $$y=2x+1, y=3x+1$$, and $$x=4$$.
On solving these question, we obtain the vertices of triangle as $$A(0, 1), B(4, 13)$$, and $$C(4, 9)$$.
It can be observed that,
$$Area (\Delta ACB)=Area (OLBAO)-Area (OLCAO)$$
$$=\displaystyle \int_0^4(3x+1)dx-\int_0^4(2x+1)dx$$
$$=\left [\displaystyle \frac {3x^2}{2}+x\right ]_0^4-\left [\dfrac {2x^2}{2}+x\right ]_0^4$$
$$=(24+4)-(16+4)$$
$$=28-20=8$$sq. units.

(i)

Given curves are $$y^2=4x$$     ....(1)

Intersecting points of the two given curves: $$y^=4x$$ and $$x^2=4y$$ is $$(0,0)$$ and $$(4,4)$$

The point of intersection of the parabola and the line is given by
$$4x-x^{2}=5-2x$$
$$x^{2}-6x+5=0$$
$$(x-1)(x-5)=0$$
Hence $$x=1$$ and $$x=5$$.
Therefore, the area bounded by the parabola and the line between $$x=1$$ and $$x=5$$ is given as
$$A=\int_{1} ^{5} (5-2x)-(4x-x^{2})dx$$
$$=\int_{1} ^{5} x^{2}-6x+5 dx$$
$$=[\dfrac{x^{3}}{3}-3x^{2}+5x]_{1} ^{5}$$
$$=\dfrac{125}{3}-75+25-\dfrac{1}{3}+3-5$$
$$=\dfrac{124}{3}-50-2$$
$$=\dfrac{124}{3}-52$$
$$=\dfrac{124-156}{3}$$
$$=\dfrac{-32}{3}$$
Hence area is $$|A|=\dfrac{32}{3}$$

Area of the rhombus enclosed between $$f(x) , g(x)$$ 

Given curves are $$y^{2}=4ax$$ and $$x^{2}=4ay$$

The given curves are $$y=x^2$$ and $$y^2=x$$ i.e. $$y=\sqrt{x}$$

(i)

$$0\le t\le 2\quad \& \quad x=t$$ 

Area enclose by $$y^2=5x+6,$$

and $$x^2=y$$

$$\Rightarrow x^4-1=5x+5\\(x^2+1)(x-1)(x+1)-5(x+1)=0\\x=-1$$

or, $$(x^2+1)(x-1)-5=0\\x^3-x^2=x-1-5=0\\x^3-x^2+6=0\\x^3-2x^2+x^2-2x+3x-6=0\\x^3(x-2)+x(x-2)+3(x-2)=0\\(x-2)(x^2+x+3)=0\\x-2$$

or $$x^2+x+3=0$$

$$\int _{ -1 }^{ 2 }{ \left( \sqrt { 5x+6 } -{ x }^{ 2 } \right) dx }$$

$$=\int _{ -1 }^{ 2 }{ \sqrt { 5x+6 }  } dx-\int _{ -1 }^{ 2 }{ { x }^{ 2 } } dx$$

$$=\left[ \cfrac { 1 }{ 5 } \cfrac { { \left( 5x+6 \right)  }^{ \cfrac { 3 }{ 2 }  } }{ \cfrac { 3 }{ 2 }  }  \right] _{ -1 }^{ 2 }-\left[ \cfrac { { x }^{ 3 } }{ 3 }  \right] _{ -1 }^{ 2 }$$

$$=\cfrac { 2 }{ 11 } \left[ { \left( 16 \right)  }^{ \cfrac { 3 }{ 2 }  }-{ 1 }^{ \cfrac { 3 }{ 2 }  } \right] -\left[ \cfrac { 8 }{ 3 } +\cfrac { 1 }{ 3 }  \right] \\ =\cfrac { 2 }{ 11 } \left[ { 4 }^{ 3 }-1 \right] -\cfrac { { 9 } }{ 3 } \\ =\cfrac { 2 }{ 11 } \times 63-3\\ =\cfrac { 126 }{ 15 } -3\\ =\cfrac { 42-15 }{ 5 } =\cfrac { 27 }{ 5 } sq\quad unit$$

 

The area will be infinite because the two curves does not meet any point

To  Prove  that   the  area  enclosed  between  two    parabolas  $$y^{2} = 4ax$$      and    $$x^{2} = 4ay$$     is     $$\dfrac{16a^{3}}{3}$$

Given:

Given Curves $$y=x^2,y=x+2$$

Given the curves $$y=x^2-2x+2\ and\ y=2+4x-x^2$$

Area of road i.e., area of shaded region 

Given the curves $$y=x^2,y=2x-x^2$$

Given the curves $$y=x^2-2x+3,y=3x-1$$

Given the curves $$y=x^2,y=1+\dfrac{1}{4}x^2$$

Area square= Area circle- shaded region
     $$=2\pi-\dfrac{2+\pi}{4} (\sqrt{2}-2) -\sqrt2$$
     $$=2\pi-\dfrac{\sqrt2}{2}-\dfrac{\pi}{4}\sqrt2 + 1+ \dfrac{\pi}{2}-\sqrt2$$
     $$=\dfrac{\pi}{4}(10-\sqrt2)+1-\dfrac{3\sqrt2}{2}$$

$$\begin{array}{l}A = \int_0^{2\pi } {ydx}  = \int_0^{\pi /2} {\sin xdx + } \int_{\pi /2}^\pi  {2 - \sin xdx + \int_\pi ^{2\pi } { - \sin xdx} } \\ =  - \left[ {\cos x} \right]_0^{^{\pi /2}} + \left[ {2x} \right]_{^{\pi /2}}^\pi  + \left[ {\cos x} \right]_{^{\pi /2}}^\pi  + \left[ {\cos x} \right]_\pi ^{2\pi }\\ =  - \left[ { - 1} \right] + 2\left[ {\frac{\pi }{2}} \right] + \cos \sqrt x  + \cos 2\pi  - \cos \pi \\ = 1 + \pi  + \left( { - 1} \right) + \left( {1 + 1} \right)\\ = 2 + \pi \end{array}$$

In the first quadrant
$$\begin{array}{l}x \ge 0,y \ge 0\\x + y \le k\end{array}$$
In the second quadrant
$$ - x + y \le k$$
square area
 $$\begin{array}{l} = {(k\sqrt 2 )^2} = 2{k^2}\\f(k) = 2{k^2}\\f(x) = 2{x^2}\\f(10) = 2 \times 100\\\frac{{f(10)}}{{100}} = 2\end{array}$$

$$y^2=4ax$$

$$x^2+y^2-2x=0$$ and y$$y=\dfrac{\sin\pi x}{2} in the upper half of the circle is $$\dfrac{pi^2a}{2\pi}$$

The point of intersection of these two parabolas are $$O(0,0)$$ and $$A(1,1)$$ as shown in the figure.

Here, we can swet $$y^2=x$$ or $$y=\sqrt x=f(x)$$ and $$y=x^2=g(x)$$

where $$f(x)\ge g(x)$$ in $$[0,1]$$

Therefore, the required area of the shaded region

$$=\int _{ 0 }^{ 1 }{ \left[ f\left( x \right) -g\left( x \right)  \right] dx } \\ =\int _{ 0 }^{ 1 }{ \left[ \sqrt { x } -{ x }^{ 2 } \right] dx } ={ \left[ \cfrac { 2 }{ 3 } { x }^{ \cfrac { 3 }{ 2 }  }-\cfrac { { x }^{ 3 } }{ 3 }  \right]  }_{ 0 }^{ 1 }\\ =\cfrac { 2 }{ 3 } -\cfrac { 1 }{ 3 } =\cfrac { 1 }{ 3 } $$

 

$$A=2\int _{ 0 }^{ \pi /4 }{ \tan { x }  } dx\\ A=-2{ [\ln { \cos { x }  } ] }_{ 0 }^{ \pi /4 }\\ A=-2{ [\ln { \cos { \pi /4-\ln { \cos { 0 }  }  }  } ] }\\ A=-2{ [\ln { \cfrac { 1 }{ \sqrt { 2 }  }  } -\ln { 1 } ] }\\ A=-2{ \ln { \cfrac { 1 }{ \sqrt { 2 }  }  }  }\\ A=2\ln { \sqrt { 2 }  } \\ A=\ln { 2 } $$

$$\left(A\right)$$It is clear from the graphs(shown below), $${y}_{1}$$ and $${y}_{2}$$ are periodic with period $$\pi$$ and area same.
and length of curves are also same.

$$\left[-{\sin}^{2}x\right]=0$$ or $$-1$$ but $${\sec}^{-1}\left(0\right)$$ is not defined.
$$\Rightarrow {\sec}^{-1}{\left[-{\sin}^{2}x\right]}={\sec}^{-1}{\left(-1\right)}=\pi$$
The required area$$=\int_{-\sqrt{\pi}}^{\sqrt{\pi}}{\left(\pi-{x}^{2}\right)dx}$$
                 $$=\left[\pi x-\dfrac{{x}^{3}}{3}\right]_{-\sqrt{\pi}}^{\sqrt{\pi}}$$
                 $$=\dfrac{4}{3}\pi\sqrt{\pi}$$ (on simplification)

let $$c>0$$

$$Area(ABC)=Area(ABED)-Area(ACED)=\displaystyle\int_{x=0}^{4} (3x+1)-0\ dx-\int_{x=0}^4 (2x+1)-0\ dx\\=\displaystyle\int_{x=0}^{4} x\ dx=\left[\dfrac {x^2}2\right]^{4}_{0}=8 $$

$$\begin{array}{l}\dfrac{1}{{1 + {x^2}}} - \log \left( {1 + {x^2}} \right)\\{y^2} = 8x\\y = \sqrt {8x} \\y = 2\sqrt {2x} \\{\rm{area}} = 2\int\limits_0^2 {ydx}  = 2\int\limits_0^2 {2\sqrt 2 \sqrt x dx} \\ = 4\sqrt 2 \int\limits_0^2 {\sqrt x dx}  = 4\sqrt 2 \left[ {\dfrac{{{x^{3/2}}}}{{3/2}}} \right]_0^2\\ = \dfrac{{8\sqrt 2 }}{3}\left[ {{2^{3/2}} - 0} \right] = \dfrac{{8\sqrt 2 }}{3}2\sqrt 2  = \dfrac{{32}}{3}\end{array}$$

Intersecting points 

$${x^2} - 4x = 6y + 14$$
$${x^2} - 4x + 4 = 6y + 18$$
$${(x - 2)^2} = 6(y + 3)$$
$${x^2} = 4ay$$
$$4a = 6$$
$$a = \frac{2}{3}$$
Focus : $$(a,0)$$
$$x = a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 0$$
$$x - 2 = \frac{2}{3}\,\,\,\,\,\,\,\,\,y + 3 = 0$$
$$x = 2 + \frac{2}{3}\,\,\,\,\,\,\,\,\,\,y =  - 3$$
$$ = \frac{8}{3}$$
$$\left( {\frac{8}{3}, - 3} \right)$$

$$area\>of\>shaded\>portion\>=\>(4x^2-2x-6)-(x^2-4x+5)\\=3x^2+2x-11$$

$$x + y = 2$$

$$y+2x^2=0$$

$$f\left( x \right) =\left\{ \cos { x } :0\le x\le \pi /4\\ \sin { x } :\pi /4\le x\le 5\pi /4\\ \cos { x } :5\pi /4\le x\le 2\pi  \right\} \\ A=\int _{ 0 }^{ 1 }{ \sqrt { 4-{ (x-2) }^{ 2 } }  } dx+\int _{ 1 }^{ 24 }{ \sqrt { 4-x^{ 2 } }  } dx+\\ A={ [\cfrac { \sqrt { 4-{ (x-2) }^{ 2 } } (x-2) }{ 2 } +\cfrac { 4 }{ 2 } \sin ^{ -1 }{ \cfrac { x-2 }{ 2 }  } ] }_{ 0 }^{ 1 }+{ [\cfrac { \sqrt { 4-{ (x) }^{ 2 } } (x) }{ 2 } +\cfrac { 4 }{ 2 } \sin ^{ -1 }{ \cfrac { x }{ 2 }  } ] }_{ 1 }^{ 2 }\\ A={ 2[(\cfrac { \sqrt { 3 } (-1) }{ 2 } +2\sin ^{ -1 }{ \cfrac { -1 }{ 2 }  } ) }-{ ((-2)(0)+2\sin ^{ -1 }{ (-1) } ) }+{ ((0)+2\sin ^{ -1 }{ (1) } ) }-{ (\cfrac { \sqrt { 3 } (1) }{ 2 } +2\sin ^{ -1 }{ \cfrac { 1 }{ 2 }  } ) }]\\ A={ 2[\cfrac { -\sqrt { 3 }  }{ 2 } -2\sin ^{ -1 }{ (\cfrac { 1 }{ 2 } ) }  }+{ 2\sin ^{ -1 }{ (1) }  }+{ 2\sin ^{ -1 }{ (1) }  }-{ \cfrac { \sqrt { 3 }  }{ 2 } -2\sin ^{ -1 }{ (\cfrac { 1 }{ 2 } ) } ] }\\ A=2[-\sqrt { 3 } -4\sin ^{ -1 }{ (\cfrac { 1 }{ 2 } ) } +4\sin ^{ -1 }{ (1) } ]\\ A=2[-\sqrt { 3 } -4(\pi /6)+4(\pi /2)]\\ A=2[-\sqrt { 3 } +4(\pi /3)]\\ A=-2\sqrt { 3 } +8(\pi /3)$$

$$0\leq y\leq x^2+1$$ (parabola)

$$A=\int _{ 0 }^{ 2x }{ \cos { x }  } dx$$

Given equations of curves are

The area bounded by the curves, $$y=x^2+2,y=x,x=0,x=3$$ is the shaded area OCBAO.

$$y^2=x\\x^2+y^2=2\\x^2+x-2=0\\x^2+2x-x-2=0\\x(x+2)-1(x+2)=0$$

$$\begin{array}{l} Given, \\ { x^{ 2 } }+{ y^{ 2 } }\le 1,\, \, \, and\, \, \, x+\frac { y }{ 2 } =2x+y\ge 2,\, \, \, \therefore \, \, \, \, y=2-2x \\ then,\, \, { x^{ 2 } }+{ \left[ { 2(1-x) } \right] ^{ 2 } }=1 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, { x^{ 2 } }+4(1+{ x^{ 2 } }-2x)=1 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, { x^{ 2 } }+4+4{ x^{ 2 } }-8x-1=0 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, 5{ x^{ 2 } }-8x+3=0 \\ \, \, \, \, \, \, \, \, \, \, \, \, 5{ x^{ 2 } }-5x-3x+3=0 \\ \, \, \, \, \, \, \, \, \, \, \, (5x-3)(x-1)=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \, \, x=\frac { 3 }{ 5 } \, \, ,\, \, \, x=1 \\ and\, \, \, y=2-2(\frac { 3 }{ 5 } \, )=\frac { 4 }{ 5 }  \\ Now \\ Area\, \, of\, region-\int _{ \frac { 3 }{ 5 }  }^{ 1 }{ y.dy-\frac { 1 }{ 2 } (1-\frac { 3 }{ 5 } \, )\times  } \frac { 4 }{ 5 } \, \, \, \,  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \int _{ \frac { 3 }{ 5 }  }^{ 1 }{ \sqrt { 1-{ x^{ 2 } } } .dx- } \frac { 4 }{ 5 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \int { \sqrt { { a^{ 2 } }-{ x^{ 2 } } } dx=\frac { { { a^{ 2 } } } }{ 2 } { { \tan   }^{ -1 } }\frac { x }{ { \sqrt { { a^{ 2 } }-{ x^{ 2 } } }  } } +\frac { x }{ 2 } \sqrt { { a^{ 2 } }-{ x^{ 2 } } }  }  } \right.  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, { \left[ { \frac { 1 }{ 2 } { { \tan   }^{ -1 } }\frac { x }{ { \sqrt { 1-{ x^{ 2 } } }  } } +\frac { x }{ 2 } \sqrt { 1-x }  } \right] _{ \frac { 3 }{ 5 }  } }^{ 1 }\, -\, \, \frac { 4 }{ 5 } \,  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \frac { 1 }{ 2 } \times \frac { \pi  }{ 2 } -\frac { 1 }{ 2 } { { \tan   }^{ -1 } }\frac { { \frac { 3 }{ 5 }  } }{ { \sqrt { 1-\frac { 9 }{ { 25 } }  }  } } -\frac { { \frac { 3 }{ 5 }  } }{ 2 } \sqrt { 1-\frac { 9 }{ { 25 } }  }  } \right] -\, \, \frac { 4 }{ 5 }  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \frac { \pi  }{ 4 } -\frac { 1 }{ 2 } { { \tan   }^{ -1 } }\frac { 3 }{ 4 } -\frac { 3 }{ { 5\times 2 } } \times \frac { 4 }{ 5 }  } \right] -\, \, \frac { 4 }{ { 25 } }  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \frac { \pi  }{ 4 } -\frac { 1 }{ 2 } { { \tan   }^{ -1 } }\frac { 3 }{ 4 } -\frac { 6 }{ { 25 } }  } \right] -\, \, \frac { 4 }{ { 25 } }  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \therefore \, \, \, \frac { \pi  }{ 4 } -\frac { 1 }{ 2 } { \tan ^{ -1 }  }\frac { 3 }{ 4 } -\frac { 2 }{ 5 }  \end{array}$$

Whenever a function is represented by $$y=\left|x\right|$$ there arises two cases:

We have,

$$ {{y}^{2}}=x\,\,.......\,\,\left( 1 \right) $$

$$ {{x}^{2}}=y\,\,.......\,\,\left( 2 \right) $$

Put the value of equation (2) in equation (1) and we get,

$$ {{y}^{2}}=x $$

$$ {{\left( {{x}^{2}} \right)}^{2}}=x $$

$$ {{x}^{4}}-x=0 $$

$$ x\left( {{x}^{3}}-1 \right)=0 $$

$$ x=0,\,\,{{x}^{3}}-1=0 $$

$$ x=0,\,\,x=1 $$

Put the value of $$x$$ in equation (2) and we get,

$$ {{x}^{2}}=y $$

$$ y=0,\,\,y=1 $$

So. Coordinates are $$\left( 0,0 \right)$$ and $$\left( 1,1 \right)$$

Let $$P\left( 0,0 \right)$$ and $$Q\left( 1,1 \right)$$

So, point $$Q\left( 1,1 \right)$$ lies in first quadrant.

Required area is

$$R.A=\int_{0}^{1}{{{y}_{1}}dx}-\int_{0}^{1}{{{y}_{2}}dx}$$

So, we know that,

By equation (1),

$$ {{y}^{2}}=x $$

$$ y=\pm \sqrt{x} $$

$$ \text{Since,} $$

$$ {{\text{y}}_{1}}=\sqrt{x} $$

By equation (2)

$$ {{x}^{2}}=y $$

$$ y={{x}^{2}} $$

$$ {{y}_{2}}={{x}^{2}} $$

Then Area Required.

$$ R.A.=\int_{0}^{1}{{{y}_{1}}dx}-\int_{0}^{1}{{{y}_{2}}dx} $$

$$ =\int_{0}^{1}{\sqrt{x}dx}-\int_{0}^{1}{{{x}^{2}}dx} $$

$$ =\int_{0}^{1}{{{x}^{\dfrac{1}{2}}}dx}-\int_{0}^{1}{{{x}^{2}}dx} $$

$$ ={{\left[ \dfrac{{{x}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right]}_{0}}^{1}-{{\left[ \dfrac{{{x}^{2+1}}}{2+1} \right]}_{0}}^{1} $$

$$ ={{\left[ \dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]}_{0}}^{1}-{{\left[ \dfrac{{{x}^{3}}}{3} \right]}_{0}}^{1} $$

$$ =\dfrac{{{1}^{\dfrac{3}{2}}}-{{0}^{^{\dfrac{3}{2}}}}}{\dfrac{3}{2}}-\dfrac{{{1}^{3}}-{{0}^{3}}}{3} $$

$$ =\dfrac{1}{\dfrac{3}{2}}-\dfrac{1}{3} $$

$$ =\dfrac{2}{3}-\dfrac{1}{3} $$

$$ =\dfrac{1}{3}\,sq.units $$

Hence, this is the answer.

The desired are $$=\int_0^{\pi/4}(cos x)dx$$

The radius of the circle $$x^2+y^2=16$$ is $$4$$

$$ (i) y^2=4ax,x^2=4ay$$

By solving for pts of intersection,

Area of triangle between the coordinate axis of line $$3x - 4y - 12 = 0$$ is

We have, $$y = 2\sqrt {x}, x = 0$$ and $$x = 1$$.
$$\therefore$$ Area of shaded region, $$A = \int_{0}^{1} (2\sqrt {x})dx$$
$$= 2.\left [\dfrac {x^{3/2}}{3} \times 2\right ]_{0}^{1}$$

$$= 2\left (\dfrac {2}{3} \times 1 - 0\right ) = \dfrac {4}{3} sq\ units$$.

Let the given equations

Point of intersection 

Given, $$y=x^2,y^2=4ax$$

$$y^2=2x$$

Line $$y=2$$ and $$y=x^2$$ intersects at $$(\pm\sqrt 2,2)$$

Required area $$\displaystyle =\int _{ a }^{ 2a }{ xdy } $$
$$\displaystyle =\int _{ a }^{ 2a }{ { \left( a{ y }^{ 2 } \right)  }^{ 1/3 }dy } ={ a }^{ 1/3 }\int _{ a }^{ 2a }{ y^{ 2/3 }dy } ={ a }^{ 1/3 }\int _{ a }^{ 2a }{ { \left[ y^{ 2/3 } \right]  }_{ 0 }^{ 2a }dy } =\dfrac { 3 }{ 5 } { a }^{ 1/3 }\left\{ { \left( 2a \right)  }^{ 5/3 }-{ a }^{ 5/3 } \right\} $$

We have, $$y = x^{3}, y = x + 6$$ and $$x = 0$$

Here, 

Common area = Area of sector

We have,

$$y=\cos x\,and\,y=\sin x$$

Now,

$$ \sin x=\cos x $$

$$ \dfrac{\sin x}{\cos x}=1 $$

$$ \Rightarrow \tan x=1 $$

$$ \Rightarrow \tan x=\tan \dfrac{\pi }{4} $$

$$ \Rightarrow x=\dfrac{\pi }{4} $$

Then,

At $$x=\dfrac{\pi }{4}$$, both are equal

So,

$$ y=\cos x=\cos \dfrac{\pi }{4} $$

$$ y=\dfrac{1}{\sqrt{2}} $$

So, Finding area

$$ \text{Requried}\,\text{area = Area of ABCO-Area of  BCO} $$

$$ \text{Requried}\,\text{area = }\int_{0}^{\dfrac{\pi }{4}}{y}dx-\int_{0}^{\dfrac{\pi }{4}}{y}dx $$

$$ Area\,ABCO\,=\,\int_{0}^{\dfrac{\pi }{2}}{\cos xdx} $$

$$ ={{\left[ \sin x \right]}_{0}}^{\dfrac{\pi }{4}} $$

$$ =\left[ \sin \dfrac{\pi }{4}-\sin 0 \right] $$

$$ =\dfrac{1}{\sqrt{2}}-0 $$

$$ =\dfrac{1}{\sqrt{2}} $$

Area $$BCO$$ $$=\int_{0}^{\dfrac{\pi }{4}}{y}dx$$

$$ =\int_{0}^{\dfrac{\pi }{4}}{\sin xdx} $$

$$ ={{\left[ -\cos x \right]}_{0}}^{\dfrac{\pi }{4}} $$

$$ =-\left[ \cos \dfrac{\pi }{4}-\cos 0 \right] $$

$$ =-\left[ \dfrac{1}{\sqrt{2}}-1 \right] $$

$$ =1-\dfrac{1}{\sqrt{2}} $$

Now,

Required area

$$ =\dfrac{1}{\sqrt{2}}-\left[ 1-\dfrac{1}{\sqrt{2}} \right] $$

$$ =\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-1 $$

$$ =\dfrac{2}{\sqrt{2}}-1 $$

$$ =\sqrt{2}-1 $$

Hence, this is the answer.

$$\begin{array}{l} y=4-{ x^{ 2 } },\, \, \, y=0,\, \, y=3 \\ Area=\, \int { ydx\, \, of\, \, \int { xdx }  }  \\ if\, \, y=0,\, \, then, \\ 0=4-{ x^{ 2 } } \\ { x^{ 2 } }=4=7x=2 \\ if\, y=3\, ,\, \, then, \\ 3=4-{ x^{ 2 } } \\ { x^{ 2 } }=1\Rightarrow x=1 \\ area=\, \, \int _{ 1 }^{ 2 }{ ydx }  \\ =\int _{ 1 }^{ 2 }{ \left( { 4-{ x^{ 2 } } } \right) dx }  \\ =\left[ { 4x } \right] _{ 1 }^{ 2 }-\dfrac { 1 }{ 3 } \left[ { { x^{ 3 } } } \right] _{ 1 }^{ 2 } \\ =4\left[ { 2-1 } \right] -\dfrac { 1 }{ 3 } \left[ { 8-1 } \right]  \\ =4-\dfrac { 7 }{ 3 } =\dfrac { 5 }{ 3 }  \end{array}$$

$$\begin{array}{l} { { Re } }quired\, area\, =\int  _{ 1 }^{ 2 }{ { x^{ 2 } }+x\, \, \, dx\, \, \,  }+\int  _{ 1 }^{ 2 }{ \sin  x\, dx } \\ =\left[ { \frac { { { x^{ 3 } } } }{ 3 } +\frac { { { x^{ 2 } } } }{ 2 }  } \right] _{ 1 }^{ 2 }+\left[ { \frac { { \cos  \pi x } }{ \pi  }  } \right] _{ 1 }^{ 2 } \\ =\frac { 8 }{ 3 } +2-\left( { \frac { 1 }{ 3 } +\frac { 1 }{ 2 }  } \right) +\left( { \frac { 1 }{ \pi  } -\frac { 1 }{ \pi  }  } \right)  \\ =\frac { { 14 } }{ 3 } -\frac { 5 }{ 6 } =\frac { { 23 } }{ 6 } \, sq.\, unit \end{array}$$

$$ y=\tan  x\, \, and\, \, y={ \tan ^{ 2 }  }x \\ $$

$$\begin{array}{l} 2d=42 \\ d=21\, \, cm \end{array}$$

$$\begin{array}{l} Area\, \, of\, \, path=area\, \, of\, \, outer\, \, rec\tan  gle-area\, \, of\, \, inner\, \, rec\tan  gle \\ =60\times 40-\left( { 60-4 } \right) \left( { 40-4 } \right)  \\ =2400-56\times 36 \\ =2400-2016 \\ =384\, \, { m^{ 2 } } \end{array}$$

$$\begin{array}{l} Let\, \, triangle\, \, are\, \, equillateral \\ \Rightarrow Area\, \, of\, \, equilateral\, \, triangle=\frac { { \sqrt { 3 }  } }{ 4 } \times { \left( { side } \right) ^{ 2 } } \\ \Rightarrow 68.2\, \, sq\, \, m\, =0.43\, \, { \left( { side } \right) ^{ 2 } } \\ \Rightarrow \sqrt { 158.6 } =side,\, \, \, \, \, side\, \, =12.6\, \, m \\ \Rightarrow Ar\, \, \Delta \, BDC\, \, =\frac { 1 }{ 2 } \times 12.6\times 4.21\, =26.51\, \, { m^{ 2 } } \\ \Rightarrow Area\, \, of\, \, shaded\, \, =\left( { 68.2-26.51 } \right) =41.69\, \, { m^{ 2 } }\, \, Ans. \end{array}$$

We have,

$$ {{x}^{2}}+{{y}^{2}}=16\,......\,\,\left( 1 \right) $$

$$ {{x}^{2}}=6y\,......\,\,\left( 2 \right) $$

From equation (1) and (2) to,

$$ {{y}^{2}}+6y=16 $$

$$ {{y}^{2}}+6y-16=0 $$

$$ {{y}^{2}}+\left( 8-2 \right)y-16=0 $$

$$ {{y}^{2}}+8y-2y-16=0 $$

$$ y\left( y+8 \right)-2\left( y+8 \right)=0 $$

$$ \left( y+8 \right)\left( y-2 \right)=0 $$

If

$$ y+8=0 $$

$$ y=-8 $$

$$ y-2=0 $$

$$ y=2 $$

$$y=-8$$ as parabola lies in first or fourth quadrant.

We get,

When, $$y=2$$ put in equation (2) and we get,

$$ {{x}^{2}}=6y $$

$$ {{x}^{2}}=6\times 2 $$

$$ {{x}^{2}}=12 $$

$$ x=2\sqrt{3} $$

By equation (1) and (2) intersect $$p\left( 2,2\sqrt{3} \right)$$

$$ =16\pi -2\int_{0}^{2}{ydx}-\int_{2}^{4}{ydx} $$

$$ =16\pi -2\int_{0}^{2}{\sqrt{6}{{x}^{\dfrac{1}{2}}}dx}-2\int_{2}^{4}{\sqrt{16-{{x}^{2}}}dx} $$

$$ =16\pi -2\sqrt{6}{{\left[ \dfrac{2}{3}{{x}^{\dfrac{3}{2}}} \right]}_{0}}^{2}-2\left[ \dfrac{x}{2}\sqrt{16-{{x}^{2}}}+\dfrac{16}{2}{{\sin }^{-1}}\dfrac{x}{4} \right] $$

$$ =16\pi -\dfrac{4}{3}\sqrt{6}{{\left( 2 \right)}^{\dfrac{3}{2}}}-2\left[ \dfrac{4}{2}\sqrt{16-16}+\dfrac{16}{2}{{\sin }^{-1}}1-\dfrac{2}{2}\sqrt{16-4}-\dfrac{16}{2}{{\sin }^{-1}}\dfrac{2}{4} \right] $$

$$ =16\pi -\dfrac{16}{\sqrt{3}}-16\times \dfrac{\pi }{2}+4\sqrt{3}+\dfrac{8\pi }{3} $$

$$ =\dfrac{-4}{\sqrt{3}}+\dfrac{32\pi }{3} $$

$$ =\dfrac{4}{3}\left( 8\pi -\sqrt{3} \right) $$

Hence, this is the answer.

Total shaded area = Area of semi-circle $$\left[ {ABD - ALC + CKG - BNE + EFG} \right]$$

$$\displaystyle \int_0^a {\left[ {\left( {x + 2} \right) - {x^2}} \right]} \,dx$$

$$\begin{array}{l} Area\, \, of\, \, shaded\, \, \, part= \\ \int _{ 0 }^{ 2 }{ \left( { x+2 } \right) dx-\int _{ 0 }^{ 2 }{ { x^{ 2 } }dx }  }  \\ =\left[ { \frac { { { x^{ 2 } } } }{ 2 } +2x } \right] _{ 0 }^{ 2 }-\left[ { \frac { { { x^{ 3 } } } }{ 3 }  } \right] _{ 0 }^{ 2 } \\ =\frac { { 10 } }{ 3 } \, \, unit\, \, sq. \end{array}$$

$$\begin{array}{l} From\, the\, figure \\ Radius\, of\, semi-circle=\cfrac { { 14 } }{ 2 } =7\ cm \\ Area\, of\, the\, shaded\, region\, =area\, of\, rec\tan  e-area\, of\, semicircle \\ =length\, \times breadth-\cfrac { { \pi { r^{ 2 } } } }{ 2 }  \\ =20\times 14-\cfrac { { 22 } }{ 7 } \times \cfrac { { 7\times 7 } }{ 2 }  \\ =280-77 \\ =203\ c{ m^{ 2 } } \\ Hence,\, area\, of\, shaded\, region\, is\, 203\ c{ m^{ 2 } }. \end{array}$$

$$\begin{array}{l} we\, have\, the\, \, given\, equation \\ 2x+{ y^{ 2 } }=48 \\ and\,  \\ x=y. \\ solving\, for\, x\, from\, \, the\, given\, equatin\, we\, get \\ x=-\frac { 1 }{ 2 } { y^{ 2 } }+24 \\ x=y \\ To\, find\, the\, po{ { int } }\, of\, { { int } }er\sec  tion\, \, we\, have \\ y=-\frac { 1 }{ 2 } { y^{ 2 } }+24 \\ 2y=-{ y^{ 2 } }+48 \\ { y^{ 2 } }+2y-48=0 \\ \left( { y+8 } \right) \left( { y-6 } \right) =0 \\ So,\, the\, po{ { ints } }\, of\, intersection\, \, are\, when\, x=-8,6. \\ since,\, the\, 2x+{ y^{ 2 } }=48\, function\, is\, \, the\, \, right\, bounded\, we\, have\, an\, area\, of\,  \\ \int _{ -8 }^{ 6 }{ \left( { -\frac { 1 }{ 2 } { y^{ 2 } }+24 } \right) -ydy } =\left[ { -\frac { 1 }{ 6 } { y^{ 3 } }+24y-\frac { 1 }{ 2 } { y^{ 2 } } } \right] _{ -8 }^{ 6 } \\ =-\frac { 1 }{ 6 } { \left( 6 \right) ^{ 3 } }+24\left( 6 \right) -\frac { 1 }{ 2 } { \left( 6 \right) ^{ 2 } }-\left( { -\frac { 1 }{ 2 } { { \left( { -8 } \right)  }^{ 3 } }+24\left( { -8 } \right) -\frac { 1 }{ 2 } { { \left( { -8 } \right)  }^{ 2 } } } \right)  \\ =90-\left( { \frac { { -416 } }{ 3 }  } \right)  \\ =\frac { { 686 } }{ 3 }  \end{array}$$

area of shaded portion = area of rectangle - area of triangle

$$y=3x+2,\quad x=-1,\quad x=1$$

$$y=x^2+x$$

$$y=x^3-2x^2$$

The area is given as $$\displaystyle \int _3^6 x^2+3dx \\\left.\dfrac{x^3}3+3x\right|_3^6\\\dfrac {6^3}{3}+3(6)-\dfrac{3^3}{3}-3(3)\\243+18-9-9=243$$

$$y^2=4x$$ $$\implies x=\dfrac{y^2}{4}$$ --- (i)

The given curves are$$y = x^2$$$$y = |2 - x^2|$$The graph of these curves is as follows$$\therefore$$ Required area $$= BCDEB$$$$= \displaystyle \int_1^{\sqrt{2}} [x^2 - (2 - x^2)] dx + \displaystyle \int_\sqrt{2}^2 [2 - (x^2 - 2)] dx$$$$= \displaystyle \int_1^{\sqrt{2}} (2x^2 - 2) dx + \int_{\sqrt{2}}^2 (4 - x^2) dx$$$$= \left[\dfrac{2x^3}{3} - 2x\right]_1^{\sqrt{2}} + \left[4x - \dfrac{x^3}{3} \right]_{\sqrt{2}}^2$$$$= \left(\dfrac{4 \sqrt{2}}{3} - 2 \sqrt{2} - \dfrac{2}{3} + 2 \right) + \left(8 - \dfrac{8}{3} - 4 \sqrt{2} + \dfrac{2 \sqrt{2}}{3} \right)$$$$= \left(\dfrac{20}{3} - 4 \sqrt{2}\right) sq. units$$

Here area is calculated on positive x-axis as $$x>0$$

Let we have the vertices of $$\triangle ABC$$ as $$A (-1, 1), B(0, 5)$$ and $$C(3, 2)$$.

$$\therefore$$ Equation of $$AB$$ is $$y - 1 = \left (\dfrac {5 - 1}{0 + 1}\right ) (x + 1)$$

$$\Rightarrow y - 1 = 4x + 4$$

We know curve is ellipse vertex $$(0,0)$$ major axis is $$'y'$$ and major $$'x'$$
Area of given area is $$=4\times$$ (Area of shaded)
$$=4\times \displaystyle\int_{0}^{2}|y|dx$$
$$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1$$
$$\Rightarrow y=\dfrac{3}{2}\sqrt{4-x^{2}}$$
$$4\times \dfrac{3}{2}\displaystyle\int_{0}^{2} \sqrt{4-x^{2}}dx=6\displaystyle\int_{0}^{2} \sqrt{2^2-x^2}dx$$
$$\boxed{\displaystyle\int \sqrt{a^2-x^2\ dx}=\dfrac{x}{2}\sqrt{a^{2}-x^{2}}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}}$$
$$=6\left[\dfrac{x}{2}\sqrt{2^{2}-x^{2}}+\dfrac{2^{2}}{2}\sin^{-1}\dfrac{x}{2}\right]^{2}_{0}$$
$$=6\left[\dfrac{2}{2}\sqrt{2^2-2^2}+2\sin^{-1}\dfrac{2}{2}-\dfrac{0}{2}\sqrt{2^{2}-0}-2\sin^{-1}\dfrac{0}{2}\right]$$
$$=6[0+2\sin^{-1}1-0-0]=6\times 2\times \dfrac{\pi}{2}=6 \pi $$ sq.units

Given equation of curves are
$$y^{2} = 4x .... (i)$$
and $$x^{2} = 4y .... (ii)$$
$$\Rightarrow \left (\dfrac {x^{2}}{4} \right ) = 4x$$
$$\Rightarrow \dfrac {x^{4}}{16} = 4x$$
$$\Rightarrow x^{4} = 64x$$
$$\Rightarrow x^{4} - 64x = 0$$
$$\Rightarrow x (x^{3} - 4^{3}) = 0$$
$$\Rightarrow x = 4, 0$$
$$\therefore$$ Area of shaded region, $$A = \int_{0}^{4}\left (\sqrt {4x} - \dfrac {x^{2}}{4}\right ) dx$$
$$= \int_{0}^{4} \left (2\sqrt {x} - \dfrac {x^{2}}{4}\right ) dx $$

$$4y = 3x^{2}$$ and $$2y = 3x + 12$$
$$4y = 3x^{2}$$
$$\Rightarrow 2y = \dfrac {3}{2} x^{2}$$
$$\Rightarrow 3x + 12 = \dfrac {3}{2}x^{2}$$
$$\Rightarrow 6x + 24 = 3x^{2}$$
$$\Rightarrow 2x + 8 = x^{2}$$
$$\Rightarrow x^{2} - 2x - 8 = 0$$
$$x = \dfrac {2\pm \sqrt {4 + 32}}{2} = \dfrac {2\pm 6}{2}$$
$$\therefore x = 4, -2$$
and $$y = 12, 3$$
$$\therefore$$ the line $$2y = 3x + 12$$ intersects the parabola $$4y = 3x^{2}$$ at point $$(-2, 3)$$ and $$(4, 12)$$
$$\therefore$$ the area enclosed between the parabola and line is shown in the figure as the shaded portion.
$$Area = \int_{a}^{b} [f(x) - g(x)] dx$$
$$a = -2\ b = 4\ f(x) = \dfrac {3x + 12}{2} g(x) = \dfrac {3x^{2}}{4}$$
$$\therefore A = \int_{-2}^{4} \left [\left [\dfrac {3x + 12}{2}\right ] - \left (\dfrac {3x^{2}}{4}\right )\right ]dx = \int_{-2}^{4} \dfrac {3x + 12}{2} dx - \int_{-2}^{4} \dfrac {3x^{2}}{4} dx$$
On integrating we get
$$A = \dfrac {1}{2} \left [\dfrac {3x^{2}}{2} + \dfrac {12x}{4}\right ]_{2}^{4} - \dfrac {3}{4} \left [\dfrac {x^{3}}{3}\right ]_{-2}^{4}$$
$$= \dfrac {1}{2} [24 + 48 - 6 + 24] - \dfrac {1}{4} [64 + 8]$$
$$= 45 - 18 = 27\ sq\ units$$.

See the diagram enclosed of $$\cos x$$ from $$x=0$$ to $$x-2\pi$$

$$x^2 + y^2 = 9, (x - 3)^2 + y^2 = 9$$

$$ \sqrt {|x|}\,+\sqrt{|y|}\,=\sqrt{a}$$ 

$$ x=0\, \Rightarrow y=\pm a $$
$$ y=0\, \Rightarrow x=\pm a $$
(a) required area

$$ \Delta = 4\displaystyle \int_{0}^{a} \sqrt{a^2-x^2}\, dx\,- 4\displaystyle \int_{0}^{a} (\sqrt{a}-\sqrt{x})^2\,dx $$

$$ =\pi a^{2}-4\displaystyle \int_{0}^{a}(\sqrt{a}-\sqrt{x})^2\,dx $$

$$ =\pi a^{2}-4\displaystyle \int_{0}^{a}[a+x-2\sqrt{a}\sqrt{x}]dx $$

$$ = \pi a^{2}-4\left [ a^{2}+ \dfrac{a^{2}}{2}-2\sqrt{a}\dfrac{2}{3}a^{\frac{3}{2}} \right ] $$

$$ = \pi a^{2} - 4\left [ \dfrac{3a^{2}}{3}-\dfrac{4}{3}a^{2} \right ]\,=\pi a^{2}-4 \dfrac{a^{2}}{6} $$

$$ = \left ( \pi - \dfrac{2}{3} \right )a^{2}\,sq.units $$

b. Area included between curves and circles in Ist quadrant 

$$ = \dfrac{1}{4} \pi a^{2} - \dfrac{1}{2}a$$ x $$a= \dfrac{(\pi -2)a^2}{4} $$

Area included between $$|x|\,+\,|y|=a $$ and curve $$ \sqrt{|x|}\,+\, \sqrt{|y|}\,=\, \sqrt{a} $$ in 1st quadrant.

$$ = \dfrac{1}{4} \left ( \pi - \dfrac{2}{3} \right )a^2\,-\,\left ( \dfrac{\pi}{4}-\dfrac{1}{2} \right )a^2\,=\, \dfrac{a^2}{3} $$

Area ratio $$ = \dfrac{4}{3(\pi -2)} $$

The given curves are

Solving the given curves $$y=\dfrac{1}{x^2}\,;\, y=\dfrac{1}{4(x-1)} $$

$$ x^2=4(x-1) \Rightarrow (x-2)^2=0 $$

$$ \Rightarrow $$ curves touch other.

$$ \therefore A=\int_{2}^{a}\left ( \dfrac{1}{4(x-1)}-\dfrac{1}{x^2} \right )dx = \dfrac{1}{a} $$

$$ \Rightarrow \left [ \dfrac{1}{4}log(x-1)+\dfrac{1}{x^2} \right ]_{2}^{a}= \dfrac{1}{a} $$

$$ \Rightarrow \dfrac{1}{4}log(a-1)+\dfrac{1}{a}-\dfrac{1}{2}=\dfrac{1}{a} $$

$$ \Rightarrow log(a-1)=2 $$

$$ \Rightarrow a=e^2+1 $$

Also, $$1- \dfrac{1}{b} = \int_{b}^{2}\left ( \dfrac{1}{4(x-1)}-\dfrac{1}{x^2} \right )dx\,\Rightarrow b=1+e^{-2} $$

$$x_1\, and x_2$$ are the roots of the equation

$$x^2+2x-3=kx+1,$$ or
$$x^2+(2-k)x-4=0$$

$$ \Rightarrow x_1+x_2=k-2\,\,\,\,and\,\,x_1x_1=-4$$

$$ A=\int_{x_1}^{x_2}\left [ (kx+1)-(x^2+2x-3) \right ]dx $$

$$ \left [ (k-2)\dfrac{x^2}{2}-\dfrac{x^3}{3}+4x \right ]_{x_1}^{x_2} $$

$$=\left [ (k-2)\dfrac{x_2^2-x_1^2}{2}-\dfrac{1}{3}(x_2^3-x_1^3)+4(x_2-x_1) \right ]$$

$$ =(x_2-x_1)\left [ \frac{(k-2)^2}{2}-\frac{1}{3}((x_2+x_1)^2-x_1x_2)+4 \right ] $$

$$ =\sqrt{(x_2+x_1)^2-4x_1x_2}\left [ \dfrac{(k-2)^2}{2}-\dfrac{1}{3}((k-2)^2+4)+4 \right ] $$

$$=\dfrac{\sqrt{(k-2)^2+16}}{6}\left [ \dfrac{1}{6}(k-2)^2+\dfrac{8}{3} \right ] $$

$$  = \dfrac{\left [(k-2)^2+16  \right ]^\frac{3}{2}}{6}   $$

which is the least when $$k=2$$ and $$A_{least}=\, \dfrac{32}{3}\,sq.units$$.

Let P(x,y) be any point on the curve $$C$$.

Required area
$$\displaystyle A=\int^3_0x\sqrt{9-x^2}dx$$; 

Let the point of the curve is $$(x,x^2+1)$$
Now, the slope of tangent at this point is $$2x$$, which is equal to the slope of the line joining $$(x,x^2+1)$$ and $$(0,0)$$
Hence $$2x=(x^2+1)/x\Rightarrow 2x^2=x^2+1$$
$$\Rightarrow x^2=1\Rightarrow x=\pm 1$$
Hence equation of tangent is $$y=\pm 2x$$
Now area $$\displaystyle 2\int^1_0 (x^2+1-2x)dx$$
$$=\displaystyle 2\int^1_0 (x-1)^2dx$$
$$=2\left[\dfrac{(x-1)^3}{3}\right]_0^1=2/3$$
$$3A=2$$

The given curves are $$y = x - bx^2$$ __(1)
and $$y = x^2/b$$  __(2)
$$\Rightarrow \left(y - \dfrac{1}{4b} \right) = -b \left(x - \dfrac{1}{2b} \right)^2 $$ and $$x^2 = by$$
Here, clearly the first curve is a downward parabola which meets x-axis at $$(0, 0)$$ and $$(1/b , 0)$$, while the second is an upward parabola with vertex at $$(0,0)$$.
Solving (1)  and (2), we get the intersection points of two curves at $$(0,0)$$ and $$\left(\dfrac{b}{1 + b^2}, \dfrac{b}{(1 + b^2)^2}\right)$$
Hence, the graph of given curves is as below
Shaded portion represents the required area
$$A = \displaystyle \int_0^{\dfrac{b}{1 + b^2}} \left(x - bx^2 - \dfrac{x^2}{b} \right) dx$$
$$= \left(\dfrac{x^2}{2} - \dfrac{bx^2}{3} - \dfrac{x^2}{3b} \right)_0^{\dfrac{b}{1 + b^2}}$$
$$= \dfrac{b^2}{2 (1 + b^2)^2} - \dfrac{b^4}{3 (1 + b^2)^3} - \dfrac{b^2}{3 (1 +b^2)^3}$$
$$= \dfrac{b^4 + b^2}{6(1 + b^2)^3} = \dfrac{b^2}{6 (1 + b^2)^2}$$
$$= \dfrac{1}{6\left(\dfrac{1}{b} + b\right)^2}$$ sq. units.
Now, $$\left(\dfrac{1}{b} + b\right) \ge 2$$ or $$\le - 2 \Rightarrow \left(\dfrac{1}{b} + b \right)^2 \ge 4$$
Here, area is max when $$\left(\dfrac{1}{b} + b\right)^2_{min} = 4$$, for which $$b = \pm 1$$
but given that $$b > 0$$
$$\therefore b = 1$$

$$f(x) =$$ Maximum $$\{x^2, (1 - x)^2 , 2x (1 - x)\}$$
we draw the graphs of 
$$y = x^2$$
$$y = (1 - x)^2$$
$$y = 2x (1 - x)$$
Solving (1) and (3), we get $$x^2 = 2x (1 - x)$$
$$\Rightarrow 3x^2 = 2x \Rightarrow x = 0$$ or $$x = 2/3$$
Solving (2) and (3) we get $$(1 - x)^2 = 2x (1 - x)$$
$$\Rightarrow x = 1/3$$ and $$x = 1$$,
Ref. image 
From fig., it is clear that
$$f(x) = \begin{cases} (1 - x)^2 \ for 0 \le x \le 1/3 \\ 2x (1 - x) \ for \ 1/3 \le x \le 2/3 \\ x^2 \ for \ 2/3 < x \le 1 \end{cases}$$ 
The required area A is given by
$$A = \displaystyle \int_0^1 f(x) dx$$
$$\displaystyle \int_0^{1/3} (1 - x)^2 dx + \int_{1/3}^{2/3} 2x (1 - x) dx + \int_{2/3}^1 x^2 dx$$
$$= \left[-\dfrac{1}{3} (1 - x)^3 \right]_0^{1/3} + \left[\left(x^2 - \dfrac{2x^3}{3}\right)\right]_{1/3}^{2/3} + \left[\dfrac{x^3}{3}\right]^t_{2/3}$$
$$= - \dfrac{1}{3} \left(\dfrac{2}{3} \right)^3 + \dfrac{1}{3} + \left(\dfrac{2}{3} \right)^2 - \dfrac{2}{3} \left(\dfrac{2}{3} \right)^3 - \left(\dfrac{1}{3} \right)^2 + \dfrac{2}{3} \left(\dfrac{1}{3} \right)^3 + \dfrac{1}{3} - \dfrac{1}{3} \left(\dfrac{2}{3} \right)^3$$
$$= \dfrac{17}{27} sq. units$$

The given curve are, 
$$x^2 = y$$    ___(1)
$$x^2 = -y$$   ___(2)
$$y^2 = 4x - 3$$  ___(3)
Clearly (1) and (2) meet at $$(0,0)$$
Solving (1) and (3), we get $$x^4 - 4x + 3 = 0$$
$$\Rightarrow (x - 1) (x^3 + x^2 + x - 3) = 0$$
$$\Rightarrow (x - 1)^2 (x^2 + 2x + 3) = 0$$
$$\Rightarrow x = 1 \Rightarrow y = 1$$
$$\Rightarrow $$ Point of intersection is $$(1, 1)$$
Similarly, point of intersection of (2) and (3) is $$(1, -1)$$ 
The graph of three curves is as follow:
Ref. image
we also observe that at $$x = 1$$ and $$y = 1, \dfrac{dy}{dx}$$ for (1) and (3) is same and hence the two curves touch each other at $$(1, 1)$$.
Same is the case with (2) and (3) at $$(1, -1)$$
Required area = shaded region in the figure
$$= 2 (Ar \ OPA)$$
$$= 2 \left[\displaystyle \int_0^1 x^2 dx - \int_{3/4}^1 \sqrt{4x - 3} dx\right]$$
$$= 2 \left[\left(\dfrac{x^3}{3} \right)_0^1 - \left(\dfrac{2 (4x -3)^{3/2}}{4 \times 3} \right)^1 _{3/4}\right] = 2 \left[\dfrac{1}{3} - \dfrac{1}{6} \right]$$
$$= \dfrac{1}{3} sq. units$$

The given curves are $$y = ex \log_e x$$    ____(1)$$y = \dfrac{\log x}{ex}$$   ____(2)The two curves intersect where $$ex \ \log x = \dfrac{\log x}{ex}$$$$\Rightarrow \left(ex - \dfrac{1}{ex} \right) \log x = 0$$$$\Rightarrow x = \dfrac{1}{e} $$ or $$x = 1$$At $$x = 1/e , y = -1$$ (from (1))At $$x = 1, y = 0$$ (from (1))So points of intersection are $$\left(\dfrac{1}{e}, -1\right) $$ and $$ (1,0)$$Curve 1Curve 2For $$0 < x < 1 , y < 0$$For $$0 < x < 1 , y < 0$$For $$x > 1 , y > 0$$For $$x > 1, y > 0$$When $$x \rightarrow 0, y \rightarrow - \infty$$When $$x \rightarrow 0 , y \rightarrow - \infty$$When $$x \rightarrow \infty, y \rightarrow \infty$$When $$x \rightarrow \infty, y \rightarrow 0$$$$\dfrac{dy}{dx} = e (\log x + 1)$$$$\dfrac{dy}{dx} = \dfrac{(1 - \log x)}{ex^2}$$$$x = \dfrac{1}{e} $$ is a point of min$$x = e$$ is a point of maxFrom the above information, the rough sketch of two curves is as shown in and shaded area is the required area.$$\therefore$$ the required area = the shaded area$$= \left|\displaystyle \int_{1/e}^1 \left[ex \log x - \dfrac{\log x}{ex} \right] dx \right|$$$$\left|e \left[\dfrac{x^2}{2} \log x - \dfrac{x^2}{4} \right]_{1/e}^1 - \dfrac{1}{e} \left[\dfrac{(\log x)^2}{2}\right]_{1/e}^1 \right|$$$$= \left|e \left[ \left(- \dfrac{1}{4} \right) - \left(- \dfrac{1}{2 e^2} - \dfrac{1}{4 e^2} \right) \right] - \dfrac{1}{e} \left[0 - \dfrac{1}{2} \right]\right|$$$$= \left|e \left[- \dfrac{1}{4} + \dfrac{3}{4e^2}\right] + \dfrac{1}{2e}\right|$$$$= \left[\dfrac{5 - e^2}{4e} \right|$$$$= \dfrac{e^2 - 5}{4e} $$sq. units.

$$f(x) = \begin{cases} x^2 + ax + b  \ ; \ x < -1 \\ 2x \ ; \ -1 \le x \le 1 \\ x^2 + ax + b \ ; \ x > 1 \end{cases}$$
$$\because f(x)$$ is continuous at $$x = -1 $$ and $$x = 1$$
$$\therefore (-1)^2 + a (-1) + b = -2 \Rightarrow b - a = -3$$
$$2 = (1)^2 + a . 1 + b \Rightarrow a + b = 1$$
On solving, we get $$a = 2, b = -1$$
$$\therefore f(x) = \begin{cases} x^2 + 2x - 1 \ ; \ x < -1\\ 2x \ ; \ -1 \le x \le 1 \\ x^2 + 2x - 1 \ ; \ x > 1 \end{cases}$$
Given curves are $$y = f(x) , x = -2y^2$$ and $$8x + 1 = 0$$
Solving $$x = -2 y^2 , y = x^2 + 2x - 1 $$ (where $$x < -1$$) , we get $$x = -2$$.
Also, $$y = 2x , x = -2y^2$$ meet at $$(0, 0)$$.
$$y = 2x$$ and $$x = -1/8$$ meet at $$\left(- \dfrac{1}{8} , \dfrac{-1}{4}\right)$$.
The required area is the shaded region in fig.
Ref. image 
$$\therefore $$ required area
$$= \displaystyle \int_{-2}^{-1} \left[- \sqrt{\dfrac{-x}{2}} - (x^2 + 2x - 1) dx \right] + \int_{-1}^{-1/8} \left[-\sqrt{\dfrac{-x}{2}} - 2x\right] dx$$
$$= \left[\dfrac{1}{\sqrt{2}} \dfrac{2(-x)^{3/2}}{3} - \dfrac{x^3}{3} - x^2 + x \right]_{-2}^{-1} + \left[\dfrac{1}{\sqrt{2}} \dfrac{2 (-x)^{3/2}}{3} - x^2 \right]_{-1}^{-1/8}$$
$$= \dfrac{257}{192} sq. units$$
$$= \left(\dfrac{\sqrt{2}}{3} + \dfrac{1}{3} - 1 - 1 \right) - \left(\dfrac{4}{3} + \dfrac{8}{3} - 4 - 2 \right) $$$$+ \left(\dfrac{\sqrt{2}}{3} \times \dfrac{1}{16 \sqrt{2}} - \dfrac{1}{64} \right) - \left(\dfrac{\sqrt{2}}{3} - 1 \right)$$
$$= \left(\dfrac{\sqrt{2} - 5}{3} \right) - \left(\dfrac{4 + 8 - 18}{3} \right) + \left(\dfrac{4 - 3}{192} \right) - \left(\dfrac{\sqrt{2} - 3}{3} \right)$$

The given curves are 
$$x = \dfrac{1}{2}$$
$$x =2$$
$$y = \log_e x$$
$$y = 2^x$$
Required area $$ = ABCDA$$
$$= \displaystyle \int_{1/2}^2 (2^x - \log x) dx$$
$$= \left[\dfrac{2^x}{\log 2} - (x \log x - x) \right]^2_{1/2}$$
$$= \left(\dfrac{4}{\log 2} - 2 \log 2 + 2 \right) - \left(\dfrac{\sqrt{2}}{\log 2} - \dfrac{1}{2} \log \dfrac{1}{2} + \dfrac{1}{2} \right)$$
$$ = \left(\dfrac{4 - \sqrt{2}}{\log 2} - \dfrac{5}{2} \log 2 + \dfrac{3}{2} \right) sq. $$ units.

We have An=∫π/40(tanx)ndxAn=∫0π/4(tan⁡x)ndx.

 The given curves are $$y = x^2$$
and $$y = \dfrac{2}{1 + x^2}$$ 
Solving (1) and (2), we have 
$$x^2 = \dfrac{2}{1 + x^2}$$
$$\Rightarrow x^2 + x^2 - 2 = 0$$
$$\Rightarrow (x^2 - 1) (x^2 + 2) = 0$$
$$\Rightarrow x = \pm 1$$
Also, $$ y = \dfrac{2}{1 + x^2}$$ is an even function.
Hence, its graph is symmetrical about y-axis.
At $$x = 0 , y = 2$$, by increasing the values of $$x, y$$ decreases and when $$x \rightarrow \infty , y \rightarrow 0$$.
$$\therefore y = 0$$ is an asymptote of the given curve.
Thus, the graph of the two curves is as follows
Ref. image 
$$\therefore$$ The required area
$$ = 2 \displaystyle \int_0^1  \left(\dfrac{2}{1 + x^2} - x^2 \right) dx$$
$$= \left(4 \tan^{-1} x - \dfrac{2x^3}{3} \right)^1_0$$
$$= \pi - \dfrac{2}{3} sq. \ units$$

We have, $$y^{2} = 2px$$ and $$x^{2} = 2py$$
$$\therefore y = \sqrt {2px}$$
$$\Rightarrow x^{2} = 2p\sqrt {2px}$$
$$\Rightarrow x^{4} = 4p^{2} . (2px)$$
$$\Rightarrow x^{4} = 8p^{3}x$$
$$\Rightarrow x^{4} - 8p^{3} x = 0$$
$$\Rightarrow x(x^{3} - 8p^{3}) = 0$$
$$\Rightarrow x = 0, 2p$$
$$\therefore$$ Required area $$= \therefore \int_{0}^{2p} \sqrt {2px}dx - \int_{0}^{2p} \dfrac {x^{2}}{2p} dx$$
$$= \sqrt {2p} \int_{0}^{2p} x^{1/2} dx - \int_{0}^{2p}\dfrac {1}{2p} x^{2} dx$$
$$= \sqrt {2p}\left [\dfrac {2(x)^{3/2}}{3} \right ]_{0}^{2p} - \dfrac {1}{2p} \left [\dfrac {x^{3}}{3}\right ]_{0}^{2p}$$
$$= \sqrt {2}p\left [\dfrac {2}{3}. (2p^{3/2} - 0\right ] - \dfrac {1}{2p} \left [\dfrac {1}{3} (2p^{3}) - 0\right ]$$
$$= \sqrt {2p} \left (\dfrac {2}{3} . 2\sqrt {2} p^{3/2}\right ) - \dfrac {1}{2p} \left (\dfrac {1}{3} 8p^{3}\right )$$
$$= \sqrt {2p}\left (\dfrac {4\sqrt {2}}{3} p^{3/2}\right ) - \dfrac {1}{2p} \left (\dfrac {8}{3} p^{3}\right )$$
$$= \dfrac {4\sqrt {2}}{3} . \sqrt {2} p^{2} - \dfrac {8}{6} p^{2}$$
$$= \dfrac {(16 - 8)p^{2}}{6} = \dfrac {8p^{2}}{6}$$
$$= \dfrac {4p^{2}}{3} sq\ unit$$.

Area $$OAB = \int_{0}^{\pi} ydx $$

$$= \int_{0}^{\pi}\sin x dx $$

$$= |-\cos x|_{0}^{\pi}$$

$$= \cos 0 - \cos \pi = 2\ sq\ units$$.

Solving the equation $$x^{2} + y^{2} = a^{2}$$ and $$x = \dfrac {a}{2}$$., we obtain their points of intersection which are $$\left (\dfrac {a}{2}, \sqrt {3} \dfrac {a}{2}\right )$$ and $$\left (\dfrac {a}{2}, -\dfrac {\sqrt {3a}}{2}\right )$$.
Hence, From Fig, we get
Required Area $$=$$
$$= 2\left [\dfrac {x}{2} \sqrt {a^{2} - x^{2}} + \dfrac {a^{2}}{2} \sin^{-1} \dfrac {x}{a}\right ]_{\frac {a}{2}}^{a}$$

$$= 2\left [\dfrac {a^{2}}{2} . \dfrac {\pi}{2} - \dfrac {a}{4} . a \dfrac {\sqrt {3}}{2} - \dfrac {a^{2}}{2} . \dfrac {\pi}{6}\right ]$$


$$= \dfrac {a^{2}}{12} (6\pi - 3\sqrt {3} - 2\pi)$$

$$= \dfrac {a^{2}}{12} (4\pi - 3\sqrt {3})sq\ unit$$.

$$\dfrac {37}{3} sq. units$$
$$Area = \int_{3}^{4} xdy$$
$$= \int_{3}^{4} y^{2} dy = \left (\dfrac {y^{3}}{3} \right )_{3}^{4} = \dfrac {64 - 27}{3} = \dfrac {37}{3} sq. units$$.

$$\dfrac {297}{6} sq. units$$
$$Area = \int_{2}^{5} ydx = \int_{2}^{5} x^{2} + xdx$$
$$= \left (\dfrac {x^{3}}{3} + \dfrac {x^{2}}{2}\right )_{2}^{5} = \left (\dfrac {125}{3} + \dfrac {25}{2} - \dfrac {8}{3} - \dfrac {4}{2} \right )$$
$$= \dfrac {297}{6} sq. units$$.

We have $$y^{2} = 9x$$ and $$y = 3x$$
$$\Rightarrow (3x)^{2} = 9x$$
$$\Rightarrow 9x^{2} - 9x = 0$$
$$\Rightarrow 9x (x - 1) = 0$$
$$\Rightarrow x = 1, 0$$
$$\therefore$$ Required area, $$A = \int_{0}^{1} \sqrt {9x} dx - \int_{0}^{1} 3xdx$$
$$= 3\int_{0}^{1} x^{1/2} dx - 3\int_{0}^{1} xdx$$
$$= 3\left [\dfrac {x^{3/2}}{3/2}\right ]_{0}^{1} - 3\left [\dfrac {x^{2}}{2}\right ]_{0}^{1}$$
$$= 3\left (\dfrac {2}{3} - 0\right ) - 3\left (\dfrac {1}{2} - 0\right )$$
$$= 2 - \dfrac {3}{2} = \dfrac {1}{2}sq\ units$$.

The intersecting points of the given parabolas are obtained by solving these equations for $$x$$ and $$y$$, which are $$O(0, 0)$$ and $$B(6, 6)$$.

Solving the given equations of curves, we have
$$x^{2} + ax = 2ax$$

Or $$x = 0, x = a$$, which give

$$y = 0, y = \pm a$$

From Fig. area $$ODAB = \int_{0}^{a} (\sqrt {2ax - x^{2}} - \sqrt {ax})dx$$

Let $$x = 2a \sin^{2}\theta$$. Then $$dx = 4a \sin \theta \cos \theta d\theta$$

and $$x = 0, \Rightarrow \theta = 0, x = a\Rightarrow \theta = \dfrac {\pi}{4}$$.
Again, $$\int_{0}^{a} \sqrt {2ax - x^{2}} dx$$

We have, $$y = -x^{2}$$ and $$x + y + 2 = 0$$
$$\Rightarrow -x - 2 = -x^{2} $$

$$\Rightarrow x^{2} - x - 2 = 0$$

$$\Rightarrow x^{2} + x - 2x - 2 = 0$$

$$\Rightarrow x (x + 1) - 2(x + 1) = 0$$

$$\Rightarrow (x - 2)(x + 1) = 0$$

$$\Rightarrow x = 2, -1$$

$$\therefore$$ Area of shaded region,
$$A = |\int_{-1}^{2} (-x - 2 + x^{2})dx| = |\int_{-1}^{2} (x^{2} - x - 2)dx|$$

$$= \left |\left [\dfrac {x^{3}}{3} - \dfrac {x^{2}}{2} - 2x \right ]\right |_{-1}^{2} = \left |\left [\dfrac {8}{3} - \dfrac {4}{2} - 4 + \dfrac {1}{3} + \dfrac {1}{2} - 2\right ]\right |$$

$$= \left |\dfrac {16 - 12 - 24 + 2 + 3 - 12}{6}\right | = \left |-\dfrac {27}{6}\right | = \dfrac {9}{2} sq\ units$$.

Given equation of the curve is $$y = \sqrt {x - 1}$$
$$\Rightarrow y^{2} = x - 1$$
$$\therefore$$ Area of shaded region, 

We have, $$y^{2} = 9x$$ and $$y = x$$
$$\Rightarrow x^{2} = 9x$$
$$\Rightarrow x^{2} - 9x = 0$$
$$\Rightarrow x(x - 9) = 0$$
$$\Rightarrow x =0, 9$$
$$\therefore$$ Area of shaded region, $$A = \int_{0}^{9} (\sqrt {9x} - x)dx = \int_{0}^{9} 3x^{1/2} dx - \int_{0}^{9} xdx$$
$$= \left (\dfrac {6x^{3/2}}{3} - \dfrac {x^{2}}{2}\right )_{0}^{9}$$
$$= \left (6 . \dfrac {9^{3/2}}{3} - \dfrac {81}{2} - 0\right )$$
$$= 54 - \dfrac {81}{2} = \dfrac {108 - 81}{2} = \dfrac {27}{2} sq\ units$$.

We have, $$y^{2} = 2x$$ and $$x^{2} + y^{2} = 4x$$

$$\Rightarrow x^{2} + 2x = 4x$$

$$\Rightarrow x^{2} - 2x = 0$$

$$\Rightarrow x (x - 2) = 0$$

$$\Rightarrow x = 0, 2$$

Also, $$x^{2} + y^{2} = 4x$$

$$\Rightarrow x^{2} - 4x = -y^{2}$$

$$\Rightarrow x^{2} - 4x + 4 = -y^{2} + 4$$

$$\Rightarrow (x - 2)^{2} - 2^{2} = -y^{2}$$

Given region is $$\left \{(x, 0) : y = \sqrt {4 - x^{2}}\right \}$$ and $$X-axis$$.
We have, $$y = \sqrt {4 - x^{2}} \Rightarrow y^{2} = 4 - x^{2} \Rightarrow x^{2} + y^{2} = 4$$
$$\therefore$$ Area of shaded region, $$A = \int_{-2}^{2} \sqrt {4 - x^{2}} dx = \int_{-2}^{2} \sqrt {2^{2} - x^{2}} dx$$
$$= \left [\dfrac {x}{2} \sqrt {2^{2} - x^{2}} + \dfrac {2^{2}}{2} . \sin^{-1} \dfrac {x}{2} \right ]_{-2}^{2}$$
$$= \dfrac {2}{2} . 0 + 2.\dfrac {\pi}{2} + \dfrac {2}{2} .0 - 2\sin^{-1} (-1) =2. \dfrac {\pi}{2} + 2.\dfrac {\pi}{2}$$
$$= 2\pi sq\ units$$.

Given equation of the curve is $$y = \sqrt {a^{2} - x^{2}}$$
$$\Rightarrow y^{2} = a^{2} - x^{2}\Rightarrow y^{2} + x^{2} = a^{2}$$
$$\therefore$$ Required area of shaded region, $$A = \int_{0}^{a} \sqrt {a^{2} - x^{2}} dx$$
$$= \left [\dfrac {x}{2} \sqrt {a^{2} - x^{2}} + \dfrac {a^{2}}{2} \sin^{-1} \dfrac {x}{2}\right ]_{0}^{a}$$
$$= \left [0 + \dfrac {a^{2}}{2} \sin^{-1}(1) - 0 - \dfrac {a^{2}}{2} \sin^{-1} 0\right ]$$
$$= \dfrac {a^{2}}{2} \cdot \dfrac {\pi}{2} = \dfrac {\pi a^{2}}{4} sq\ units$$.

Given equation of are $$y = \sqrt {x}$$ and $$y = x$$
$$\Rightarrow x = \sqrt {x} \Rightarrow x^{2} = x$$
$$\Rightarrow x^{2} - x = 0\Rightarrow x (x - 1) = 0$$
$$\Rightarrow x = 0, 1$$
$$\therefore$$ Required area of shaded region,
$$A = \int_{0}^{1} (\sqrt {x})dx - \int_{0}^{1} xdx$$

$$= \left [2 . \dfrac {x^{3/2}}{3} \right ]_{0}^{1} - \left [\dfrac {x^{2}}{2}\right ]_{0}^{1}$$

$$= \dfrac {2}{3} . 1 - \dfrac {1}{2} = \dfrac {2}{3} - \dfrac {1}{2} = \dfrac {1}{6} sq\ units$$.

We have, $$y^{2} = 6ax$$ and $$x^{2} + y^{2} = 16a^{2}$$

$$\Rightarrow x^{2} + 6ax = 16a^{2}$$

$$\Rightarrow x^{2} + 6ax - 16a^{2} = 0$$

$$\Rightarrow x^{2} + 8ax - 2ax - 16a^{2} = 0$$

$$\Rightarrow x(x + 8a) - 2a(x + 8a) = 0$$

$$\Rightarrow (x - 2a)(x + 8a) = 0$$

$$\Rightarrow x = 2a, -8a$$

$$\therefore$$ Area of required region $$= 2\left [\int_{0}^{2a} \sqrt {6ax} dx + \int_{2a}^{4a} \sqrt {(4a)^{2} - x^{2}}dx\right ]$$

$$= 2\left [\int_{0}^{2a} \sqrt {6a} x^{1/2} dx + \int_{2a}^{4a} \sqrt {(4a^{2}) - x^{2}}dx\right ]$$

$$= 2\left [\sqrt {6a}\left [\dfrac {x^{3/2}}{3/2}\right ]_{0}^{2a} + \left (\dfrac {x}{2} \sqrt {(4a)^{2} - x^{2}} + \dfrac {(4a)^{2}}{2} \sin^{-1} \dfrac {x}{4a}\right )_{2a}^{4a}\right ]$$

We have, $$y = 1 + |x + 1|, x = -3, x = 3, y = 0$$
$$\therefore y = \left\{\begin{matrix} -x,& if & x < -1\\ x + 2, & if & x\geq -1\end{matrix}\right.$$
$$\therefore$$ Area of shaded region,
$$A = \int_{-3}^{-1} -xdx + \int_{-1}^{3} (x + 2)dx$$

$$= -\left [\dfrac {x^{2}}{2} \right ]_{-3}^{-1} + \left [\dfrac {x^{2}}{2} + 2x\right ]_{-1}^{3}$$

$$= -\left [\dfrac {1}{2} - \dfrac {9}{2}\right ] + \left [\dfrac {9}{2} + 6 - \dfrac {1}{2} + 2\right ]$$

$$= -[-4] + [8 + 4]$$
$$= 12 + 4 = 16\ sq\ units$$.

Given equations of lines are
$$y = 4x + 5 .... (i)$$
$$y = 5 - x .... (ii)$$ and
$$4y = x + 5 .... (iii)$$
On solving Eqs. (i) and (ii), we get
$$4x + 5 = 5 - x$$
$$\Rightarrow x = 0$$
On solving Eqs. (i) and (iii)
$$4(4x + 5) = x + 5$$
$$\Rightarrow 16x + 20 = x + 5$$
$$\Rightarrow 15x = -15$$
$$\Rightarrow x = -1$$
On solving Eqs. (ii) and (iii), we get
$$4(5 - x) = x + 5$$
$$\Rightarrow 20 - 4x = x + 5$$
$$\Rightarrow x = 3$$
$$\therefore$$ Required area $$= \int_{-1}^{0} (4x + 5)dx + \int_{0}^{3} (5 - x)dx - \dfrac {1}{4} \int_{-1}^{3} (x + 5)dx$$

CircleCircle                   

The equation of curves are $$y^2=4x$$.........(i) and 

Considering the curve $$y = x^2 +1$$ and the lines
$$y = x + 1$$
$$y=0$$
$$x=0$$
and $$x = 2$$
according to Question the required region is the shaded region
$$Required \: Area =\overset { 1}{ \underset { 0 }{ \int }  } (x^2+1)dx+area\:  of\:  BMNC$$
$$A=\frac{1}{3}+1+\frac{1}{2}(2+3).1=\frac{23}{6}\Rightarrow      \left [ \frac{23}{A} \right ]=6$$

$${ D }_{ 1 }={ D }_{ 2 }\\ \Rightarrow \int _{ 1 }^{ c }{ \left( \cfrac { 1 }{ x } -\ln { x }  \right) dx } =\int _{ c }^{ a }{ \left( \ln { x } -\cfrac { 1 }{ x }  \right) dx } \\ \Rightarrow \int _{ 1 }^{ a }{ \left( \cfrac { 1 }{ x } -\ln { x }  \right) dx } =0\\ \Rightarrow { \left[ x-\left( x-1 \right) \ln { x }  \right]  }_{ 1 }^{ a }=0\\ \Rightarrow a-\left( a-1 \right) \ln { a } -1-0=0\\ \Rightarrow \left( a-1 \right) \ln { a } =a-1\\ \Rightarrow a=1$$

Roots of $$x^2-x=0$$ are 0,1. and that of $$4x-x^2$$ are 0,4.
intersection of both parabola : 0, $$\dfrac{5}{2}$$
So,
$$\dfrac{\int _{ 0 }^{ 1 }{ (4x-x^2 )dx } + \int _{ 1 }^{ \frac{5}{2} }{[(4x-x^2)-(x^2-x)]dx  } }{\int _{ 0 }^{ 1 }{ -(x^2-x )dx} }=\dfrac{a}{b}$$
$$ \dfrac{a}{b} = \dfrac{\dfrac{5}{3}+\dfrac{27}{8}}{\dfrac{1}{6}}=\dfrac{121}{4}$$
$$ab=121 \times 4= 484$$

The equation $$\displaystyle y=2x-{ x }^{ 2 }$$ represents a opening downwards having vertex at $$(1,1)$$ and crossing the $$x$$-axis at $$(0,0)$$ and $$(2,0)$$.

A)Let $$g(x)= -x $$ and $$ f (x)=x^{2}+2 $$ Then $$g(x) \leq f (x)$$ for all $$x$$ in$$ [0,1]$$.
Moreover $$ y= g(x)$$ represents a straight line and $$y =f (x)$$ represents a parabola.So the required area is
$$\displaystyle \int_{0}^{1} [(x^{2}+2)-(-x)] dx=\displaystyle \int_{0}^{1} (x^{2}+2+x) dx =\frac{x^{3}}{3} +2x+\left. \frac{x^{2}}{2} \right|_{0}^{1}=\frac{17}{6}$$

B) $$ y = x^{2}$$ is a parabola.The curve $$y =2(1 +x^{2})$$ passes through $$ (0,2)$$ and as $$ x\rightarrow \infty y\rightarrow 0$$
Moreover y is never greater than $$2$$ (if $$ y>2 $$ then $$ x^{2}<0).$$ so the shape of $$y= 2/(1+x^{2})$$ is a bell-shaped curve. The points of intersection of these curves is given by $$x^{2}=2/(1+x^{2}), i.e.,(x^{2})^{2}+x^{2}-2=0 $$ which implies $$ x^{2}=-2, 1 $$ so $$ x=\pm 1,$$ and hence $$ y(1)=1,y(-1)=1$$ Required area is
$$\displaystyle =0 \left[ \int_{0}^{1} \frac{2}{1+x^{2}} dx -\int_{0}^{1} x^{2} dx \right] $$
$$\displaystyle =2 \left[ 2 [\tan^{-1} 1]-\frac{1}{3} \right] =2 \left( \frac{\pi}{2}-\frac{1}{3} \right) =\pi-\frac{2}{3}$$

C) The points of intersection of $$ f(x)=\sin x $$ and $$ g(x)=\cos x $$ are given by $$ x=\pi/4$$ and $$ 5\pi/4 $$ Since $$ \sin x \geq \cos x$$ on the interval
$$[\pi/4, 5\pi/4],$$ the area of the required region is
$$\displaystyle \int_{\pi/4}^{5\pi/4} (\sin x-\cos x) dx=\left. -\cos x -\sin x \right|_{\pi/4}^{5\pi/4} $$
$$ \displaystyle = \left [ \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\left( -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\right) \right] =2\sqrt{2}.$$

D) solving the system of equations we get the points of intersection $$ y_{1}=-1 ,y_{2}=1$$
Since $$ 1-3y^{2} \geq -2y^{2} $$ for $$-1\geq y\leq 1,$$ the required area is equal to 
$$\displaystyle \int_{-1}^{1}[(1-3y^{2}) -(-2y)^{2}]dy=\frac{4}{3}$$

$$y=\left| x-1 \right| \Rightarrow y=x-1$$ and $$y=-(x-1)$$

The curves $${ y }^{ 2 }=4x$$ and $${ x }^{ 2 }=4y$$ intersect at the point $$(4,4).$$ 

Given equations to the curves are
$$y=6x-{x}^{2}.........(i)$$
$$y={x}^{2}-2x.........(ii)$$
Solving equation $$(i)$$ and $$(ii)$$ , we get $$x=0,4$$
Required area $$=\displaystyle \int _{ 0 }^{ 4 }{ \left( 6x-{ x }^{ 2 } \right) dx-\int _{ 0 }^{ 4 }{ \left( { x }^{ 2 }-2x \right)  } dx } $$
$$=\displaystyle \int _{ 0 }^{ 4 }{ \left( 6x-{ x }^{ 2 }-{ x }^{ 2 }+2x \right) dx } $$
$$=\displaystyle \int _{ 0 }^{ 4 }{ \left( 8x-2{ x }^{ 2 } \right)  } dx$$
$$={ \left[ 4{ x }^{ 2 }-\cfrac { 2{ x }^{ 3 } }{ 3 }  \right]  }_{ 0 }^{ 4 }=\left( 64-\cfrac { 128 }{ 3 }  \right) =\cfrac { 64 }{ 3 } =21\cfrac { 1 }{ 3 }$$ sq.units

The given curves are
$$x^{2} + y^{2} = 9 $$ ..... (i)
and $$y^{2} = 8x $$ ...... (ii)
To find the points of Intersection of the curves, we solve (i) and (ii), we get
$$x^{2} + 8x = 9$$
$$x^{2} + 8x - 9 = 0$$
$$x^{2} + 9x - x - 9 = 0$$
$$x(x + 9) - 1(x + 9) = 0$$
$$(x + 9)(x - 1) = 0$$,
$$x + 9 = 0\Rightarrow x = -9, x - 1 = 0\Rightarrow x = 1$$
If $$x = -9$$, then from the equation (ii)
$$y^{2} = 8(-9) = -72 < 0$$ which is not possible
When $$x = 1, y^{2} = 8(1) = 8$$
$$\Rightarrow y = \pm 2\sqrt {2}$$
$$\therefore$$ the points of Intersection are $$P(1, 2\sqrt {2})$$ and $$Q(1, -2\sqrt {2})$$ equation (i) represents a circle with center at $$(0, 0)$$ and radius $$3$$.
Equation (ii) represents a parabola with vertex at $$(0, 0)$$.
The required area is shown shaded
$$\therefore$$ Required area $$=$$ Area of the region $$OQAPO$$
$$= 2$$ (Area of the region $$OMAPO)$$
$$= 2 [(\text {Area of the region OMPA)} + (\text {Area of the region PMAP}]$$
Now, Area of the region $$OMPO$$
$$=$$ Area under the parabola $$y^{2} = 8x \Rightarrow y = 2\sqrt {2}\sqrt {x}$$
$$= \displaystyle \int_{0}^{1} 2\sqrt {2} \sqrt {x} dx$$
$$= 2\sqrt {2}\left [\dfrac {x^{\tfrac {3}{2}}}{\dfrac32}\right ]_{0}^{1}$$
$$= 2\sqrt {2} \times \dfrac {2}{3} [(1)^{\tfrac32} - 0] = \dfrac {4\sqrt {2}}{3}$$ .... (iii)
and Area of the region $$PMAP$$
$$=$$ Area under the circle $$x^{2} + y^{2} = 9$$, i.e., $$y = \sqrt {9 - x^{2}}$$
$$=\displaystyle  \int_{1}^{3} \sqrt {9 - x^{2}}dx$$
$$= \left [\dfrac {x}{2}\sqrt {9 - x^{2}} + \dfrac {9}{2} \sin^{-1} \left (\dfrac {x}{3}\right )\right ]_{1}^{3}$$
$$= \left (\dfrac {3}{2}\sqrt {9 - 9} + \dfrac {9}{2} \sin^{-2}(1) \right ) - \left (\dfrac {1}{2}\sqrt {9 - 1} + \dfrac {9}{2} \sin^{-1} \dfrac {1}{3}\right )$$
$$= 0 + \dfrac {9}{2}\cdot \dfrac {\pi}{2} - \sqrt {2} - \dfrac {9}{2} \sin^{-1}\dfrac {1}{3}$$ .... (iv)
$$\therefore$$ required area $$= 2\left (\dfrac {4\sqrt {2}}{3} + \dfrac {9\pi}{4} - \sqrt {2} - \dfrac {9}{2} \sin^{-1}\dfrac {1}{3}\right )$$
$$= \left (\dfrac {8\sqrt {2}}{3} + \dfrac {9\pi}{2} - 2\sqrt {2} - 9\sin^{-1}\dfrac {1}{3}\right )$$ square units

Red graph indicates the equation $$4y^2=9x$$ &  blue curve indicates$$3x^2=16y$$

Solving the equation
$$4{ y }^{ 2 }=9x$$ ...(1)
and $$3{ x }^{ 2 }=16y$$ ...(2)
$$\Rightarrow x=\cfrac { 4 }{ 9 } { y }^{ 2 }$$
Substitute the value of $$x$$ in equation (2), we get
$$3\left( \cfrac { 16{ y }^{ 4 } }{ 81 }  \right) =16y$$ ($$\div$$ by $$16$$)
$$\Rightarrow 3\left( \cfrac { { y }^{ 4 } }{ 81 }  \right) =y$$
$$\Rightarrow 3{ y }^{ 4 }-81y=0$$
$$\Rightarrow 3y({y}^{3}-27)=0$$
When $$y=0,x=0$$
When $$y=3,x=4$$
$$\therefore$$ the point of intersection are $$(0,0)$$ and $$(4,3)$$.
Required area $$=\displaystyle \int _{ 0 }^{ 4 }{ \left( { y }_{ 1 }-{ y }_{ 2 } \right) dx } $$
$$=\displaystyle \int _{ 0 }^{ 4 }{ \left( \cfrac { 3 }{ 2 } { x }^{ 1/2 }-\cfrac { 3 }{ 16 } { x }^{ 3 } \right) dx } $$
$$=\displaystyle \int _{ 0 }^{ 4 }{ { \left[ \cfrac { 3{ x }^{ 3/2 } }{ 2\cfrac { 3 }{ 2 }  } -3\cfrac { { x }^{ 3 } }{ 3(16) }  \right]  }_{ 0 }^{ 4 } } $$
$$={ \left[ { x }^{ 3/2 }-\cfrac { { x }^{ 3 } }{ (16) }  \right]  }_{ 0 }^{ 4 }$$
$$=(8-4)-0$$
$$=4$$ sq.units

Required area $$\displaystyle =\int^{\pi/4}_0\cos xdx+\int^{5\pi /6}_{\pi/4}\sin xdx+\int^{5\pi/3}_{5\pi/5}\dfrac{1}{2}dx+\int^{2\pi}_{5\pi/3}\cos x dx$$

$$\\y_1=x-[x]=\{x\}=fractional\ part\ function \\ y_2 = [x] \\area =((\frac{1}{2})\cdot 1\cdot 1)\cdot 2= (\frac{1}{2})\cdot 2=1\\\therefore m =1$$

$$A=L^2=\pi R^2$$

We have,

$$y={{\sin }^{-1}}x$$

$$y={{\cos }^{-1}}x$$

$$y=0$$

 

Area enclosed by these curves,

$$ =\int_{0}^{1/\sqrt{2}}{{{\sin }^{-1}}xdx}+\int_{1/\sqrt{2}}^{1}{{{\cos }^{-1}}xdx} $$

$$ =\left[ x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}} \right]_{0}^{1/\sqrt{2}}+\left[ x{{\cos }^{-1}}x-\sqrt{1-{{x}^{2}}} \right]_{1/\sqrt{2}}^{1} $$

$$ =\left[ \left( \dfrac{1}{\sqrt{2}}{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)+\sqrt{1-\dfrac{1}{2}} \right)-\left( 0+\sqrt{1-0} \right) \right]+\left[ \left( {{\cos }^{-1}}\left( 1 \right)-\sqrt{1-1} \right)-\left( \dfrac{1}{\sqrt{2}}{{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)-\sqrt{1-\dfrac{1}{2}} \right) \right] $$

$$ =\left[ \left( \dfrac{1}{\sqrt{2}}{{\sin }^{-1}}\left( \sin \dfrac{\pi }{4} \right)+\sqrt{\dfrac{1}{2}} \right)-1 \right]+\left[ \left( {{\cos }^{-1}}\left( \cos 0 \right) \right)-\left( \dfrac{1}{\sqrt{2}}{{\cos }^{-1}}\left( \cos \dfrac{\pi }{4} \right)-\sqrt{\dfrac{1}{2}} \right) \right] $$

$$ =\dfrac{\pi }{4\sqrt{2}}+\dfrac{1}{\sqrt2}-1-\dfrac{\pi }{4\sqrt{2}}+\dfrac{1}{\sqrt{2}}$$

$$ =\sqrt{2}-1 $$

$$=0.41$$sq.units

 

Hence, this is the answer.

Consider the following expression.

$$ \left( i \right)\ 0\le y\le {{x}^{2}}+1 $$

$$ \left( ii \right)\ 0\le y\le x+1 $$

$$ \left( iii \right)\ 0\le x\le 2 $$

 

Therefore, required region will be as shown in the figure.

 

We can see that, we have

$$y={{x}^{2}}+1$$ and $$y=x+1$$

 

Therefore,

$$ {{x}^{2}}+1=x+1 $$

$$ {{x}^{2}}-x=0 $$

$$ x\left( x-1 \right)=0 $$

$$ x=0,x=1 $$

 

Therefore, points of intersection will be,

$$P\left( 0,1 \right)$$ and $$Q\left( 1,2 \right)$$

 

Thus,

Area required $$=$$ Area $$OPQRST$$ $$=$$ Area $$OPQT$$ $$+$$ Area $$QRST$$

 

So, area $$OPQT$$ will be,

$$ \Rightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}+1 \right)dx} $$

$$ \Rightarrow \left[ \dfrac{{{x}^{3}}}{3}+x \right]_{0}^{1} $$

$$ \Rightarrow \dfrac{4}{3} $$

 

Also, area $$QRST$$ will be,

$$ \Rightarrow \int\limits_{1}^{2}{\left( x+1 \right)dx} $$

$$ \Rightarrow \left[ \dfrac{{{x}^{2}}}{2}+x \right]_{1}^{2} $$

$$ \Rightarrow \dfrac{5}{2} $$

 

Therefore,

Area required $$=\dfrac{4}{3}+\dfrac{5}{2}=\dfrac{23}{6}\ sq.\ units$$

 

Hence, this is the required result.

$$y \ge |x + 2|$$

For $$p$$   $$y=\sqrt x$$

first find intersection point in first quadrant 

$$x^2+y^2=1$$                   $$(x-1)^2+(y-0)^2=1$$

$$  {\textbf{Step -1: Find intersecting points}} $$

               $$  {x^2} + {y^2} = 4\;{\text{is a circle having centre at }}\left( {0,0} \right){\text{ and radius 2 units}} $$

               $$  {\text{and the line is }}x = \sqrt 3 y. $$

                                                       

               $$  {\text{For intersecting point of given circle and line}} $$

               $$  {\text{Putting }} x = \sqrt 3 y\;{\text{ in }}{x^2} + {y^2} = 4{\text{ we get }} $$

               $$   \Rightarrow {\left( {\sqrt 3 y} \right)^2} + {y^2} = 4 $$

               $$   \Rightarrow 3{y^2} + {y^2} = 4 $$

               $$   \Rightarrow 4{y^2} = 4 $$

               $$   \Rightarrow y =  \pm 1 $$

               $$  {\text{So, }}x = \sqrt 3 y =  \pm \sqrt 3  $$

               $$  {\text{Intersecting points are }}\left( {\sqrt 3 ,1} \right),\left( { - \sqrt 3 , - 1} \right) $$

$$  {\textbf{Step -2: Find the area}} $$

               $$  {\text{Shaded region is required region}} $$

               $$  {\text{Now required area  =  }}\int\limits_0^{\sqrt 3 } {\dfrac{x}{{\sqrt 3 }}dx + \int\limits_{\sqrt 3 }^2 {\sqrt {4 - {x^2}} } }  $$

               $$   = \dfrac{1}{{\sqrt 3 }}\left[ {\dfrac{{{x^2}}}{2}} \right]_0^{\sqrt 3 } + \left[ {\dfrac{{x\sqrt {4 - {x^2}} }}{2} + \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{x}{2}} \right]_{\sqrt 3 }^2 $$

               $$   = \dfrac{1}{{2\sqrt 3 }}\left( {3 - 0} \right) + \left[ {2{{\sin }^{ - 1}}1 - \left( {\dfrac{{\sqrt 3 }}{2} + 2{{\sin }^{ - 1}}\dfrac{{\sqrt 3 }}{2}} \right)} \right] $$

               $$   = \dfrac{{\sqrt 3 }}{2} + \left[ {2\dfrac{\pi }{2} - \dfrac{{\sqrt 3 }}{2} - \dfrac{{2\pi }}{3}} \right] $$

               $$   = \dfrac{{\sqrt 3 }}{2} + \pi  - \dfrac{{\sqrt 3 }}{2} - \dfrac{{2\pi }}{3} $$

               $$   = \pi  - \dfrac{{2\pi }}{3} $$

               $$   = \dfrac{\pi }{3}\;{\text{sq}}{\text{.unit}}{\text{.}} $$

$$  {\textbf{Hence, the required area is }}\dfrac{\pi }{3}\;{\textbf{sq}}{\textbf{.unit}}{\text{.}} $$ 

Area $$=\int _{ 0 }^{ 4 }{ \sqrt { 16x+{ x }^{ 2 } }  } =\int _{ 0 }^{ 4 }{ \sqrt { { \left( x+8 \right)  }^{ 2 }-64 }  } $$

$$AO = OB = \dfrac{AB}{2}$$

$$\begin{array}{l} 2x+y=6 \\ put\, \, x=0 \\ y=6 \\ put\, x=1 \\ y=4 \\ put\, \, x=2 \\ y=2 \\ and\, \, 2x-y+2=0 \\ put\, \, y=0 \\ x=1 \\ put\, \, x=0 \\ y=2 \\ put\, x=2 \\ y=6 \\ and\, \, x-y=0 \\ x=0,\, \, y=0 \\ x=1,y=1 \\ x=2,\, y=2 \end{array}$$

$$\begin{array}{l} \Rightarrow Area\, \, of\, \, square\, \, =14\times 14=196\, \, c{ m^{ 2 } } \\ \Rightarrow Area\, \, of\, \, two\, \, Semicircles=\pi { r^{ 2 } }=\frac { { 22 } }{ 7 } \times 7\times 7 \\ \Rightarrow Area\, \, of\, \, two\, \, Semicircles=154\, \, c{ m^{ 2 } } \\ Then,\, \, shaded\, \, area=196-154\, \, \, =42\, \, c{ m^{ 2 } }\, \, \, \, \, Ans. \end{array}$$

According to the given conditions

$$\int_{0}^{t}[f(x)-(x^4-4x^2)]dx\,=\,k\int_{0}^{t}[(2x^2-x^3)-f(x)]dx $$

Differentiate both sides w.r.t $$'t'$$, we get

$$f(t)-(t^4-4t^2)\,=\, k(2t^2-t^3-f(t))$$ or

$$(1+k)f(t)\,=\, k2t^2-kt^3+t^4-4t^2 $$

$$ \Rightarrow f(t)=\dfrac{1}{k+1} [t^4-kt^3+(2k-4)t^2] $$

Hence,required $$f$$ is given by $$f(x)\,=\dfrac{1}{k+1} (x^4-k^3+2(k-2)x^2) $$.