MCQ Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 4

Find the total area between the curve

$y=x^3$ and the x-axis between

$x=-2$ and

$x=2$ .

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

Required Area =

$\int_{0}^{2} x^3 \ dx +\int_{-2}^{0} x^3 \ dx$

$=\left [\dfrac{x^4}{4}\right ]_{0}^{2}+\left [\dfrac{x^4}{4}\right ]_{-2}^{0}$

$=\dfrac{16}{4}+\left (-\dfrac{16}{4}\right )$

We would take

$-\dfrac{16}{4}$ as positive as it is area.

Thus, required area =

$4+4=8 \ sq. units$

1461738_292620_ans_746133c9fcbe46499e3f54f6cc2efbec.PNG

The area bounded by the parabola

$y^2=x$ and the straight line

$2y=x$ is

0%
$\displaystyle\frac{4}{3}$ sq. units
0%
$1$ sq. units
0%
$\displaystyle\frac{2}{3}$ sq. units
0%
$\displaystyle\frac{1}{3}$ sq. units

Explanation

Given curves are

$y^2=x$ and

$2y=x$

$\Rightarrow y^2=2y\Rightarrow y=2$

$\therefore\displaystyle A=\int _{ 0 }^{ 2 }{ \left( { y }^{ 2 }-2y \right) dy } ={ \left[ \frac { { y }^{ 3 } }{ 3 } -{ y }^{ 2 } \right] }_{ 0 }^{ 2 }$

$\displaystyle =\frac { 4 }{ 3 }$ sq. units

$[$ neglect negative sign as area will be positive

$]$

The area of the smaller portion between curves

$x^2 + y^2 = 8$ and

$y^2 = 2x$ is

0%
$\displaystyle \pi + \frac{2}{3}$
0%
$\displaystyle 2\pi + \frac{2}{3}$
0%
$\displaystyle 2 \pi + \frac{4}{3}$
0%
$\displaystyle \pi + \frac{4}{3}$

Explanation

Given curves

$x^2 + y^2 = 8$ ....(1)
Center is (0,0) and radius

$r=2\sqrt{2}$

$y^2 = 2x$ ....(2)
Solving (1) and (2), we get

$x^2+2x=8$

$\Rightarrow x=2,-4$
At

$x=2 \Rightarrow y=2,-2$
Hence, the point of intersection of the curves is

$(2,2)$ and

$(2,-2)$

Now, required area

$=2(area OPA)$

$=2[ar(OPB)+ar(PBA)]$

$= \displaystyle 2 \int_0^2 \sqrt{2x} dx + 2 \int_{2}^{2 \sqrt 2} \sqrt{8 - x^2} dx$

$=2\sqrt{2}[\dfrac{x^{3/2}}{3/2}]_{0}^2 +2 [\dfrac{x}{2}\sqrt{8 - x^2}+4\sin^{-1}{\dfrac{x}{2\sqrt{2}}}]_{2}^{2\sqrt{2}}$

$=\dfrac{4\sqrt{2}}{3} (2\sqrt{2})+2(0+{2}{\pi}-2-\pi)$

Required Area

$= \displaystyle \left( 2 \pi + \frac{4}{3} \right )$ sq.units

Lines

$7x-2y+10=0$ and

$7x+2y-10=0$ forms an isosceles triangle with the line

$y=2$ . Area of this triangle is equal to.

0%
$\cfrac {15}{7}$
0%
$\cfrac {10}{7}$
0%
$\cfrac {18}{7}$
0%
$\cfrac {8}{7}$

Explanation

Line

$L_1$ and line

$L_2$ intersect at

$y=5,x=0$

$L_1, x=\dfrac{2y}{7}-\dfrac{10}{7}$

and

$L_2, x=\dfrac{10}{7}-\dfrac{2y}{7}$

hence area bounded by both line and

$y=0$ is

$A=\int_{2}^{5}(\dfrac{2y}{7}-\dfrac{10}{7})+\int_{2}^{5}(\dfrac{10}{7}-\dfrac{2y}{7})$

$A=2\times [\int_{2}^{5}(\dfrac{2y}{7}-\dfrac{10}{7})]$

on solving and putting upper and lower limit we will get

$A=\dfrac{18}{7}$

Area lying in the first quadrant and bounded by the circle

$x^2 + y^2 = 4$ and the lines

$x = 0$ and

$x = 2$ is

0%
$\pi$
0%
$\dfrac {\pi}{2}$
0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{4}$

Explanation

The area bounded by the circle and the lines,

$x=0$ and

$x=2$ , in the first quadrant is represented as shaded region in the plot.

$\therefore$ Area

$OAB=\int_0^2y dx$

$=\displaystyle \int_0^2\sqrt {4-x^2}dx$

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

$=2\left (\dfrac {\pi}{2}\right )=\pi$ sq. units
Thus, the correct answer is A.

Smaller area enclosed by the circle

$x^2 + y^2 = 4$ and the line

$x + y = 2$ is

0%
$2(\pi -2)$
0%
$\pi -2$
0%
$2\pi -1$
0%
$2(\pi +2)$

Explanation

The smaller area enclosed by the circle,

$x^2+y^2=4$ and the line,

$x+y=2$ is represented by the shaded area

$ACBA$ as shown in the diagram.
It can be observed that,

$Area\ ACBA = Area\ OACBO - Area\ (\Delta OAB)$

$=\displaystyle \int_0^2\sqrt {4-x^2}dx-\int_0^2(2-x)dx$

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

$=\left [2\cdot \dfrac {\pi}{2}\right ]-[4-2]$

$=(\pi -2)$ sq. units
Thus, the correct answer is

$B$ .

Area bounded by the curves

$y=x^3$ , the

$x$ -axis and the ordinates

$x=-2$ and

$x=1$ is

0%
$-9$
0%
$-\dfrac {15}{4}$
0%
$\dfrac {15}{4}$
0%
$\dfrac {17}{4}$

Explanation

Required area

$=-\displaystyle \int_{-2}^{0}x^3 dx+\int_{0}^{1}x^3 dx$

$=-\left [\dfrac {x^4}{4}\right ]_{-2}^{0}+\left [\dfrac {x^4}{4}\right ]_{0}^{1}$

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

$=\left (\dfrac {1}{4}+4\right )=\dfrac {17}{4}$ sq. units
Thus, the correct answer is D.

The area bounded by

${y}^{2}=4x$ and

${x}^{2}=4y$ is

0%
$\cfrac { 20 }{ 3 }$ sq. units
0%
$\cfrac { 16 }{ 3 }$ sq. units
0%
$\cfrac { 14 }{ 3 }$ sq. units
0%
$\cfrac { 10 }{ 3 }$ sq. units

Explanation

The intersection point of a curves

${y}^{2}=4x$ and

${x}^{2}=4y$ are

$O(0,0)$ and

$A(4,4)$

$\therefore$ Required area

$\displaystyle \int _{ 0 }^{ 4 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) dx }$

$=\displaystyle \int _{ 0 }^{ 4 }{ \left( \sqrt { 4x } -\cfrac { { x }^{ 2 } }{ 4 } \right) dx }$

$={ \left[ \cfrac { 2{ x }^{ 3/2 } }{ 3/2 } -\cfrac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }\quad$

$=\cfrac { 4 }{ 3 } \left[ { 4 }^{ 3/2 }-0 \right] -\cfrac { 1 }{ 12 } \left[ { 4 }^{ 3 }-0 \right]$

$=\cfrac { 32 }{ 3 } -\cfrac { 16 }{ 3 } =\cfrac { 16 }{ 3 }$ sq. units

The area of the region bounded by the y-axis,

$y = \cos x$ and

$y = \sin x, 0\leq x \leq \dfrac {\pi}{2}$ is

0%
$2(\sqrt {2} - 1)$
0%
$\sqrt {2} - 1$
0%
$\sqrt {2}+ 1$
0%
$\sqrt {2}$

Explanation

Refer image 1.

Finding point of Intersection

$B$

Solving

$y=\cos x$ and

$y=\sin x$

$\cos x=\sin x$

Refer image 2.

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

Also,

$y=\cos x=\cos \dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$

So,

$B=\left(\dfrac{\pi}{4},\dfrac{1}{\sqrt{2}}\right)$

Refer image 3.

Refer image 4.

Area

$ABCO$

Area ABCO

$\displaystyle \int^{\pi/4}_0 ydx$

Here,

$y=\cos x$

Thus,

Area ABCO=

$\displaystyle \int^{\pi/4}_0 \cos x dx$

$=[\sin x]^{\pi/4}_0$

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

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

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

Refer image 5.

Area

$BCO$

Area BCO

$\displaystyle \int^{\pi/4}_0 ydx$

Here,

$y=\sin x$

Thus,

Area BCO =

$\displaystyle \int^{\pi/4}_0 \sin x dx$

$=-[\cos x]^{\pi/4}_0$

$=-[\cos\dfrac{\pi}{4}-\cos(0)]$

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

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

Therefore

Area Required = Area ABCD - Area BCO

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

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

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

$=\sqrt{2}-1$

$\therefore$ Option

$B$ is correct.

1521667_428566_ans_d0ca9efb62ec4cc7a2bba45b0ea202ff.png

The area of the region bounded by the curve

$y=x^3$ , its tangent at

$(1, 1)$ and

$x$ -axis, is

0%
$\dfrac{1}{12}$ sq unit
0%
$\dfrac{1}{6}$ sq unit
0%
$\dfrac{2}{17}$ sq unit
0%
$\dfrac{2}{15}$ sq unit

Explanation

We have,

$y=x^3$ and

$A(1,1)$

$\therefore \dfrac{dy}{dx}=3x^2$

On putting

$x=1$ in Eq. (i), we get

$\dfrac{dy}{dx}=3(1)^2=3$

$\therefore$ Equation of tangent at

$A(1,1)$ is

$y-1=3(x-1)\Rightarrow y=3x-2$

$\therefore$ Required area

$=\displaystyle \int^1_0x^3dx-\int^1_{2/3}(3x-2)dx$

$=\displaystyle \left[\frac{x^4}{4}\right]^1_0-\left[\frac{3x^2}{2}-2x\right]^2_{2/3}$

$=\displaystyle \frac{1}{4}\left[\left(\frac{3}{2}-2\right)-\left(\frac{2}{3}-\frac{4}{3}\right)\right]$

$=\displaystyle \frac{1}{12}$ sq unit

The area enclosed between

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

$y=x$

0%
$\cfrac{2}{3}$ sq unit
0%
$\cfrac{1}{2}$ sq unit
0%
$\cfrac{1}{3}$ sq unit
0%
$\cfrac{1}{6}$ sq unit

Explanation

Intersection point of two curves are

$O(0,0)$ and

$A(1,1)$

$\therefore$ Required area

$=\displaystyle \int _{ 0 }^{ 4 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) dx }$

$\displaystyle \int _{ 0 }^{ 1 }{ \left( \sqrt { x } -x \right) dx }$

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

$=\cfrac { 2 }{ 3 } \left[ 1-0 \right] -\cfrac { 1 }{ 2 } \left( 1-0 \right)$

$=\cfrac { 2 }{ 3 } -\cfrac { 1 }{ 2 } =\cfrac { 1 }{ 6 }$ unit

$f(x)={x}^{2/3}, x \ge 0$ . Then,the area of the region enclosed by the curve

$y=f(x)$ and the three lines

$y=x, x=1$ and

$x=8$ is

0%
$\cfrac{63}{2}$
0%
$\cfrac{93}{5}$
0%
$\cfrac{105}{7}$
0%
$\cfrac{129}{10}$

Explanation

Given

$f(x)={x}^{2/3}, x \ge 0$ and line

$y=x$

$\therefore$ Required area

$A=\displaystyle \int _{ x=1 }^{ 8 }{ \left( x-{ x }^{ 2/3 } \right) dx }$

$={ \left[ \cfrac { { x }^{ 2 } }{ 2 } -\cfrac { 3 }{ 5 } { x }^{ 5/3 } \right] }_{ 1 }^{ 8 }$

$=\left( 32-\cfrac { 3 }{ 5 } \times 32 \right) -\left( \cfrac { 1 }{ 2 } -\cfrac { 3 }{ 5 } \right) =32\times \cfrac { 2 }{ 5 } -\cfrac { (5-6) }{ 10 }$

$=\cfrac { 64 }{ 5 } +\cfrac { 1 }{ 10 } =\cfrac { 128+1 }{ 10 } =\cfrac { 129 }{ 10 }$

The area enclosed between the curve

$\displaystyle y=1+{ x }^{ 2 }$ , the y-axis and the straight line

$\displaystyle y=5$ is given by

0%
$\displaystyle \frac { 14 }{ 3 }$ sq unit
0%
$\displaystyle \frac { 7 }{ 3 }$ sq unit
0%
$\displaystyle 5$ sq unit
0%
$\displaystyle \frac { 16 }{ 3 }$ sq unit

Explanation

Given curve is

$\displaystyle y=1+{ x }^{ 2 }$ and line

$\displaystyle y=5$

$\displaystyle \therefore$ Required area

$\displaystyle =\int _{ 1 }^{ 5 }{ x } dy=\int _{ 1 }^{ 5 }{ \sqrt { y-1 } dx }$

$\displaystyle ={ \left[ \frac { { \left( y-1 \right) }^{ { 3 }/{ 2 } } }{ { 3 }/{ 2 } } \right] }_{ 1 }^{ 5 }=\frac { 2 }{ 3 } \left[ { \left( 4 \right) }^{ { 3 }/{ 2 } }-0 \right]$

$\displaystyle =\frac { 16 }{ 3 }$ sq unit.

The area bounded by the parabolas

$y=4x^2,\,y=\dfrac{x^2}{9}$ and line

$y=2$ is

0%
$\dfrac{5\sqrt{2}}{3}$ sq units
0%
$\dfrac{10\sqrt{2}}{3}$ sq units
0%
$\dfrac{15\sqrt{2}}{3}$ sq units
0%
$\dfrac{20\sqrt{20}}{3}$ sq units

Explanation

$y=4x^2\;\;\;\dots(i)$

$y=\dfrac{x^2}{4}\;\;\;\dots(ii)$

$\therefore\;A=\int_0^2\begin{bmatrix}\dfrac{\sqrt{y}}{2}-3\sqrt{y}\end{bmatrix}dy$

$=\begin{pmatrix}\dfrac{1}{2}-3\end{pmatrix}\int_0^2\sqrt{y}dy$

$=\begin{pmatrix}\dfrac{-5}{2}\end{pmatrix}.\dfrac{2}{3}\begin{bmatrix}y^{{3}/{2}}\end{bmatrix}_0^2$

$=-\dfrac{5}{3}(2\sqrt{2}-0)=\begin{vmatrix}\dfrac{-10\sqrt{2}}{3}\end{vmatrix}$

$=\dfrac{10\sqrt{2}}{3}\Rightarrow\text{Area of bounded Area}=2A=\dfrac{20\sqrt{20}}{3}$

The area of the circle

$x^2+y^2=16$ exterior to the parabola

$y^2=6x$ is

0%
$\dfrac {4}{3}(4\pi -\sqrt 3)$
0%
$\dfrac {4}{3}(4\pi +\sqrt 3)$
0%
$\dfrac {4}{3}(8\pi -\sqrt 3)$
0%
$\dfrac {4}{3}(8\pi +\sqrt 3)$

Explanation

The given equations are

$x^2+y^2=16........ (1)$

$y^2=6x ....... (2)$

Area bounded by the circle and parabola

$=2[Area (OADO)+Area (ADBA)]$

$=2[\displaystyle \int_{0}^{2}\sqrt {16x}dx+\int_{2}^{4}\sqrt {16-x^2}dx]$

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

$=2\sqrt 6\times \dfrac {2}{3}\left [x^{\dfrac {3}{2}}\right ]_{0}^{2}+2\left [8\cdot \dfrac {\pi}{2}-\sqrt {16-4}-8 \sin^{-1}\left (\dfrac {1}{2}\right )\right ]$

$=\dfrac {4\sqrt 6}{3}(2\sqrt 2)+2\left [4\pi-\sqrt {12}-8\dfrac {\pi}{6}\right ]$

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

$=\dfrac {4}{3}[4\sqrt 3+6\pi - 2\sqrt 3-2\pi ]$

$=\dfrac {4}{3}[\sqrt 3+4\pi]$

$=\dfrac {4}{3}[4\pi +\sqrt 3]$ sq. units

Area of circle

$=\pi(r)^2$

$=\pi(4)^2$

$=16 \pi$ sq. units

$\therefore$ Required area

$=16\pi -\dfrac {4}{3}[4\pi +\sqrt 3]$

$=\dfrac {4}{3}[4\times 3\pi -4\pi -\sqrt 3]$

$=\dfrac {4}{3}(8\pi -\sqrt 3)$ sq. units

Thus, the correct answer is C.

The area bounded by the curve

$y=x|x|$ ,

$x$ -axis and the ordinates

$x=-1$ and

$x=1$ is given by

0%
$0$
0%
$\dfrac {1}{3}$
0%
$\dfrac {2}{3}$
0%
$\dfrac {4}{3}$

Explanation

$y=x^2$ if

$x > 0$ and

$y=-x^2$ if

$x < 0$
Required area

$=\displaystyle \int_{-1}^{1}y dx$

$=\displaystyle \int_{-1}^{1}x|x|dx$

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

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

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

$=\dfrac {2}{3}$ sq. units

Thus, the correct answer is C.

The area of the region bounded by the curves

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

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

0%
$\dfrac {1}{3}$
0%
$\dfrac {1}{2}$
0%
$\dfrac { 1 }{ 4 }$
0%
$3$

Explanation

Given curves are

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

$x={y}^{2}$ , which is the form of parabola.
The point of intersection,

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

$\Rightarrow x={ x }^{ 4 }\Rightarrow x\left( 1-{ x }^{ 3 } \right) =0$

$\Rightarrow x=0$ and

$1={ x }^{ 3 }$

$\Rightarrow x=0$ and

$x=1$
When

$x=0$ , then

$y=0$
When

$x=1$ , then

$y={ 1 }^{ 2 }=1$

$\therefore$ The point of intersection is

$\left( 0,0 \right)$ and

$\left( 1,1 \right)$ .

$\therefore$ Area of shaded region

$=\displaystyle\int _{ 0 }^{ 1 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) dx }$

$=\displaystyle\int _{ 0 }^{ 1 }{ \left[ \sqrt { x } -{ x }^{ 2 } \right] dx } ={ \left[ \dfrac { { x }^{ { 3 }/{ 2 } } }{ { 3 }/{ 2 } } -\dfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 }$

$=\dfrac { 2 }{ 3 } { \left( 1 \right) }^{ { 3 }/{ 2 } }-\dfrac { { \left( 1 \right) }^{ 3 } }{ 3 } -0-0$

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

Area of the region bounded by

$y = |x|$ and

$y = |x| + 2$ , is

0%
$4\ sq. units$
0%
$3\ sq. units$
0%
$2\ sq. units$
0%
$1\ sq. units$

Explanation

The graph is shown in image

$1^{st}$ quadrant , intersection point of

$y=x$ and

$y=-x+2$ is

$(1,1)$

similarly in

$2^{nd}$ quadrant , intersection point of

$Y=X$ and

$y=-x+2$ is

$(-1,1)$

now

$A=2\Bigg[\int_{0}^{1}\bigg[-x+2\bigg]-\int_{0}^{1}\bigg[x\bigg]\Bigg]$

$A=2\times{1}$

$A=2 sq.unit$

The area included between the parabolas

$x^2=4y$ and

$y^2=4x$ is (in square units)

0%
$\dfrac{4}{3}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{16}{3}$
0%
$\dfrac{8}{3}$

Explanation

One intersection point of the two parabolas is the origin, while the other can be found out by substituting either equation into the other.

Thus we have

$\left (\dfrac{x^2}{4}\right)^2 = 4x$

which yields

$x^4 = 64x$ i.e.

$x = 0,4$

Area included between the two curves can be found out by

$\displaystyle \int_0^4 \left (2\sqrt{x} - \dfrac{x^2}{4}\right) dx$

$=\left [ \dfrac{4x^{1.5}}{3} - \dfrac{x^3}{12}\right] _0^4$

$= \dfrac{32}{3} - \dfrac{64}{12}$

$= \dfrac{32}{6}$

The area included between the parabolas

$\displaystyle { y }^{ 2 }=4x$ and

$\displaystyle { x }^{ 2 }=4y$ is

0%
$\displaystyle \frac { 8 }{ 3 }$ sq unit
0%
$\displaystyle 8$ sq unit
0%
$\displaystyle \frac { 16 }{ 3 }$ sq unit
0%
$\displaystyle 12$ sq unit

Explanation

We know that, the area of region bounded by the parabolas

$\displaystyle { y }^{ 2 }=4ax$ and

$\displaystyle { x }^{ 2 }=4by$ is

$\displaystyle \frac { 16 }{ 3 } ab$ sq.unit.
Therefore,

$\displaystyle { y }^{ 2 }=4ax$ and

$\displaystyle { x }^{ 2 }=4y$ is

$\displaystyle \frac { 16 }{ 3 }$ sq.unit.

$\displaystyle \left( \because a=1,b=1 \right)$

The area enclosed by

$y=\sqrt{5-x^2}$ and

$y=|x-1|$ is

0%
$\begin{pmatrix}\dfrac{5\pi}{4}-2\end{pmatrix}$ sq. units
0%
$\dfrac{5\pi - 2}{2}$ sq. units
0%
$\begin{pmatrix}\dfrac{5\pi}{4}-\dfrac{1}{2}\end{pmatrix}$ sq. units
0%
$\begin{pmatrix}\dfrac{\pi}{2}-5\end{pmatrix}$ sq. units

Explanation

Finding the intersection points

$\sqrt { 5-{ x }^{ 2 } } =|x-1|$

$5-{ x }^{ 2 }={ x }^{ 2 }-2x+1$

$\Rightarrow 2{ x }^{ 2 }-2x-4={ 0 }$

$x=2,-1$

So area

$=\int _{ -1 }^{ 2 }{ (\sqrt { 5-{ x }^{ 2 } } -|x-1|)dx }$

$=\int _{ -1 }^{ 2 }{ (\sqrt { 5-{ x }^{ 2 } } )dx-\int _{ -1 }^{ 2 }{ |x-1| } dx }$

$=\int _{ -1 }^{ 2 }{ (\sqrt { 5-{ x }^{ 2 } } )dx-[\int _{ -1 }^{ 1 }{ (1-x) } dx+\int _{ 1 }^{ 2 }{ (x-1) } dx } ]$

$=\left[\dfrac { x }{ 2 } \sqrt { 5-{ x }^{ 2 } } +\dfrac { 5 }{ 2 } \sin ^{ -1 }{ \dfrac { x }{ \sqrt { 5 } } } \right]_{-1}^{2}-\left(2-0+\dfrac { 3 }{ 2 } -1\right)$

$=\left[\sqrt { 1 } +\dfrac { 5 }{ 2 } \sin ^{ -1 }{ \dfrac { 2 }{ \sqrt { 5 } } } -\left(-1+\dfrac { 5 }{ 2 } \sin ^{ -1 }{ \dfrac { -1 }{ \sqrt { 5 } } } \right)\right]-\dfrac { 5 }{ 2 }$

$=2+\dfrac { 5 }{ 2 } \times \dfrac { \pi }{ 2 } -\dfrac { 5 }{ 2 }$

$=\dfrac { 5\pi }{ 4 } -\dfrac { 1 }{ 2 }$

Area of the region satisfying

$x \le 2, y \le |x|,x-axis$ and

$x\ge 0$ is:

0%
$4$ sq unit
0%
$1$ sq unit
0%
$2$ sq unit
0%
None of these

Explanation

Required area = Area of shaded region OAB

$=^2_oydx=^2_0xdx=\frac{x^2}{2}^2_0$

$=$ 2 sq unit
Alternate Solution
Required area = Area of

$\Delta$ OAB

$=\frac{1}{2}\times 2\times 2$

$=$ 2 sq unit

The area of the region bounded by the curves

$y = x^{3}, y = \dfrac {1}{x}, x = 2$ is

0%
$4 - \log_{e}2$
0%
$\dfrac {1}{4} + \log_{e}2$
0%
$3 - \log_{e}2$
0%
$\dfrac {15}{4} - \log_{e}2$

Explanation

First of all we draw the graph,

$y = x^{3}, y = \dfrac {1}{x}, x = 2$

$\therefore$ Required area i.e.,

$(OMPNO)$

$=\displaystyle \int_{0}^{1} x^{3} dx + \int_{1}^{2} \dfrac {1}{x}dx$

$=\left [\dfrac {x^{4}}{4}\right ]_{0}^{1} + [\log_{e}x]_{1}^{2} = \dfrac {1}{4} + \log_{e} 2$

The area (in square units) bounded by the curves

$y^{2} = 4x$ and

$x^{2} = 4y$ is

0%
$\dfrac {64}{3}$
0%
$\dfrac {16}{3}$
0%
$\dfrac {8}{3}$
0%
$\dfrac {2}{3}$

Explanation

$y^{2} = 4x$

$x^{2} = 4y$

solving two equations

$\dfrac {y^{2}}{4} = x$

$\left (\dfrac {y^{2}}{4}\right )^{2} = 4y$

$\dfrac {y^{4}}{4\times 4} = 4y$

$y^{4} = 64y$

$y(y^{2} - 64) = 0$

$y = 0, y^{3} = 64$

$y = 4$

$\int_{0}^{4} \left (\dfrac {y^{2}}{4} - 2\sqrt {y}\right )dy$

$= \left [\dfrac {y^{3}}{4\times 3} - \dfrac {y^{\dfrac {B}{2}}}{3}\right ]_{0}^{4}$

$= \left |\dfrac {-16}{3}\right | = \dfrac {16}{3}$

The area of the region enclosed between parabola

$y^2 = x$ and the line

$y = mx$ is

$148$ . The value of

$m$ is:

0%
$-2$
0%
$-1$
0%
$1$
0%
$2$

Explanation

The graph is shown in the image.

The intersection points of

$y^2=x$ and

$y=mx$ are

$(0,0)$ and

$\left(\dfrac{1}{m^2},\dfrac{1}{m}\right)$

Given: area enclosed by

$f(x)$ ,

$y=mx$ is

$\dfrac{1}{48}$

Now

$A=\displaystyle \int_{0}^{\tfrac{1}{m^2}}\Bigg[\sqrt{x}\Bigg]-\displaystyle \int_{0}^{\tfrac{1}{m^2}}\Bigg[mx \Bigg]$

$\Rightarrow \dfrac{1}{48}=\Bigg[\dfrac{2x^{\tfrac{3}{2}}}{3}\Bigg]\Bigg|_0^{\tfrac{1}{m^2}} -\Bigg[\dfrac{mx^2}{2}\Bigg]\Bigg|_0^{\tfrac{1}{m^2}}$

$\Rightarrow \dfrac{1}{48}=\dfrac{2}{3m^2}-\dfrac{1}{2m^3}$

$\Rightarrow \dfrac{1}{48}=\dfrac{1}{6m^3}$

$\Rightarrow 8=m^3$

$\Rightarrow m=2$

The area of the region bounded by the parabola

$y = x^2-4x+5$ and the straight line

$y = x + 1$ is:

0%
$\cfrac{1}{2}$
0%
$2$
0%
$3$
0%
$\cfrac{9}{2}$

Explanation

The graph is shown in the image.

The intersection points of

$y=x+1$ and

$y=x^2-4x+5$

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

$x(x-4)-1(x-4)=0\Rightarrow (x-1)(x-4)=0\Rightarrow x=1,4$

$x=1\Rightarrow y=1+1=2$ and

$x=4\Rightarrow y=4+1=5$

$\therefore$ The intersection points are

$(1,2)$ and

$(4,5)$ .

Now

$A=\displaystyle \int_{1}^{4}\Bigg[x+1\Bigg]-\displaystyle \int_{1}^{4}\Bigg[x^2-4x+5\Bigg]$

$A=\Bigg[\cfrac{x^2}{2}+x \Bigg]\Bigg|_1^4 -\Bigg[\dfrac{x^3}{3}-2x^2+5x \Bigg]\Bigg|_1^4$

$A=\left[ \dfrac {16}{2}+4-\dfrac 12 -1\right]-\left[\dfrac {64}{3}-32+20-\dfrac 13+2-5\right]$

$A=\cfrac{21}{2}-6$

$A=\cfrac{9}{2}$ sq.unit

The area of the region bounded by the graph of

$y = \sin x$ and

$y = \cos x$ between

$x = 0$ and

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

0%
$\sqrt {2} + 1$
0%
$\sqrt {2} - 1$
0%
$2\sqrt {2} - 2$
0%
$2\sqrt {2} + 2$

Explanation

Area bounded by the curves

$y=\sin x$ and

$y=\cos x$ is given by

$A=\displaystyle \int _{0} ^{\tfrac{\pi}{4}} (\cos x-\sin x) dx$

$=[\sin x+\cos x]_{0} ^{\tfrac{\pi}{4}}$

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

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

$=\sqrt{2}-1$ sq. units.

The area bounded by

$y = xe^{|x|}$ and lines

$|x| = 1, y = 0$ is

0%
4 sq units
0%
6 sq units
0%
1 sq units
0%
2 sq units

Explanation

$|x| = 1$

$\therefore x = \pm 1$

$\therefore y = xe^{|x|} = \begin{Bmatrix} x.e^{-x}, & -1 <x < 0\\ x.e^x, & 0 < x < 1\end{Bmatrix}$

$\therefore$ Required area

$= \left| \int_{-1}^0 xe^{-x} dx + \int_0^1 xe^x dx \right |$

$= \left| \left[ -x.e^{-x} - e^{-x} \right ]_{-1}^0 + \left\{x.e^x - e^x \right \}_0^1 \right |$

$= \left| \left\{ (0 - 1) - (1.e - e) \right \} \right | + \left| \left\{ (e - e) - (0 - 1) \right \} \right |$

$= 1 + 1 = 2$ sq units

The area in the first quadrant between

$x^2 + y^2 = \pi^2$ and

$y = sin x$ is

0%
$\dfrac{\pi^3 - 8}{4}$
0%
$\dfrac{\pi^3}{4}$
0%
$\dfrac{\pi^3 - 16}{4}$
0%
$\dfrac{\pi^3 - 8}{2}$

Explanation

$x^2 + y^2 = \pi^2$ is a circle of radius

$\pi$ and centre at origin.

Required area

= Area of circle (1st quadrant)

$- \int_0^{\pi} sin x dx$

$= \dfrac{\pi. \pi^2}{4} - [-cos x]_{0}^{\pi} = \dfrac{\pi^3}{4} + (cos \pi - cos 0)$

$= \dfrac{\pi^3}{4} + (-1 - 1) = \dfrac{\pi^3}{4} - 2 = \dfrac{\pi^3 - 8}{4}$

Hence, option A is correct.

For which of the following values of

$m$ , the area of the region bounded by the curve

$y = x - x^{2}$ and the line

$y = mx$ equals

$\dfrac {9}{2}$

0%
$-4$
0%
$-2$
0%
$2$
0%
$4$

Explanation

The equation of curve is

$y = x - x^{2}$

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

$\Rightarrow \left (x - \dfrac {1}{2}\right )^{2} = -\left (y - \dfrac {1}{4}\right )$
which is a parabola whose vertex is

$\left (\dfrac {1}{2}, \dfrac {1}{4}\right )$
Hence, finding the point of intersection of the curve and the line,

$x - x^{2} = mx \Rightarrow x (1 - x - m) = 0$
i.e.,

$x = 0$ or

$x = 1 - m$

$\therefore \dfrac {9}{2} = \int_{0}^{1 - m}(x - x^{2} - mx)dx$

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

$= (1 - m) \dfrac {(1 - m)^{2}}{2} - \dfrac {(1 - m)^{3}}{3} = \dfrac {(1 - m)^{3}}{6}$

$\therefore (1 - m)^{3} = \dfrac {6\times 9}{2} = 27$

$\Rightarrow 1 - m = (27)^{1/3} = 3$
Also,

$(1 - m)^{3} - (3)^{3} = 0$

$\therefore (1 - m)^{3} = 3^{3} \Rightarrow 1 - m = 3$
or

$m = -2$

The area of the region bounded by the curves

$x^{2} + y^{2} = 9$ and

$x + y = 3$ is

0%
$\dfrac {9\pi}{4} + \dfrac {1}{2}$
0%
$\dfrac {9\pi}{4} - \dfrac {1}{2}$
0%
$9\left (\dfrac {\pi}{4} - \dfrac {1}{2}\right )$
0%
$9\left (\dfrac {\pi}{4} + \dfrac {1}{2}\right )$

Explanation

Area of required region

$= \dfrac {1}{4}\times \text {Area of circle} - \text {Area of} \triangle OAB$

$= \dfrac {1}{4}\times \pi \times (3)^{2} - \dfrac {1}{2}\times 3\times 3$

$= 9\left (\dfrac {\pi}{4} - \dfrac {1}{2}\right )$

Hence, option C is correct.

The area bounded by the curves

$y = \cos x$ and

$y = \sin x$ between the ordinates

$x = 0$ and

$x = \dfrac {3\pi}{2}$ is

0%
$(4\sqrt {2} - 2)sq\ units$
0%
$(4\sqrt {2} + 2)sq\ units$
0%
$(4\sqrt {2} - 1)sq\ units$
0%
$(4\sqrt {2} + 1)sq\ units$

Explanation

Required area

$= \int_{0}^{\pi /4} (\cos x - \sin x) dx + \int_{\pi/4}^{5\pi/4}(\sin x - \cos x) dx + \int_{5\pi/4}^{3\pi/2} (\cos x - \sin x)dx$

$= [\sin x + \cos x]_{0}^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{5\pi/4} + [\sin x + \cos x]_{5\pi/4}^{3\pi/2}$

$=\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-1 \right)-\left( -\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} \right) + \left( -1+0+\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right)$

$=\dfrac{8}{\sqrt{2}} - 2$

$= (4\sqrt {2} - 2)sq\ units$

Hence, option A is correct.

The area bounded by the curve x = f(y), the y-axis and the two lines y = a and y = b is equal to :

0%
$\int _ {a} ^{b} {y dx}$
0%
$\int _ {a} ^{b} {y^2 dx}$
0%
$\int _ {a} ^{b} {x dy}$
0%
None of the above

Explanation

The above is an example of area of a function and bounded by

$y=c$ to

$y=d$

So area will be

$\int _{ c }^{ d }{ f(y)dy }$

In the given question

area of a function and bounded by

$y=c$ to

$y=d$

So area will be

$\int _{ a }^{ b }{ f(y).dy }=\int _{ a }^{ b }{ x.dy }$

Consider the line

$x = \sqrt {3}y$ and the circle

$x^{2} + y^{2} = 4$ .

What is the area of the region in the first quadrant enclosed by the y-axis, the line

$x = \sqrt {3}$ and the circle?

0%
$\dfrac {2\pi}{3} + \dfrac {\sqrt {3}}{2}$
0%
$\dfrac {\pi}{2} - \dfrac {\sqrt {3}}{2}$
0%
$\dfrac {\pi}{3} - \dfrac {1}{2}$
0%
None of the above

Explanation

Equation of line

$BC$ is

$x=\sqrt { 3 }$

$B(\sqrt { 3 } ,1) ; C(\sqrt { 3 } ,-1)$

The equation of circle in first quadrant is

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

Area enclosed in first quadrant,

$A=\int _{ 0 }^{ \sqrt { 3 } }{ \sqrt { 4-{ x }^{ 2 } } } dx\quad =_{ 0 }^{ \sqrt { 3 } }{ (\frac { x }{ 2 } \sqrt { 4-{ x }^{ 2 } } +\frac { { 2 }^{ 2 } }{ 2 } { sin }^{ -1 }(\frac { x }{ 2 } )) }\\ =\frac { \sqrt { 3 } }{ 2 } (1)+2(\frac { \pi }{ 3 } )-0\\ =\frac { \sqrt { 3 } }{ 2 } +\frac { 2\pi }{ 3 } \\$

Consider the curves

$y = \sin x$ and

$y = \cos x$ .

What is the area of the region bounded by the above two curves and the lines

$x = 0$ and

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

0%
$\sqrt {2} - 1$
0%
$\sqrt {2} + 1$
0%
$\sqrt {2}$
0%
$2$

Explanation

The area enclosed by

$y=sinx, y=cosx,x=0,x=\frac { \pi }{ 4 }$ is given by

$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ (\cos x-\sin x)dx }$

$=|\sin x+\cos x|_{0}^{\frac{\pi}{4}}$

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

$=\sqrt { 2 } -1$

Consider the curves

$y = \sin x$ and

$y = \cos x$ .

What is the area of the region bounded by the above two curves and the lines

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

$x = \dfrac {\pi}{2}$ ?

0%
$\sqrt {2} - 1$
0%
$\sqrt {2} + 1$
0%
$2\sqrt {2}$
0%
$2$

Explanation

The area enclosed by

$y=\sin x, y=\cos x, x=\dfrac { \pi }{ 2}, x=\dfrac { \pi }{ 4 }$ is given by

$\displaystyle \int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2} }{ (-\cos x+\sin x)dx }$

$=|-\sin x-\cos x|_{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }$

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

$=\sqrt { 2 } -1$

The figure shows a portions of the graph

$y=2x-4x^3$ . The line y = c is such that the areas of the regions marked I and II are equal. If a,b are the x-coordinates of A,B respectively, then a + b equals

0%
$\frac{2}{\sqrt{7}}$
0%
$\frac{3}{\sqrt{7}}$
0%
$\frac{4}{\sqrt{7}}$
0%
$\frac{5}{\sqrt{7}}$

Explanation

$\int_{a}^{b}(2x-4x^3)dx=2(b-a)C$

$(x^2-x^4)_a^b=2(b-a)C$

$(b+a)[1-(a^2+b^2)]=2C$

$(a+b)[1-(a+b)^2+2ab]=2C$ ......(i)
Again

$2x-4x^3=C$

$4x^3-2x+C=0$
Roots, a, b,

$\alpha$

$4x^3+0.x^2-2x+C=0$

$a+b+\alpha =0$ ........(ii)

$ab +(a+b)\alpha =-\frac{1}{2}$ ......(iii)

$7 \alpha ^3=4\alpha \Rightarrow \alpha =\frac{2}{\sqrt{7}}$

The area of the region bounded by the curve

$y=2x-{x}^{2}$ and x-axis is

0%
$\cfrac { 2 }{ 3 }$ sq. units
0%
$\cfrac { 4 }{ 3 }$ sq. units
0%
$\cfrac { 5 }{ 3 }$ sq. units
0%
$\cfrac { 8 }{ 3 }$ sq. units

Explanation

To get intersection with x-axis, put

$y=0$

$2x-x^2=0$

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

$x=0$ and

$2$

Now making the graph we see that the curve makes an intercept between

$(0,0)$ and

$(2,0)$ .

So area

$=\int _{ 0 }^{ 2 }{ (2x-x^2).dx }$

$={ [x^{ 2 }-x^{ 3 }/3] }_{ 0 }^{ 2 }$

$=4-\dfrac { 8 }{ 3 } -0$

$=\dfrac { 4 }{ 3 }$

answer (b)

The line

$2y=3x+12$ cuts the parabola

$4y=3x^2$ . What is the area enclosed by the parabola and the line?

0%
$27$ square unit
0%
$36$ square unit
0%
$48$ square unit
0%
$54$ square unit

Explanation

The graph is shown in the image.

The intersection point of

$2y=3x+12$ and

$4y=3x^2$

$2\Big(\dfrac{3x^2}{4}\Big)=3x+12$

$x^2-2x-8=0$

$x=-2,4$

Hence, intersection point are

$(-2,3)$ and

$(4,12)$

Now,

$A=\int_{-2}^{4}\Bigg[\dfrac{3x+12}{2}\Bigg]-\int_{-2}^{4}\Bigg[\dfrac{3x^2}{4}\Bigg]$

$A=\Bigg[\dfrac{3x^2}{4}+6x \Bigg]\Bigg|_{-2}^4 -\Bigg[\dfrac{x^3}{4}\Bigg]\Bigg|_{-2}^4$

$A=9+36-18$

$A=27$

$sq.unit$

What is the area of the parabola

${ x }^{ 2 }=y$ bounded by the line

$y=1$ ?

0%
$\dfrac { 1 }{ 3 }$ square unit
0%
$\dfrac { 2 }{ 3 }$ square unit
0%
$\dfrac { 4 }{ 3 }$ square unit
0%
$2$ square unit

Explanation

Here, the area is symmetric about y-axis, we can find the area on one side and then multiply it by

$2$ , we will get the area,

So required area

$=2\int _{ 0 }^{ 1 }{ (1-x^{ 2 }).dx }$

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

$=\dfrac { 4 }{ 3 }$

Hence, C is correct.

The area of the region bounded by the curve

$y = x^{2}$ and the line

$y = 16$ is

0%
$\dfrac {128}{3} sq. units$
0%
$\dfrac {64}{3} sq. units$
0%
$\dfrac {32}{3} sq. units$
0%
$\dfrac {256}{3} sq. units$

Explanation

Area

$=\int _{ 0 }^{ 16 }{ x } dy$

$=2\int _{ 0 }^{ 16 }{ \sqrt { y } } dy$

$=2\cdot \cfrac { 2 }{ 3 } { \left| { y }^{ { 3 }/{ 2 } } \right| }_{ 0 }^{ 16 }$

$=2\cdot \cfrac { 2 }{ 3 } \left[ { 16 }^{ { 3 }/{ 2 } }-0 \right]$

$=2\cdot \cfrac { 2 }{ 3 } \cdot { \left[ 4 \right] }^{ 3 }$

$=\cfrac { 256 }{ 3 }$ sq. units

Consider an ellipse

$\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$ What is the area included between the ellipse and the greatest rectangle inscribed in the ellipse?

0%
$ab(\pi -1)$
0%
$2ab(\pi -1)$
0%
$ab(\pi -2)$
0%
None of the above

Explanation

As the rectangle lies inside a standard ellipse, the vertices of the rectangle can be assumed to be

$(h,k),(h,-k),(-h,k)$ and

$(-h,-k)$ without any loss of generality.

$\therefore \dfrac{h^2}{a^2}+\dfrac{k^2}{b^2}=1\implies k=b\sqrt{1-\dfrac{h^2}{a^2}}$

Clearly, the sides of the rectangle based on the coordinates chosen can be determined as

$2h$ and

$2k$ .

$\therefore$ area of rectangle

$=A(h)=2h\times 2k=4h b\sqrt{1-\dfrac{h^2}{a^2}}\implies 4b\sqrt{h^2-\dfrac{h^4}{a^2}}$

Now, let

$P(h)=h^2-\dfrac{h^4}{a^2}$ . Thus, the maximization of

$A(h)$ boils down just to the maximization of

$P(h)$ .

Differentiating

$P(h)$ wrt

$h$ , we have

$2h-\dfrac{4h^3}{a^2}=0\implies h=\dfrac{2h^3}{a^2}\implies 2h^2=a^2\implies h=\dfrac{a}{\sqrt{2}}\implies k=\dfrac{b}{\sqrt{2}}$

Hence,

$A_{max}=2ab$

$\therefore$ area included between this ellipse and rectangle

$=\pi ab-2ab=ab(\pi-2)$ .

this is the required answer.

The area of the figure bounded by the parabolas

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

$x = 1 - 3y^{2}$ is

0%
$\dfrac {4}{3}$ square units
0%
$\dfrac {2}{3}$ square units
0%
$\dfrac {3}{7}$ square units
0%
$\dfrac {6}{7}$ square units

Explanation

Given curves are

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

$x=1-3y^{2}$

$-2y^{2}=1-3y^{2}$

$\implies y^{2}=1 \implies y=\pm 1$

$\therefore$ Curves intersect at

$(-2, \pm 1)$
Area

$=\int_{-1}^{1} (1-3y^{2}-(-2y^{2}))\ dy$

$=\int_{-1}^{1} (1-y^{2})\ dy$

$=\int_{-1}^{0} (1-y^{2})\ dy +\int_{0}^{1} (1-y^{2})\ dy$

Substituting

$y=-y$ in first integral we get,

Area

$=\int_{0}^{1} (1-y^{2})\ dy +\int_{0}^{1} (1-y^{2})\ dy$

$= 2\int_{0}^{1} (1 - y^{2}) dy$

$=2\left|y-\dfrac{y^{3}}{3}\right|_{0}^{1}$

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

Hence, area bounded by the given curves is

$\dfrac{4}{3}$ square units.

What is the area of the portion of the curve

$y=\sin {x}$ , lying between

$x=0$ ,

$y=0$ and

$x=2\pi$ ?

0%
$1$ square unit
0%
$2$ square units
0%
$4$ square units
0%
$8$ square units

Explanation

The graph of $y=\sin x$ is shown in image.

$A=\int_{0} ^{2\pi}\bigg(\sin x\bigg)$

$A=\int_{0} ^{\pi}\bigg(\sin x\bigg)+\int_{\pi} ^{2\pi}\bigg(\sin x\bigg)$

Since, $\Bigg|\int_{0} ^{\pi}\bigg(\sin x\bigg)\Bigg| =\Bigg|\int_{0} ^{2\pi}\bigg(\sin x\bigg)\Bigg|$

Hence, $A=2\Bigg(\int_{0} ^{\pi}\bigg(\sin x\bigg) \Bigg)$

$A=-2\bigg[\cos x\bigg]\Bigg|^0_{\pi}$

$A=2\times 2$

$A=4$ sq. unit

Tangents are drawn to the ellipse

$\dfrac {x^{2}}{9} + \dfrac {y^{2}}{5} = 1$ at the ends of both latus rectum. The area of the quadrilateral so formed is

0%
$27$ sq.units
0%
$\dfrac {13}{2}$ sq.units
0%
$\dfrac {15}{4}$ sq.units
0%
$45$ sq.units

Explanation

At the end point

$\left (\sqrt{a^{2}-b^{2}},\dfrac {b^{2}}{a}\right)$ equation of tangent is

$\dfrac{xx_{1}}{a^{2}}+\dfrac{yy_{1}}{b^{2}}=1$

So, the intercepts are

$\left (\dfrac {a^{2}}{x_{1}},0\right)$ and

$\left (0,\dfrac {b^{2}}{y_{1}}\right)$

Now, the area of quadrilateral

$=4\times$ area of triangle formed by the intercepts and origin.

$\Rightarrow 4\times \dfrac{1}{2}\times \dfrac{a^{2}b^{2}}{\sqrt{a^{2}-b^{2}}\times \dfrac{b^{2}}{a}}=2\times \dfrac {27}{2}=27$

So, option A is correct.

The area of the region bounded by the lines

$y = 2x + 1, y = 3x + 1$ and

$x = 4$ is

0%
$16\ sq.unit$
0%
$\dfrac {121}{3}\ sq.unit$
0%
$\dfrac {121}{6}\ sq.unit$
0%
$8\ sq.unit$

Explanation

$A$ (Shaded region)

$= A(\triangle ABD) - A(\triangle ABC) = \dfrac {1}{2} [4\times 12 - 4 \times 8] = \dfrac {1}{2} (48 - 32) = 8\ sq.units$ .
OR

$A (\triangle ACD) = \dfrac {1}{2}\begin{vmatrix} 0& 1 & 1\\ 4 & 9 & 1\\ 4 & 13 & 1\end{vmatrix}$

$= \dfrac {1}{2}\times 16 = 8$

The area bounded by the curves

$y=\cos x$ and

$y=\sin x$ between the ordinates

$x=0$ and

$x=\displaystyle\frac{3\pi}{2}$ is?

0%
$4\sqrt{2}-1$
0%
$4\sqrt{2}+1$
0%
$4\sqrt{2}-2$
0%
$4\sqrt{2}+2$

Explanation

We have,

$y=\cos x$ and

$y=\sin x$ between

$x=0$ and

$x=\displaystyle\frac{3\pi}{2}$

Required area A is given by

$A=\displaystyle\int^{3\pi /2}_0|\cos x-\sin x|dx$

$\Rightarrow A=\displaystyle\int^{\pi /4}_0|\cos x-\sin x|dx+\int^{5\pi/4}_{\pi /4}|\cos x -\sin x|dx+\displaystyle \int^{3\pi /2}_{5\pi/4}|\cos x - \sin x|dx$

$\Rightarrow \displaystyle A=\int^{\pi /4}_0(\cos x-\sin x)dx+\displaystyle \int^{5\pi/4}_{\pi /4}(\sin x -\cos x)dx+ \displaystyle \int^{3\pi /2}_{5\pi /4}(\cos x -\sin x)dx$

$\Rightarrow A=\displaystyle [\sin x+\cos x]^{\pi /4}_0+[-\cos x -\sin x]^{5\pi /4}_{\pi/4}+[\sin x+ \cos x]^{3\pi /2}_{\pi /4}$

$\Rightarrow A=(\sqrt{2}-1)+(\sqrt{2}+\sqrt{2})+(-1+\sqrt{2})$

$\Rightarrow A=4\sqrt{2}-2$ .

The line

$x=\dfrac{\pi}{4}$ divide the area of the region bounded by

$y=\sin x, y = \cos x$ and X-axis

$\left(0 \le x \le \frac{\pi}{2}\right)$ into two regions of areas

$A_1$ and

$A_2$ . Then,

$A_1:A_2$ equals

0%
$4:1$
0%
$3:1$
0%
$2:1$
0%
$1:1$

Explanation

Area

$A_1=\displaystyle \int_0^{\frac{\pi}{4}}\sin x\,dx$

$=-[\cos x]_0^{\frac{\pi}{4}}$

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

and Area

$A_2=\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos x\,dx$

$=\displaystyle [\sin x]^{\frac{\pi}{2}}_{\frac{\pi}{4}}=\left[1-\dfrac{1}{\sqrt{2}}\right]=\dfrac{\sqrt{2}-1}{\sqrt{2}}$

$\therefore A_1:A_2 = \dfrac{\sqrt{2}-1}{\sqrt{2}}: \dfrac{\sqrt{2}-1}{\sqrt{2}}=1:1$

Area bounded by the curves

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

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

0%
$\cfrac { 8 }{ 3 }$ sq. units
0%
$\cfrac { 3 }{ 8 }$ sq. units
0%
$\cfrac { 3 }{ 2 }$ sq. units
0%
None of these

Explanation

$\textbf{Step-1: Finding the bounded Area}$

$\text{On plotting these curves we can easily see area bounded by them}$

$\textbf{Step-2: And Area bounded by them is given as}$

$\Rightarrow 2\int\limits_{0}^{1}{[(2-{{x}^{2}})-{{x}^{2}}]}dx$

$\Rightarrow 2\int\limits_{0}^{1}{[2-2{{x}^{2}}]}dx$

$\Rightarrow 2\times \left| 2x-\dfrac{2{{x}^{3}}}{3} \right|_{0}^{1}$

$\Rightarrow 2\left[ \left( 2-\dfrac{2}{3} \right)-0 \right]$

$\Rightarrow \dfrac{8}{3}$

$\textbf{Therefore, area bounded by these curves is (A) }$ $\mathbf {\dfrac{8}{3}}$

The area bounded by the parabola

${ y }^{ 2 }=4a(x+a)$ and

${ y }^{ 2 }=-4a(x-a)$ is

0%
$\cfrac { 16 }{ 3 } { a }^{ 2 }$
0%
$\cfrac { 8 }{ 3 } { a }^{ 2 }$
0%
$\cfrac { 4 }{ 3 } { a }^{ 2 }$
0%
None of these

Explanation

Required area

$=4\int _{ 0 }^{ 4 }{ \sqrt { 4a(a-x) } } dx$

$=4\times2\sqrt { a } { \left[ \cfrac { -2{ (a-x) }^{ 3/2 } }{ 3 } \right] }_{ 0 }^{ 4 }=\cfrac { 16 }{ 3 } { a }^{ 2 }$

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 4 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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