CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 10 - MCQExams.com

Find the area enclosed between the curves $$y^2- 2ye^{sin^{- 1}x} + x^2- 1 +[x] +e^{2sin^{- 1}x} = 0$$ 
and line x = 0 and $$x=\frac{1}{2}$$ is (where [.] denotes greatest integer function)
  • $$\dfrac{\sqrt{3}}{4}+\dfrac{\pi }{6}$$
  • $$\dfrac{\sqrt{3}}{2}+\dfrac{\pi }{6}$$
  • $$\dfrac{\sqrt{3}}{4}-\dfrac{\pi }{6}$$
  • $$\dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6}$$
If $$A_1$$ is the area bounded by $$y= \cos x, y = \sin x$$ &  $$x=0$$  and $$A_2$$ the area bounded by $$y = \cos x , y = \sin x , y = 0$$ in $$(0,\frac{\pi}{2})$$ then $$\displaystyle  \dfrac{A_1}{A_2}$$ equals to :
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{\sqrt2}$$
  • $$1$$
  • None of these
The area bounded by the curves $$y = sin^{-1} |sin  x|$$ and $$y = (sin^{-1} | sin  x|)^{2},$$ where $$0\leq x\leq 2\pi ,$$ is:
  • $$\dfrac{1}{3}+\dfrac{\pi ^{2}}{4}$$ sq. units
  • $$\dfrac{1}{6}+\dfrac{\pi ^{3}}{8}$$ sq. units
  • $$2$$ sq. units
  • $$\displaystyle \frac{4}{3}+\pi ^{2}\frac{\pi -3}{6}  sq.units$$
If $$\left| z- (4 + 4i) \right| \geq  4$$, then area of the region bounded by the locii of $$z,\; iz,\; - z$$ and $$-iz$$ is:
  • $$4(4-\pi )$$
  • $$16(4-\pi )$$
  • $$16(\pi -1)$$
  • $$4(\pi -1)$$
If the area bounded by the curve $$y = f(x)$$, the coordinate axes and the line $$x = x_1$$ is given by $$x_1e^{x_1}$$. Then $$f(x)$$ equals
  • $$e^x$$
  • $$xe^x$$
  • $$xe^x-e^x$$
  • $$xe^x+e_x$$
If the area bounded by the curve $$|y|=sin^{-1}|x|$$ and  $$x=1$$ is $$a(\pi+b)$$, then the value $$a-b$$ is:
  • 1
  • 2
  • 3
  • 4
The parabola $$y^{2} = 4x$$ and $$x^{2}  = 4y$$ divide the square region bounded by the lines $$x = 4, y = 4$$ and the coordinate axes. If $$S_{1}, S_{2}, S_{3}$$ are the areas of these parts numbered from top to bottom respectively, then
  • $$S_{1} : S_{2}\equiv 1 : 1$$
  • $$S_{2} : S_{3}\equiv 1 : 2$$
  • $$S_{1} : S_{3}\equiv 1 : 1$$
  • $$S_{1} : (S_{1} + S_{2})\equiv 1 : 2$$
Area of the region bounded by the curve $$y = x^{2}$$ and $$y = sec^{-1} [sin^{2}x]$$ (where [ . ] denotes the greatest integer function) is
  • $$\frac{\pi}{3}\sqrt{\pi}$$
  • $$\frac{2\pi\sqrt{\pi}}{3}$$
  • $$\frac{4\pi\sqrt{\pi}}{3}$$
  • $$\frac{6\pi\sqrt{\pi}}{3}$$
  • $$\frac{3\pi\sqrt{\pi}}{2}$$
The area bounded by $$y=\sec^ {-1}{x}, y= \text{cosec}^{-1}{x}$$ and the line $$x-1=0$$ is:
  • $$\ln(3+2\sqrt{2})-\displaystyle \frac{\pi}{2}$$
  • $$\displaystyle \frac{\pi}{2}+\ln(3+2\sqrt{2})$$
  • $$\pi - \ln3$$
  • $$\pi + \ln3$$
The area enclosed by $$x^2 + y^2 = 4, y = x^2 + x + 1,$$  $$y=\left [ \sin^{2}\displaystyle \frac{x}{4}+\cos\displaystyle \frac{x}{4} \right ]$$ and $$x$$-axis (where $$[.]$$ denotes the greatest integer function) is:
  • $$\displaystyle \frac{2\pi}{3}+\sqrt{3}-\displaystyle \frac{1}{6}$$
  • $$\displaystyle \frac{2\pi}{3}+2\sqrt{3}-\displaystyle \frac{1}{6}$$
  • $$2\sqrt3-\frac{1}{3}$$
  • $$\displaystyle \frac{\pi}{3}+\sqrt3$$
The area bounded by the function $$f(x)=x^{2}:R^{+}\rightarrow R^{+}$$ and its inverse function is:
  • $$\dfrac{1}{2}$$sq.units
  • $$\dfrac{1}{3}$$sq.units
  • $$\dfrac{2}{3}$$sq.units
  • $$\dfrac{1}{6}$$sq.units
Find the area of the region bounded by the curves $$y= log_{e}x $$, $$ y=\sin ^{4} \pi x $$, $$x=0 $$
  • $$\displaystyle\ \frac{11}{8}sq.units$$
  • $$\displaystyle\ \frac{9}{8}sq.units$$
  • $$\displaystyle\ \frac{13}{8}sq.units$$
  • $$\displaystyle\ \frac{15}{8}sq.units$$
State the following statement is True or False
The area bounded by the circle $$x^2 + y^2 = 1, x^2 + y^2 = 4$$ and the pair of lines $$\sqrt{3} (x^2 + y^2) = 4xy$$, is equal to $$\dfrac{\pi}{2}$$. The statement is true or false.
  • True
  • False
Find the area bounded by the curves $$\displaystyle\ y=\sqrt{1-x^{2}}$$ and $$\displaystyle\ y=x^{3}-x $$. Also find the ratio in which the y-axis divide this area
  • $$\displaystyle\ \frac{\pi }{2}$$ , $$\displaystyle\ \frac{\pi -1}{\pi +1} $$
  • $$\displaystyle\ \frac{\pi }{4}$$ , $$\displaystyle\ \frac{\pi -1}{\pi +1} $$
  • $$\displaystyle\ \frac{\pi }{2}$$ , $$\displaystyle\ \frac{\pi +1}{\pi -1} $$
  • None of these
Find the area of the region enclosed between the two circles $$\displaystyle\ x^{2}+y^{2}=1$$ & $$(x-1)^{2}+y^{2}=1$$
  • $$\displaystyle\ \frac{\pi }{6}-\frac{\sqrt{3}}{2}$$ sq.units
  • $$\displaystyle\ \frac{\pi }{3}-\frac{\sqrt{3}}{2}$$ sq.units
  • $$\displaystyle\ \frac{\pi }{6}-\frac{\sqrt{3}}{4}$$ sq.units
  • $$\displaystyle\ \frac{\pi }{3}-\frac{\sqrt{3}}{4}$$ sq.units
A polynomial function f(x) satisfies the condition $$f(x+1) =f(x)+2x+1$$. Find $$f(x)$$ if $$f(0)=1$$. Find also the equations of the pair of tangents from the origin on the curve $$y=f(x)$$ and compute the area enclosed by the curve and the pair of tangents.
  • $$f(x)=x^2+1;$$, $$y=\pm 2x;$$ , $$\displaystyle A=\frac{2}{3}$$ sq.units
  • $$f(x)=x^2-1;$$, $$y=\pm 2x;$$ , $$\displaystyle A=\frac{2}{3}$$ sq.units
  • $$f(x)=x^2+1;$$, $$y=\pm 2x;$$ , $$\displaystyle A=\frac{3}{2}$$ sq.units
  • $$f(x)=x^2-1;$$, $$y=\pm 2x;$$ , $$\displaystyle A=\frac{3}{2}$$ sq.units
Find the area enclosed the curves : $$y=ex\log { x } $$ and $$\displaystyle y=\frac { \log { x }  }{ ex } $$ where $$\log { e } =1$$
  • $$\displaystyle\frac { { e }^{ 2 }-5 }{ 4e } $$
  • $$\displaystyle\frac { { e }^{ 2 }+5 }{ 4e } $$
  • $$\displaystyle\frac { { e }^{ 2 }-3 }{ 2e } $$
  • $$\displaystyle\frac { { e }^{ 2 }+3 }{ 2e } $$
The area included between the curve $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$ and $$\displaystyle \sqrt { \left| x \right|  } +\sqrt { \left| y \right|  } =\sqrt { a } \left( a>0 \right) $$ is:
  • $$ \displaystyle \left( \pi +\frac { 2 }{ 3 }  \right) { a }^{ 2 }$$
  • $$ \displaystyle \left( \pi -\frac { 2 }{ 3 }  \right) { a }^{ 2 }$$
  • $$ \displaystyle \frac { 2 }{ 3 } { a }^{ 2 }$$
  • $$\displaystyle \frac { 2\pi }{ 3 } { a }^{ 2 }$$
Sketch the region bounded by the curves $$ \displaystyle y=x^{2}$$ & $$ \displaystyle\ y= 2/(1+x^{2})$$. Find the area:
  • $$\displaystyle\ \pi -\frac{2}{3}$$
  • $$\displaystyle\ \pi -\frac{1}{3}$$
  • $$\displaystyle\ \pi -\frac{5}{3}$$
  • $$\displaystyle\ \pi -\frac{7}{3}$$
The ratio in which the area bounded by the curves $$y^{2}=4x$$ and $$x^{2}=4y$$ is divided by the line $$x=1$$ is
  • $$64 : 49$$
  • $$15 : 34$$
  • $$15 : 49$$
  • none of these
If f(x) be an increasing function defined on [a, b] then
max {f(t) such that $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x)  & min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a),
min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x).
On the basis of above information answer the following questions.
$$\int_{0}^{\pi }max\left \{ \sin x, \cos x \right \}dx$$ is equal to
  • $$\sqrt{2}-1$$
  • $$\displaystyle 1-\frac{1}{\sqrt{2}}$$
  • $$\displaystyle 1+\frac{1}{\sqrt{2}}$$
  • None of these
The ratio of the area's bounded by the curves $$ \displaystyle y^{2}=12x $$ and $$ \displaystyle x^{2}=12y $$ is divided by the line $$ \displaystyle x=3 $$ is
  • 15 : 49
  • 9 : 15
  • 7 : 15
  • 7 : 5
The function $$ \displaystyle f\left ( x \right )=\max \left \{ x^{2},\left ( 1-x \right )^{2},2x\left ( 1-x \right ) \forall 0\leq x\leq 1\right \} $$ then area of the region bounded by the curve $$ \displaystyle y=f\left ( x \right ) $$, x-axis and $$ \displaystyle x=0,x=1 $$ is equals,
  • $$ \displaystyle \frac{27}{17} $$
  • $$ \displaystyle \frac{9}{17} $$
  • $$ \displaystyle \frac{18}{17} $$
  • None of these
Compute the area of the curvilinear triangle bounded by the y-axis & the curve, $$\displaystyle\ y=\tan x$$ & $$ \displaystyle\ y=(2/3) \cos x$$
  • $$\displaystyle\ \frac{1}{3}+ ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units$$
  • $$\displaystyle\ \frac{1}{3}- ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units$$
  • $$\displaystyle\ \frac{2}{3}+ ln\left [ \frac{\sqrt{3}}{2} \right ]sq.units$$
  • $$\displaystyle\ \frac{1}{3}+ ln\left [ \frac{\sqrt{1}}{2} \right ]sq.units$$
The area bounded by $$ \displaystyle x=a\cos ^{3}\theta,y=a\sin ^{3}\theta $$ is:
  • $$ \displaystyle \frac{3\pi a^{2}}{16} $$
  • $$ \displaystyle \frac{3\pi a^{2}}{8} $$
  • $$ \displaystyle \frac{3\pi a^{2}}{32} $$
  • $$ \displaystyle 3\pi a^{2} $$
The area lying in the first quadrant inside the circle $${ x }^{ 2 }+{ y }^{ 2 }=12$$ and bounded by the parabolas $${ y }^{ 2 }=4x,{ x }^{ 2 }=4y$$ is:
  • $$\displaystyle 2\left( \frac { \sqrt { 2 }  }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 }  }  \right) $$
  • $$\displaystyle 4\left( \frac { \sqrt { 2 }  }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 }  }  \right) $$
  • $$\displaystyle \left( \frac { \sqrt { 2 }  }{ 3 } +\frac { 3 }{ 2 } \sin ^{ -1 }{ \frac { 1 }{ 3 }  }  \right) $$
  • none of these
If f(x) be an increasing function defined on [a, b] then
max {f(t) such that $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x)  & min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a),
min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x).
On the basis of above information answer the following questions.
Let $$\displaystyle f\left ( x \right )=min \left \{ 1, 1-\cos x, 2\sin x \right \}$$ then $$\displaystyle \int_{0}^{\pi}f\left ( x \right )dx$$ is
  • $$\displaystyle \frac{\pi }{3}+1-\sqrt{3}$$
  • $$\displaystyle \frac{2\pi }{3}-1+\sqrt{3}$$
  • $$\displaystyle \frac{5\pi }{6}+1-\sqrt{3}$$
  • $$\displaystyle \frac{\pi }{6}+1-\sqrt{3}$$
The area of the plane region bounded by the curves $$x+2y^{2}=0$$ and $$x+3y^{2}=1$$ is
  • $$\displaystyle \frac{1}{3}$$
  • $$\displaystyle \frac{2}{3}$$
  • $$\displaystyle \frac{4}{3}$$
  • $$\displaystyle \frac{5}{3}$$
Find the area bounded by the curves $$\displaystyle x = y^{2}$$ and $$\displaystyle x = 3-2y^{2}$$
  • $$2$$ sq. units
  • $$4$$ sq. units
  • $$6$$ sq. units
  • $$8$$ sq. units
The area bounded by the curves $$y=\log x$$, $$y=\log \left | x \right |$$, $$y=\left | \log x \right |$$ and $$y=\left | \log \left | x \right | \right |$$
  • 4 sq. units
  • 6 sq. units
  • 10 sq. units
  • None of these
Let y=f(x) be the given curve and $$x=a$$, $$x=b$$ be two ordinates then area bounded by the curve $$y=f(x)$$, the axis of x between the ordinates $$x=a$$ & $$x=b$$, is given by definite integral
$$\int_{a}^{b}ydx$$ or $$\int_{a}^{b}f\left ( x \right )dx$$ and the area bounded by the curve $$x=f(y)$$, the axis of y & two abscissae $$y=c$$ & $$y=d$$ is given by $$\int_{c}^{d}xdy$$ or $$\int_{c}^{d}f\left ( x \right )dy$$. Again if we consider two curves $$y=f(x)$$, $$y=g(x)$$ where $$f\left ( x \right )\geq g\left ( x \right )$$ in the interval [a, b] where $$x=a$$ & $$x=b$$ are the points of intersection of these two curves Shown by the graph given
Then area bounded by these two curves is given by
$$\int_{a}^{b}\left [ f\left ( x \right )-g\left ( x \right ) \right ]dx$$
On the basis of above information answer the following questions.

The area bounded by parabolas $$y=x^{2}+2x+1$$ & $$y=x^{2}-2x+1$$ and the line $$\displaystyle y=\frac{1}{4}$$ is equal to

161838_6c80fc7958864f1f961bdcd5221bb036.png
  • $$\displaystyle \frac{2}{3}$$ square unit
  • $$\displaystyle \frac{1}{3}$$ square unit
  • $$\displaystyle \frac{3}{2}$$ square unit
  • $$\displaystyle \frac{1}{2}$$ square unit
If f(x) be an increasing function defined on [a, b] then
max {f(t) such that $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x)  & min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a) and if f(x) be decreasing function defined on [a, b] then
max {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(a),
min {f(t), $$a\leq t\leq x$$, $$a\leq x\leq b$$}=f(x).
On the basis of above information answer the following questions.
$$\int_{0}^{\pi /2}min\left \{ \sin x, \cos x \right \}dx$$ equals
  • $$2\left ( \sqrt{3}-1 \right )$$
  • $$\sqrt{2}\left ( \sqrt{2}-1 \right )$$
  • $$\left ( \sqrt{3}-1 \right )$$
  • $$\sqrt{2}\left ( \sqrt{2}+1 \right )$$
The parabolas $$y^{2}=4x$$ and $$x^{2}=4y$$ divide the square region bounded by the lines $$x=4, y=4$$ and the coordinate axes. If $$S_{1}$$, $$S_{2}$$, $$S_{3}$$ are respectively the areas of these parts numbered from top to bottom; $$S_{1}: S_{2}: S_{3}$$ is
  • $$1: 2: 3$$
  • $$1: 2: 1$$
  • $$1: 1: 1$$
  • $$2: 1: 2$$
The  area bounded by $${ y }^{ 2 }+8x=16$$ and $${ y }^{ 2 }-24x=48$$ is $$\displaystyle \frac { a\sqrt { 6 }  }{ c } $$, then $$a+c=$$
  • $$30$$
  • $$32$$
  • $$35$$
  • None
The area enclosed between the curves $$y=x^3$$ and $$y=\sqrt{x}$$ is (in square units)
  • $$\displaystyle\frac{5}{3}$$
  • $$\displaystyle\frac{5}{4}$$
  • $$\displaystyle\frac{5}{12}$$
  • $$\displaystyle\frac{12}{5}$$
Find the area bounded by $$\displaystyle y = \cos ^{-1}x,y=\sin ^{-1}x$$ and $$y-$$axis
  • $$\displaystyle \left ( 2-\sqrt{2} \right )$$ sq. units
  • $$\displaystyle \left ( \sqrt{2}-{2} \right )$$ sq. units
  • $$2 \sqrt{2}$$ sq. units
  • $$\sqrt {2}$$ sq. units
Consider two curves $$\displaystyle C_{1}:y=\frac{1}{x}$$ and $$\displaystyle C_{2}$$ : $$y = \displaystyle lnx$$ on the xy plane Let $$\displaystyle D_{1}$$ denotes the region surrounded by $$\displaystyle C_{1}$$, $$\displaystyle C_{2}$$ and the line $$x = 1$$ and $$\displaystyle D_{2}$$ denotes the region surrounded by $$\displaystyle C_{1}$$, $$\displaystyle C_{2}$$ and the line $$x = a$$ If $$\displaystyle D_{1}$$=$$\displaystyle D_{2}$$ then the value of 'a':
  • $$\displaystyle \frac{e}{2}$$
  • $$e$$
  • $$e-1$$
  • $$2(e-1)$$
Suppose $$y = f(x)$$ and $$y = g(x)$$ are two functions whose graphs intersect at three points $$(0, 4), (2, 2)$$ and $$(4, 0)$$ with $$f(x) > g(x)$$ for $$0 < x < 2$$ and $$f(x) < g(x)$$ for $$2 < x < 4 $$. if $$\displaystyle \int_{0}^{4}\left ( f(x)-g(x) \right )dx=10$$ and $$\displaystyle \int_{2}^{4}\left ( g(x)-f(x) \right )dx=5$$, the area between two curves for $$0 < x < 2$$, is:
  • 5
  • 10
  • 15
  • 20
The area bounded by the curves $$\displaystyle y=-\sqrt{-x}$$ and $$\displaystyle x=-\sqrt{-y}$$ were $$\displaystyle x,y\leq 0$$ 
  • Can not be determined
  • is 1/3
  • is 2/3
  • is same as that of the figure by the curves $$\displaystyle y=\sqrt{-x};x\leq 0$$ and $$\displaystyle x=\sqrt{-y};y\leq 0$$
Area of the region enclosed between the curves $$\displaystyle x=y^{2}-1$$ and $$\displaystyle x = \left | y \right |\sqrt{1-y^{2}}$$ is
  • $$1$$
  • $$4/3$$
  • $$2/3$$
  • $$2$$
Find the area of the region bounded by the curves $$\displaystyle x=\frac{1}{2},x=2,y=logx$$ and $$y=2^{x}$$
  • $$\displaystyle \frac{4-\sqrt{2}}{\log2}-\frac{5}{2} \log2+\frac{3}{2}sq.units$$
  • $$\displaystyle \frac{4+\sqrt{2}}{\log2}-\frac{3}{2} \log2+\frac{5}{2}sq.units$$
  • $$\displaystyle \frac{4-\sqrt{2}}{\log2}-\frac{3}{2} \log2+\frac{5}{2}sq.units$$
  • $$\displaystyle \frac{4+\sqrt{2}}{\log2}-\frac{5}{2} \log2+\frac{3}{2}sq.units$$
 Area bounded by $$\displaystyle y=2x-{ x }^{ 2 }$$ & $$\displaystyle (x-1{ ) }^{ 2 }+{ y }^{ 2 }=1$$ in first quadrant, is: 
  • $$\displaystyle \frac { \pi }{ 2 } -\frac { 4 }{ 3 } $$
  • $$\displaystyle \frac { \pi }{ 2 } -\frac { 2 }{ 3 } $$
  • $$\displaystyle \frac { \pi }{ 2 } +\frac { 4 }{ 3 } $$
  • $$\displaystyle \frac { \pi }{ 2 } +\frac { 2 }{ 3 } $$
In what ratio does the x-axis divide the area of the region bounded by the parabolas $$\displaystyle y=4x-x^{2}$$ and $$\displaystyle y=x^{2}-x$$?
  • 4 : 121
  • 4 : 144
  • 4 : 169
  • 4 : 100
For what value of 'a' is the area of the figure bounded by $$\displaystyle y=\frac{1}{x}, y=\frac{1}{2x-1}$$ $$x = 2$$ & $$x = a$$ equal to $$\displaystyle ln\frac{4}{\sqrt{5}}$$?
  • $$\displaystyle a=4\:$$
  • $$\displaystyle a=8\:$$
  • $$\displaystyle a=4\: or \frac{2}{5}\left ( 6-\sqrt{21} \right )$$
  • none of these
If the area enclosed by the parabolas $$\displaystyle y=a-x^{2}$$ and $$\displaystyle y=x^{2}$$ is $$\displaystyle 18\sqrt {2}$$ sq. units Find the value of 'a'
  • $$a = -9$$
  • $$a= 6$$
  • $$a =9$$
  • $$a=-6$$
Area enclosed between the curves $${ y }^{ 2 }=x$$ and $${ x }^{ 2 }=y$$ is equal to
  • $$\displaystyle 2\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 2 } \right) dx } $$
  • $$\displaystyle \frac { 1 }{ 3 } $$
  • area of region $$\left\{ \left( x,y \right) :{ x }^{ 2 }\le y\le \left| x \right|  \right\} $$
  • $$\displaystyle \frac { 2 }{ 3 } $$
Find the area enclosed between the curves $$\displaystyle y=\log_{e}\left ( x+e \right ), x=\log_{e}\left ( 1/y \right )$$ & the x-axis
  • 1 sq. units
  • 2 sq. units
  • 3 sq. units
  • 4 sq. units
Let $$f(x)$$ be a continuous function given by $$\displaystyle f\left ( x \right )=2x$$ for $$\displaystyle \left | x \right |\leq 1$$ for $$\displaystyle f\left ( x \right )=x^{2}+ax+b$$ for $$\displaystyle \left | x \right |> 1$$. Find the area of the region in the third quadrant bounded by the curves $$\displaystyle x=-2y^{2}$$ and $$y = f(x)$$ lying on the left of the line $$8x + 1 = 0$$
  • $$\dfrac{235}{192} ; a = 2 ; b = - 1$$
  • $$\dfrac{235}{192} ; a = -1 ; b = 2$$
  • $$\dfrac{257}{192} ; a = -1 ; b = 2$$
  • $$\dfrac{257}{192} ; a = 2 ; b = - 1$$
Let $$\displaystyle C_{1}$$ & $$\displaystyle C_{2}$$ be two curves passing through the origin as shown in the figure A  curve C is said to "bisect the area" the region between $$\displaystyle C_{1}$$ & $$\displaystyle C_{1}$$ if for each point P of C the two shaded regions A & B shown in the figure have equal areas Determine the upper curve $$\displaystyle C_{2}$$ given that the bisecting curve C has the equation $$\displaystyle y=x^{2}$$ & that the lower curve $$\displaystyle C_{1}$$ has the equation $$\displaystyle y=x^{2}/2$$ 
261710_21f627180e0e4e76a665804f347bc987.png
  • $$\displaystyle \left ( 16/9 \right )x^{2}$$
  • $$\displaystyle \left ( 25/9 \right )x^{2}$$
  • $$\displaystyle \left ( 25/16 \right )x^{2}$$
  • $$\displaystyle \left ( 9/25 \right )x^{2}$$
Find the area bounded by $$y = x + sinx$$ and its inverse between $$x = 0$$ and $$x = \displaystyle 2\pi$$
  • $$2$$
  • $$4$$
  • $$6$$
  • $$8$$
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