Application of Integrals Test

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Application of Integrals Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

The area enclosed by the curves $$xy^{2}=a^{2}(a-x)$$ and $$(a-x)y^{2}=a^{2}x$$ is

A
$$(\pi-2)a^{2}\ sq.units$$

B
$$(4-\pi)a^{2}\ sq.units$$

C
$$(\pi a^{2}/3\ sq.units$$

D
$$\dfrac{\pi+a^{2}}{4}\ sq.units$$

Solution
Question 2
1 / -0

The area bounded by the curved $${ y }^{ 2 }=16x$$ and the line x=4 is ___________________________.

A
$$\frac { 128 }{ 3 } sq-units$$

B
$$\frac { 64 }{ 3 } squnits$$

C
$$\frac { 32 }{ 3 } sq-units$$

D
$$\frac { 16 }{ 3 } sq-units$$

Solution
Question 3
1 / -0

If $$A_{m}$$ represents the area bounded by the curve $$y=\ln x^{m}$$., the $$x-$$axis and the lines $$x=1$$ and $$x=2$$, then $$A_{m}+m\ A_{m-1}$$ is

A
$$m$$

B
$$m^{2}$$

C
$$m^{2}/2$$

D
$$m^{2}-1$$

Solution
Question 4
1 / -0

The area of the region bounded by the curve $$y=x^{2}-3x$$ with $$y \le 0$$ is

A
$$3$$

B
$$\dfrac {9}{2}$$

C
$$\dfrac {5}{2}$$

D
$$none\ of\ these$$

Solution

REF.Image
when x=0 $$\Rightarrow $$ y0
when y=0 $$\Rightarrow $$ x(x-3)=0
x=0 and x=3
$$y=x^{2}-3x$$
$$\frac{dy}{dx}=2x-3\Rightarrow x=1.5$$
min value at x=1.5
$$y=\frac{3}{2}(\frac{3}{2}-3)$$
$$=\frac{3}{2}(\frac{-3}{2})=\frac{-9}{4}=-2.25$$
Area of shaded region = $$\int y.dx$$
$$=\int_{0}^{3}(x^{2}-3x)dx$$
$$=(\frac{x^{3}}{3}-\frac{3x^{2}}{2})^{3}_{0}$$
$$=\frac{3^{3}}{3}-3\times \frac{3^{2}}{2}=9-\frac{9\times 3}{2}=-4.5$$
considering magnitude = 4.5
Question 5
1 / -0

If a curve $$y = a \sqrt { x } +$$ bx passes through the point $$( 1,2 )$$ and the area bounded by the curve, line $$x = 4$$ and $$x$$ axis is $$8$$ square units, then

A
$$a = 3 , b = - 1$$

B
$$a = 3 , b = 1$$

C
$$a = - 3 , b = 1$$

D
$$a = - 3 , b = - 1$$

Solution
Question 6
1 / -0

The area bounded by the circle $$x^{2}+y^{2}=8$$, the parabola $$x^{2}=2y$$ and the line $$y=x$$ in first quadrant is $$\dfrac{2}{3}+k\pi$$, where $$k=$$

A
$$\dfrac{5}{7}$$

B
$$2$$

C
$$\dfrac{3}{5}$$

D
$$3$$
Question 7
1 / -0

The area enclosed between the curve $$y=\log_{e}\left(x+e\right)$$ and the coordinate axes is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

A]$$ A=\int_{1-e}^{0} ln(x+e)dx$$
$$=\int_{1-e}^{0} x^{0} ln (x+e)dx$$
= $$(x ln(x+e))-\int_{1-e}^{0}\frac{x}{x+e}dx$$
= $$-\int_{1-e}^{0}\frac{x+e-e}{x+e}dx$$
$$=\int_{1-e}^{0}(\frac{e}{x+e})dx$$
$$ =[e ln(x+e)-x]_{1-e}^{0}=1$$
Question 8
1 / -0

The area of the region formed by $${ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12\le 0,y\le x\quad and\quad x\quad \le \quad \dfrac { 5 }{ 2 } is$$

A
$$\frac { \pi }{ 6 } -\frac { \sqrt { 3}+1 }{ 8 } $$

B
$$\frac { \pi }{ 6 } +\frac { \sqrt { 3}+1 }{ 8 } $$

C
$$\frac { \pi }{ 6 } -\frac { \sqrt { 3}-1 }{ 8 } $$

D
none of these

Solution
Question 9
1 / -0

Area bounded by $$y=2x^{2}$$ and $$y=\dfrac{4}{(1+x^{2})}$$ will be (in sq units)

A
$$(2\pi+4/3)$$

B
$$(2\pi-4/3)$$

C
$$4/3-2\tan^{-1}2+\pi/2$$

D
$$4/3-8\tan^{-1}2+2\pi$$

Solution

$$\begin{array}{l} Po{ { int } }\, of\, intersection\, : \\ 2{ x^{ 2 } }=\dfrac { 4 }{ { 1+{ x^{ 2 } } } } \\ \Rightarrow { x^{ 4 } }+{ x^{ 2 } }-2=0 \\ \Rightarrow { x^{ 4 } }+2{ x^{ 2 } }-{ x^{ 2 } }-2=0 \\ \Rightarrow \left( { { x^{ 2 } }-1 } \right) \left( { { x^{ 2 } }+2 } \right) =0 \\ Then \\ Area\, bounded\, by\, y=2{ x^{ 2 } }\, and\, y=\dfrac { 4 }{ { 1+{ x^{ 2 } } } } \\ 2\int _{ 0 }^{ 1 }{ \left( { \dfrac { 4 }{ { 1+{ x^{ 2 } } } } -2{ x^{ 2 } } } \right) }dx=2\left[ { 4{ { \tan }^{ -1 } }x-\dfrac { { 2{ x^{ 3 } } } }{ 3 } } \right] _{ 0 }^{ 1 } \\ =2\left[ { 4\times \dfrac { \pi }{ 4 } -\dfrac { 2 }{ 3 } } \right] \\ =2\left[ { \pi -\dfrac { 2 }{ 3 } } \right] =2\pi -\dfrac { 4 }{ 3 } \\ Hence,\, option\, \, \, B\, \, is\, the\, correct\, answer. \end{array}$$
Question 10
1 / -0

$$Letf(x)={ sin }^{ -1 }(sin\quad x)+{ cos }^{ -1 }(\quad cos\quad x),\quad g(x)=mx\quad and\quad h(x)=x\quad $$ are three functions. Now g(x) is divided area between f(x),x=$$\pi $$ and y=0 into two equal parts.
The area bounded by the curve y=f(x), x=$$\pi $$ and y=0 is:

A
$$\dfrac { { \pi }^{ 2 } }{ 4 } sq.\quad units$$

B
$${ \pi }^{ 2 }sq.units$$

C
$$\dfrac { { \pi }^{ 2 } }{ 8 } sq.\quad units$$

D
$$2{ \pi }^{ 2 }sq.units$$

Solution

Given,

$$f(x)=\sin^{-1}\sin x+\cos^{-1}\cos x=x+x=2x$$

$$x=\pi$$

$$y=0$$

The above 3 forms a right angled triagle, with center as one vertex.

$$Area=\dfrac{1}{2} \times \pi times 2\pi$$

$$=\pi ^2sq.units$$