Application of Integrals Test

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Application of Integrals Test - 10

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Question 1
1 / -0

The area of the region bounded by the parabola $$(\mathrm{y}-2)^{2}=\mathrm{x}-1$$, the tangent to the parabola at the point $$(2,\ 3)$$ and the $$\mathrm{x}$$-axis is

A
$$3$$

B
$$6$$

C
$$9$$

D
$$12$$

Solution

Equation of tangent at $$(2,3)$$ $$:$$ $$2y=x+4$$

$$\displaystyle \int_{0}^{3}[(y-2)^{2}+1]-[2y-4] \text{ }dy$$

$$=\displaystyle \int_{0}^{3}[(y-2)^{2}-2y+5]\text{ }dy$$

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

$$=9$$
Question 2
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The area (in square units) of the region bounded by the curves $$y + 2x^2 = 0$$ and $$y + 3x^2 = 1$$, is equal to

A
$$\displaystyle \frac{1}{3}$$

B
$$\displaystyle \frac{4}{3}$$

C
$$\displaystyle \frac{3}{5}$$

D
$$\displaystyle \frac{3}{4}$$

Solution

Solving both the equations we get point of intersection both the curves as $$(1,-2)$$ and $$(-1,-2)$$
Thus required are is given by,
$$\displaystyle A =\int_{-1}^{1}(1-3x^2-(-2x^2))dx=\cfrac{4}{3}$$
Question 3
1 / -0

The area (in sq. units) of the region $$\{(x, y):x \geq 0, x+y \leq 3, x^2 \leq 4y$$ and $$y\leq 1+\sqrt{x}\}$$ is.

A
$$\displaystyle\frac{59}{12}$$

B
$$\displaystyle\frac{3}{2}$$

C
$$\displaystyle\frac{7}{3}$$

D
$$\displaystyle\frac{5}{2}$$

Solution

$$C_1: x≥0$$
$$C_2: x+y≤3$$
$$C_3: x^2≤4y$$
$$C_4: y≤1+\sqrt x$$

Solving these curves, we get their intersection points as shown in the figure.
Now, we have to find the area of the shaded region, which is formed using the constraint.
$$Area = \displaystyle\int_0^1(1+\sqrt x) \ dx + \int_1^2(3-x)\ dx - \int_0^2(x^2)\ dx$$
$$\Rightarrow Area = \left[x+\cfrac23x^{3/2}\right]_0^1+[3x-x^2]_1^3 - \left[\cfrac{x^3}3\right]_0^2 = \cfrac52$$
Question 4
1 / -0

The area of the region described by $$ A= (x,y):x^{2}+y^{2}\leq 1$$ and $$ y^{2}\leq 1-x$$ is:

A
$$ \displaystyle \frac{\pi }{2}+\displaystyle \frac{4}{3}$$

B
$$ \displaystyle \frac{\pi }{2}-\displaystyle \frac{4}{3}$$

C
$$ \displaystyle \frac{\pi }{2}-\displaystyle \frac{2}{3}$$

D
$$ \displaystyle \frac{\pi }{2}+\displaystyle \frac{2}{3}$$

Solution

$$ A=\displaystyle \frac{1}{2}\times \pi +2\int_{0}^{1}\sqrt{1-x}dx$$
$$ =\displaystyle \frac{\pi }{2}+\displaystyle \frac{4}{3}$$
Question 5
1 / -0

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(Example: $$S_1$$ is the area bounded by $$y=4$$ and $$x^{2}=4y$$ ); then $$S_{1},S_{2},S_{3}$$ is

A
$$1 : 2 : 1$$

B
$$1 : 2 : 3$$

C
$$2 : 1 : 2$$

D
$$1 : 1 : 1$$

Solution

Since   $$y^{2} = 4x$$    and   $$x^{2} = 4y$$
are symmetric   about the line   y = x
Now, Area   bounded   between   $$y^{2} = 4x   and    y =x   $$   is
$$\int_{0}^{4}\left ( 2\sqrt{x} - x \right )dx$$
$$ = \dfrac{8}{3}$$
$$A_{s_{2}} = \dfrac{16}{3}$$          and
$$A_{s_{1}} = A_{s_{3}} = \dfrac{16}{3}$$
$$ A_{s_{1}}:A_{s_{2}}:A_{s_{3}} :: 1:1:1 $$
Question 6
1 / -0

The area of the region bounded by the curves $$x+2y^{2}=0$$ and $$x+3y^{2}=1$$ is equal to

A
$$\displaystyle \frac{2}{3}$$

B
$$\displaystyle \frac{4}{3}$$

C
$$\displaystyle \frac{5}{3}$$

D
$$\displaystyle \frac{1}{3}$$

Solution

Solving the equations $$x+2{y}^{2}=0$$ and $$x+3{y}^{2}=1$$
we get points of intersection
$$(-2,1)$$ and $$(-2,-1)$$
The bounded region is show in figure.
The required area
$$\displaystyle=2\int _{ 0 }^{ 1 }{ \left( 1-{ 3y }^{ 2 } \right) -\left( -{ 2y }^{ 2 } \right) } $$
$$\displaystyle =2\int _{ 0 }^{ 1 }{ \left( 1-{ y }^{ 2 } \right) } dy=2\left[ y-\frac { { y }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }=2\times \frac { 2 }{ 3 } =\frac { 4 }{ 3 } .$$
Question 7
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The area bounded by the curves $$\mathrm{y}=$$ cosx and $$\mathrm{y}=$$ sinx between the ordinates $$\mathrm{x}=0$$ and $$\displaystyle \mathrm{x}=\frac{3\pi}{2}$$:

A
$$4\sqrt{2}+2$$

B
$$4\sqrt{2}-1$$

C
$$4\sqrt{2}+1$$

D
$$4\sqrt{2}-2$$

Solution

Required area$$=\displaystyle \int_0^{\frac{3\pi}{2}}|sinx-cosx|dx$$

$$= \displaystyle \int_0^{\frac{\pi}{4}}(\cos x-\sin x)dx+ \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}(\sin x-\cos x)dx+ \int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}}(\cos x-\sin x)dx$$

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

$$= 4\sqrt{2}-2$$
Question 8
1 / -0

The area (in square units) bounded by the curves $$y=\sqrt{x},\ 2y-x+3=0$$, $$X-$$axis, and lying in the first quadrant is:

A
$$36$$

B
$$18$$

C
$$\dfrac{27}{4}$$

D
$$9$$

Solution

Given, $$y=\sqrt{x},\ 2y-x+3=0$$
from above we have,
$$2\sqrt x-x+3=0$$
$$
2\sqrt{x}=x-3
$$
$$
4x=x^{2}-6x+9
$$
$$
x^{2}-10x+9=0
$$
$$
x=9,\ x=1
$$

now,
$$\displaystyle \int_{0}^{3}[(2y+3)-y^{2}] dy$$
$$
=\left[y^{2}+3y-\frac{y^{3}}{3}\right]_{0}^{3}=9+9-9=9
$$
Question 9
1 / -0

The area bounded between the parabolas $$4 x^{2}=y$$ and $$x^{2}=9y$$, and the straight line $$\mathrm{y}=2$$ is:

A
$$20\sqrt{2}$$

B
$$\displaystyle \frac{10\sqrt{2}}{3}$$

C
$$\displaystyle \frac{20\sqrt{2}}{3}$$

D
$$10\sqrt{2}$$

Solution

The given curves $$y=4x^2$$ and $$\displaystyle y=\dfrac {1}{9}x^2$$

$$\displaystyle Area(\text{yellow region})=2\int_0^2\left(3\sqrt y-\frac {\sqrt y}{2}\right)dy=2\left[\frac {5y\sqrt y}{3}\right]_0^2$$

$$\displaystyle =2\times \frac {5}{3}\times 2\sqrt 2=\frac {20\sqrt 2}{3}$$
Question 10
1 / -0

The area (in sq. units) of the region $$\left\{ \left( x,y \right) \in { R }^{ 2 }:{ x }^{ 2 }\le y\le 3-2x \right\} $$, is:

A
$$\cfrac { 29 }{ 3 } $$

B
$$\cfrac { 34 }{ 3 } $$

C
$$\cfrac { 31 }{ 3 } $$

D
$$\cfrac { 32 }{ 3 } $$

Solution

From given data in question,
Point of intersection of given two curve $$y = x^2$$ and $$y = -2 x + 3$$
$$x^2 = -2 x + 3$$

$$x = -3 , 1$$

Area of shaded region
$$= \displaystyle \int_{-3}^1 ((-2x + 3) - x^2 ) dx$$

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

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

$$= 12 + 8 - \dfrac{28}{3} = \dfrac{32}{3}$$.