Application of Integrals Test

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Application of Integrals Test - 16

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Question 1
1 / -0

The area bounded by tangent, normal and x-axis at $$\mathrm{P}(2,4)$$ to the curve $$y=x^{2}$$

A
$$34$$

B
$$32$$

C
$$36$$

D
$$24$$

Solution

Slope of the tangent at $$(2,4)$$
$$=\dfrac{dy}{dx}|_{x=2}$$

$$=4$$.

Hence slope of normal at $$(2,4)$$ is $$\dfrac{-1}{4}$$.

Therefore equation of tangent will be
$$y-4=4(x-2)$$
$$y-4=4x-8$$
$$4x-y=4$$

Hence it cuts the x axis at $$(1,0)$$.

Equation of normal will be $$y-4=\dfrac{-1}{4}(x-2)$$
$$4y-16=-x+2$$
$$x+4y=18$$.

Hence it cuts the x axis at $$(18,0)$$.

Now the normal and tangent meet each other at $$(2,4)$$.

Since normal is perpendicular to the tangent it is a right angled triangle.
The coordinates are $$(1,0),(18,0),(2,4)$$.

Hence
Length of hypotenuse= 17.
Length of base =$$\sqrt{17}$$
Length of height =$$\sqrt{272}=4\sqrt{17}$$.

Hence
$$Area=\dfrac{1}{2}(b\times h)$$

$$\dfrac{4\sqrt{17}\times \sqrt{17}}{2}$$

$$=2\times17$$
$$=34$$ sq units.
Question 2
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The area in square units bounded by the curves $$y=x^{3},\ y=x^{2}$$ and the ordinates $${x}=1, {x}=2$$ is

A
$$\displaystyle \frac{17}{12}$$

B
$$\displaystyle \frac{12}{13}$$

C
$$\displaystyle \frac{2}{7}$$

D
$$\displaystyle \frac{7}{2}$$

Solution

The given curves are $$y=x^3$$ and $$y=x^2$$
And the limits of $$x$$ are $$x=1, x=2$$
Then, the required area is
$$A=\displaystyle \int_{1}^{2}(x^3-x^2)dx$$
$$=\displaystyle \int_{1}^{2} x^3\:dx-\int_{1}^{2}x^2\:dx$$
$$=\left[\dfrac{x^4}{4}\right]_{1}^{2}-\left[\dfrac{x^3}{3}\right]_{1}^{2}$$
$$=\left ( \dfrac{16}{4}-\dfrac{1}{4} \right )-\left ( \dfrac{8}{3}-\dfrac{1}{3} \right )$$
$$=\dfrac{15}{4}-\dfrac{7}{3}=\dfrac{17}{12}$$ sq units
Question 3
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Area of the region bounded by $$y=|x|$$ and $$y=1-|x|$$ is

A
$$\displaystyle \frac{1}{3}$$ sq. units

B
1 sq. units

C
$$\displaystyle \frac{1}{2}$$ sq. unit

D
2 sq. units

Solution
Question 4
1 / -0

The area, in square units of the region bounded by the parabolas $$y^{2}=4x$$ and $$x^{2}=4y$$ is

A
$$\dfrac{16}{3}$$

B
$$\dfrac{32}{3}$$

C
$$\dfrac{8}{3}$$

D
$$\dfrac{4}{3}$$

Solution

The given parabolas are $$y^{2}=4x$$ and $$x^{2}=4y$$
The point of intersection of $$x^2=4y$$ and $$y^2=4x$$ will be
$$x=4$$ and $$x=0$$
Then, the required area will be
$$\overset{4}{\underset{0}{\displaystyle \int}}\left (\sqrt{4x} -\dfrac{x^2}{4} \right )dx$$
$$=\left[\sqrt{4}\times\dfrac{2}{3}x^{3/2}-\dfrac{x^3}{12}\right]_{0}^{4}$$
$$=\dfrac{4\times 8}{3}-\dfrac{4^3}{12}$$

$$=\dfrac{32}{3}-\dfrac{16}{3}$$

$$=\dfrac{16}{3}\:sq\:units$$
Question 5
1 / -0

Area of the figure bounded by Y-axis, $$y=Sin^{-1}x,\ y=Cos^{-1}x$$ and the first point of intersection from the origin is

A
$$2\sqrt{2}$$

B
$$2\sqrt{2}+1$$

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

D
$$\sqrt{2}+1$$

Solution

The curve $$y=\sin ^{ -1 }{ x } $$ and $$y=\cos ^{ -1 }{ x } $$ intesect at $$\displaystyle \left( \frac { 1 }{ \sqrt { 2 } } ,\frac { \pi }{ 4 } \right) $$
Thus area $$\displaystyle =\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \left( \cos ^{ -1 }{ x } -\sin ^{ -1 }{ x } \right) dx } $$
$$\displaystyle ={ \left[ x\cos ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }-\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } dx } +{ \left[ x\sin ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }+\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx } \\ =\sqrt { 2 } -1$$
Question 6
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The area bounded by the two curves $$y=\sqrt{x}$$ and $$x=\sqrt{y}$$ is:

A
$$\displaystyle \frac{1}{3}$$ sq, units

B
$$\displaystyle \frac{2}{3}$$ sq. units

C
$$\displaystyle \frac{1}{5}$$ sq. units

D
$$\displaystyle \frac{1}{7}$$ sq. units

Solution

$$y=\sqrt{x}$$ and $$x=\sqrt{y}$$
Area $$=\overset{1}{\underset{0}{\int}}(\sqrt{x}-x^2)dx$$
$$\quad\quad=\frac{2}{3}(1)-1/3\\\quad\quad=1/3\:sq\:units$$
Question 7
1 / -0

The area of the region bounded by $$x^{2}=8y,\ x=4$$ and the $$\mathrm{x}$$-axis is

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

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

C
$$\displaystyle \frac{8}{3}$$

D
$$\displaystyle \frac{10}{3}$$

Solution

$$y=\dfrac{x^2}{8};x=4$$

Area $$=\overset{4}{\underset{0}{\int}}\dfrac{x^2}{8}\:dx$$
$$=\dfrac{x^3}{24}|_{0}^{4}$$
$$=\dfrac{4^3}{24}\\=\dfrac{64}{24}\\=\dfrac83\:sq\:units$$
Question 8
1 / -0

The area of the region bounded by $$y=x,\ y=x^{3}$$ is:

A
$$\displaystyle \frac{1}{4}$$ sq. units

B
$$\displaystyle \frac{1}{12}$$ sq. units

C
$$\displaystyle \frac{1}{3}$$ sq. units

D
$$\displaystyle \frac{1}{2}$$ sq. units

Solution

The curve intersect at $$x=-1,y=-1$$; $$x=0,y=0$$ and $$x=1,y=1$$
Therefore area $$\displaystyle =-\int _{ -1 }^{ 0 }{ \left( { x }^{ 3 }-x \right) dx } +\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 3 } \right) dx } $$
$$\displaystyle =-{ \left[ \frac { { x }^{ 4 } }{ 4 } -\frac { { x }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 0 }+{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 4 } }{ 4 } \right] }_{ 0 }^{ 1 }$$
$$\displaystyle =-\left( \frac { 1 }{ 4 } -\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 2 } -\frac { 1 }{ 4 } \right) =-\frac { 1 }{ 2 } +1=\frac { 1 }{ 2 } $$
Question 9
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The area bounded by the parabola $$x=y^{2}$$ and the line $$y=x-6$$ is

A
$$\displaystyle \frac{125}{3}$$ sq. units

B
$$\displaystyle \frac{125}{6}$$ sq. units

C
$$\displaystyle \frac{125}{4}$$ sq. units

D
$$\displaystyle \frac{115}{3}$$ sq. units

Solution

$$x=y^{2}$$ and the line $$y=x-6$$
area in $$QI$$+ area in QIV
$$|\overset {3 }{ \underset { -2}{ \int } } [(y+6)-y^2]dy|$$
$$=|\left | \dfrac{y^2}{2} +6y-y^3/3\right |_{-2}^{3}$$
$$=\left |\dfrac{9}{2} -\dfrac{4}{2}+6(5)-\left [ \dfrac{27}{3} +8/3\right ] \right |$$
$$\left | \dfrac{5}{2}+30-\left [ \dfrac{35}{3} \right ] \right | = \dfrac{15+180-70}{6} = \dfrac{125}{6}sq\:units.$$
Question 10
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The area between the curve $$y=x^{2}$$ and $$y=x+2$$ is:

A
$$\displaystyle \frac{9}{2}$$ sq. units

B
$$\displaystyle \frac{3}{2}$$ sq. units

C
9 sq. units

D
6 sq. units

Solution

point of intersection $$x^2-x-2=0$$
2,-1
$$\overset{2}{\underset{-1}{\int}}(x+2)\:dx -\overset{2}{\underset{-1}{\int}}x^2\:dx$$
$$3/2+2(3)-\left [ \frac{8}{3}+1/3 \right ]$$
$$3/2+6-3$$
$$6-3/2=9/2\:sq\:units$$