Application of Integrals Test

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Application of Integrals Test - 11

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Question 1
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The area(in sq. units) of the smaller portion enclosed between the curves, $$x^2+y^2=4$$ and $$y^2=3x$$, is.

A
$$\displaystyle\frac{1}{\sqrt{3}}+\frac{4\pi}{3}$$

B
$$\displaystyle\frac{1}{2\sqrt{3}}+\frac{\pi}{3}$$

C
$$\displaystyle\frac{1}{2\sqrt{3}}+\frac{2\pi}{3}$$

D
$$\displaystyle\frac{1}{\sqrt{3}}+\frac{2\pi}{3}$$

Solution

Here, $$x^2+y^2=4$$ and $$y^2=3x$$
So finding the co-ordinates of $$x$$
$$\Rightarrow$$$$x^2+3x-4=0$$

$$(x+4)(x-1)=0$$

$$x=-4, x=1$$

Area$$=\left(\displaystyle\int^1_0\sqrt{3}\cdot \sqrt{x}dx+\displaystyle\int^2_1\sqrt{4-x^2}\cdot dx\right ) \times 2$$

$$=\left(\displaystyle \sqrt{3}\left(\displaystyle\frac{x^{3/2}}{3/2}\right)^1_0+\left(\displaystyle\frac{x}{2}\sqrt{4-x^2}+2sin^{-1}\frac{x}{2}\right)^2_1\right)\times 2$$

$$=\left(\sqrt{3}\left(\displaystyle\frac{2}{3}\right)+\left\{2\cdot \displaystyle\frac{\pi}{2}-\left(\displaystyle\frac{\sqrt{3}}{2}+\frac{\pi}{3}\right)\right\}\right)\times 2$$

$$=\left(\displaystyle\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2}+\frac{2\pi}{3}\right)\times 2$$

$$=\left(\displaystyle\frac{1}{2\sqrt{3}}+\frac{2\pi}{3}\right)\times 2=\displaystyle\frac{1}{\sqrt{3}}+\frac{4\pi}{3}$$.
Question 2
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For $$a > 0$$, let the curves $$C_1 : y^2 = ax$$ and $$C_2 : x^2 = ay$$ intersect at origin $$O$$ and a point $$P$$. Let the line $$x = b (0 < b < a)$$ intersect the chord $$OP$$ and the x-axis at points $$Q$$ and $$R$$, rspectively. If the line $$x=b$$ bisects the area bounded by the curves, $$C_1$$ and $$C_2$$, and the area of $$\Delta OQR = \dfrac{1}{2}$$, then '$$a$$' satisfies the equation:

A
$$x^6-12x^3 + 4 = 0$$

B
$$x^6 + 6x^33 - 4 = 0$$

C
$$x^6 - 12x^3 + 4 = 0$$

D
$$x^6 - 6x^3 + 4 = 0$$

Solution

area $$\Delta OQR = \dfrac{1}{2}.....given$$

For coordinates of $$p$$

Curves : $$x^2 = ay, y^2 = ax$$

$$\Rightarrow y^2 = \left(\dfrac{x^2}{a} \right)^2 = ax$$

$$\dfrac{x^4}{a^2} = ax$$

$$x^4 = a^3x$$

$$x(x^3 - a^3) = 0$$

$$\Rightarrow x = 0, x=a....(i)$$

$$\Rightarrow y^2 = ax$$

$$\therefore y = 0, a....(from \ i)$$

$$\therefore$$ equation of chord $$OP$$

$$O(0,0)\equiv(x_1,y_1) \ and \ P(a,a)\equiv(x_2,y_2)$$

$$y - y_1 = \dfrac{y_2-y_1}{x_2-x_1} (x-x_1)$$

$$y - 0 = \dfrac{a-0}{a-0} (x-o)$$

$$y = x$$

$$\therefore$$ At point $$Q$$

$$x = b$$ and $$y = b$$

$$\Rightarrow Q \equiv (b,b)$$

$$\therefore$$ ar $$\Delta OQR = \dfrac{1}{2} \times b \times b = \dfrac{b^2}{2} = \dfrac{1}{2}$$

$$\Rightarrow b^2 = 1 \Rightarrow b=1$$

Now, area bounded by the curves

$$= \displaystyle \int_0^a \left(\sqrt{ax} - \dfrac{x^2}{a}\right)dx = \int_0^a a^{1/2} x^{1/2}dx - \int_0^a \dfrac{x^2}{a} dx$$

$$=a^{1/2}\left[\dfrac{x^{3/2}}{3/2}\right]_0^a-\dfrac{1}{a}\left[\dfrac{x^3}{3}\right]_0^a$$

$$=\dfrac{2}{3}a^{1/2}. a^{3/2} - \dfrac{1}{a} \dfrac{a^3}{3} = \dfrac{2}{3} a^2 - \dfrac{1}{3}a^2$$

$$= \dfrac{1}{3}a^2$$

$$\therefore \displaystyle \int_0^b \left(\sqrt{ax} - \dfrac{x^2}{a}\right)dx = \dfrac{1}{2} \times \dfrac{1}{3}a^2 = \dfrac{a^2}{6}$$

$$\dfrac{2}{3} a^{1/2} [x^{3/2}]_0^b - \dfrac{1}{3a} [x^3]_0^b = \dfrac{a^2}{6}$$

$$\Rightarrow \dfrac{2}{3}a^{1/2} b^{3/2} - \dfrac{1}{3a}b^3 = \dfrac{a^2}{6}$$

Now after putting value of $$b$$.

we get,
$$\dfrac{2}{3}a^{1/2} - \dfrac{1}{3a} = \dfrac{a^2}{6}$$

$$2 a^{3/2} -1 = \dfrac{3a^3}{6} = \dfrac{a^3}{2}$$

$$4a^{3/2} - 2 =a^3 \Rightarrow 4a^{3/2} = a^3 + 2$$

Now squaring on both the side

$$\Rightarrow a^6 + 4a^3 + 4 = 16a^3$$

$$\Rightarrow a^6 - 12a^3 + 4 = 0$$

$$\therefore$$ a will satisfy

$$x^6 - 12x^3 + 4 = 0$$

$$\therefore$$ $$\boxed{x^6 - 12x^3 + 4 = 0}....Answer$$

Hence option $$'C'$$ is the answer.
Question 3
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Given $$f(x)=\begin{cases} x,0\le x<\dfrac { 1 }{ 2 } \\ \dfrac { 1 }{ 2 } ,x=\dfrac { 1 }{ 2 } \\ 1-x,\dfrac { 1 }{ 2 } <x\le 1 \end{cases}$$ and $$g(x)=\left(x-\dfrac{1}{2}\right)^{2},x\epsilon R$$, Then the area (in sq.units) of the region bounded by the curves $$y=f(x)$$ and $$y=g(x)$$ between the lines $$2x=1$$ and $$2x=\sqrt{3}$$, is:

A
$$\dfrac{1}{3}+\dfrac{\sqrt{3}}{4}$$

B
$$\dfrac{1}{2}-\dfrac{\sqrt{3}}{4}$$

C
$$\dfrac{1}{2}+\dfrac{\sqrt{3}}{4}$$

D
$$\dfrac{\sqrt{3}}{4}-\dfrac{1}{3}$$

Solution

$$\left\{\begin{matrix}x &,0<n<\dfrac{1}{2} \\\dfrac{1}{2} &,n=\dfrac{1}{2} \\ 1-n& ,\dfrac{1}{2}<n<1\end{matrix}\right.$$

and $$g(x)=(x-\dfrac{1}{2})^2$$
graph of the current is as
refer above image.
Required image =are of trapezium $$\displaystyle ABCD-\int^{\dfrac{\sqrt{3}}{2}}_{\dfrac{1}{2}}\left(x-\dfrac{1}{2}\right)^2dx$$

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

$$=\dfrac{\sqrt{3}}{4}-\dfrac{1}{3}$$
Question 4
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If the area enclosed between the curves $$y=kx^2$$ and $$x=ky^2$$, $$(k > 0)$$, is $$1$$ square unit. Then $$k$$ is?

A
$$\dfrac{1}{\sqrt{3}}$$

B
$$\dfrac{2}{\sqrt{3}}$$

C
$$\dfrac{\sqrt{3}}{2}$$

D
$$\sqrt{3}$$

Solution

Area bounded by $$y^2=4ax$$ & $$x^2=4by$$, a, b $$\neq 0$$ is $$\left|\dfrac{16ab}{3}\right|$$
by using formula: $$4a=\dfrac{1}{k}=4b, k > 0$$
Area $$=\left|\dfrac{16\cdot \dfrac{1}{4k}\cdot \dfrac{1}{4k}}{3}\right|=1$$
$$\Rightarrow k^2=\dfrac{1}{3}$$
$$\Rightarrow k=\dfrac{1}{\sqrt{3}}$$.
Question 5
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If the area of the region bounded by the curves, $$y=x^2, y=\displaystyle\frac{1}{x}$$ and the lines $$y=0$$ and $$x=t(t > 1)$$ is $$1$$ sq. unit, then t is equal to?

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

B
$$e^{{2}/{3}}$$

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

D
$$e^{{3}/{2}}$$

Solution

Given, $$y=x^2, y=\displaystyle\frac{1}{x}$$

Area bounded by the curves is the region ABCD.

Therefore, area $$=\displaystyle \int_{0}^{1}x^2dx + \int_{1}^{t}\dfrac{1}{x}dx $$
$$=\left [ \dfrac{x^3}{3} \right ]_{0}^{1} +\left [ \ln (x) \right ]_{1}^{t}$$

$$ =\dfrac{1}{3}+\ln(t) $$

It is given that area enclosed is $$1$$

$$\Rightarrow \dfrac{1}{3}+\ln(t)=1 $$
$$\Rightarrow \ln(t)=\dfrac{2}{3} $$

$$ \Rightarrow t = e^{\frac{2}{3}}$$

Hence, answer is option (B)
Question 6
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The area (in sq. units) of the region $$\{(x, y) \in R^2 |4x^2 \le y \le 8x + 12\}$$ is :

A
$$\dfrac{128}{3}$$

B
$$\dfrac{127}{3}$$

C
$$\dfrac{125}{3}$$

D
$$\dfrac{124}{3}$$

Solution

Shaded are is the area between curve

$$Y = 4x^2 $$ __(I)
and
$$Y = 8x + 12$$ __(II)

on solving equation (I) & (II) (to find intersection point of the curves).

$$4x^2 - 8x - 12 = 0$$

$$4(x +1 )(x - 3) = 0$$

$$x = -1 $$ or $$x = 3$$

$$\therefore$$ shaded Area $$= \displaystyle \int_{-1}^3 [(8x + 12) - (4x^2)] dx$$

$$= \displaystyle \int_{-1}^3 8x dx + \int_{-1}^3 12 dx - 4 \int_{-1}^3 x^2 dx$$

$$= \left[4x^2 + 12 x - \dfrac{4}{3} x^3 \right]_{-1}^3$$

$$=4\cdot 9+12\cdot 3-\dfrac 43\cdot 27-4+12-\dfrac 43$$

$$= \dfrac{128}{3}$$
Question 7
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The area enclosed between the curves $$\mathrm{y}=\mathrm{a}\mathrm{x}^{2}$$ and $$\mathrm{x}=\mathrm{a}\mathrm{y}^{2} (\mathrm{a}>0)$$ is 1 sq. unit, then the value of a is

A
$$1/\sqrt{3}$$

B
$$1/2$$

C
$$1$$

D
$$1/3$$

Solution

Points of intersection of $$y=\mathrm{a}\mathrm{x}^{2}$$ and

$$\mathrm{x}=\mathrm{a}\mathrm{y}^{2}$$ are $$(0, 0)$$ and

$$\left(\displaystyle \frac{1}{\mathrm{a}}, \displaystyle

\frac{1}{\mathrm{a}}\right)$$.
Hence area is, $$\displaystyle

\int_{0}^{1/\mathrm{a}}\left(\sqrt{\frac{\mathrm{x}}{\mathrm{a}}}- ax^{2}

\right)\mathrm{dx} =1\Rightarrow \mathrm{a}=\frac{1}{\sqrt{3}}

(\mathrm{a}\mathrm{s} \ \mathrm{a}>0)$$.
Question 8
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The area of the region $$\left \{(x, y) : xy \leq 8, 1 \leq y\leq x^{2}\right \}$$ is

A
$$16\log_{6} 2 - 6$$

B
$$8\log_{6} 2 - \dfrac {7}{3}$$

C
$$16\log_{6} 2 - \dfrac {14}{3}$$

D
$$8\log_{6} 2 - \dfrac {14}{3}$$

Solution

$$xy \leq 8$$
$$1\leq y\leq x^{2}$$
$$x^{2} . x = 8$$
$$x = 2$$
Require Area $$= \int_{1}^{4} \left (\dfrac {8}{y} - \sqrt {y}\right )dy = \left [8ln y - \dfrac {y^{3/2}}{3/2}\right ]_{1}^{4} = 8\ ln 4 - \dfrac {2}{3} . 8 - 0 + \dfrac {2}{3} = 16 ln2 - \dfrac {14}{3}$$.
Question 9
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The area of the region between the curves $$\mathrm{y}=\sqrt{\dfrac{1+\sin \mathrm{x}}{\cos \mathrm{x}}}$$ and $$\mathrm{y}=\sqrt{\dfrac{1-\sin \mathrm{x}}{\cos \mathrm{x}}}$$ bounded by the lines $$\mathrm{x}=0$$ and $$\displaystyle \mathrm{x}=\frac{\pi}{4}$$ is

A
$$\displaystyle \int_{0}^{\sqrt{2}-1}\frac{\mathrm{t}}{(1+t^{2})\sqrt{1-\mathrm{t}^{2}}} \mathrm{d}\mathrm{t}$$

B
$$\displaystyle \int_{0}^{\sqrt{2}-1}\frac{4\mathrm{t}}{(1+t^{2})\sqrt{1-\mathrm{t}^{2}}} \mathrm{d}\mathrm{t}$$

C
$$\displaystyle \int_{0}^{\sqrt{2}+1}\frac{4\mathrm{t}}{(1+t^{2})\sqrt{1-\mathrm{t}^{2}}} dt$$

D
$$\displaystyle \int_{0}^{\sqrt{2}+1}\frac{\mathrm{t}}{(1+t^{2})\sqrt{1-\mathrm{t}^{2}}} dt$$

Solution

$$\displaystyle \int_{0}^{\pi/4}(\sqrt{\dfrac{1+\sin \mathrm{x}}{\cos \mathrm{x}}}-\sqrt{\dfrac{1-\sin \mathrm{x}}{\cos \mathrm{x}}})\mathrm{d}\mathrm{x}$$

$$=\int _{0}^{\pi /4}(\sqrt{\dfrac{1+tan\dfrac{x}{2}}{1-tan\dfrac{x}{2}}}-\sqrt{\dfrac{1-tan\dfrac{x}{2}}{1+tan\dfrac{x}{2}}})dx=\int \dfrac{(1+tan\dfrac{x}{2})-(1-tan\dfrac{x}{2})}{\sqrt{1-tan^{2}\dfrac{x}{2}}}dx$$

$$=\displaystyle \int_{0}^{\pi/4}\dfrac{2\tan\dfrac{\mathrm{x}}{2}}{\sqrt{1-\tan^{2}\frac{\mathrm{x}}{2}}} dx =\displaystyle \int_{0}^{\sqrt{2}-1}\dfrac{4\mathrm{t}}{(1+\mathrm{t}^{2})\sqrt{1-\mathrm{t}^{2}}} dt$$ as $$\displaystyle \tan\dfrac{\mathrm{x}}{2}=\mathrm{t}$$.
Question 10
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Let the functions $$f:R\rightarrow R$$ and $$𝑔:R\rightarrow R$$ be defined by $$f(x)=e^{ x-1 }-e^{ -|x-1| }$$ and $$g(x)=\dfrac{1}{2}(e^{x-1}+e^{1-x})$$. Then the area of the region in the first quadrant bounded by the curves $$𝑦=𝑓(𝑥), 𝑦=𝑔(𝑥) $$ and $$x=0$$ is

A
$$(2-\sqrt{3})+\dfrac{1}{2}(e-e^{-1})$$

B
$$(2+\sqrt{3})+\dfrac{1}{2}(e-e^{-1})$$

C
$$(2-\sqrt{3})+\dfrac{1}{2}(e+e^{-1})$$

D
$$(2+\sqrt{3})+\dfrac{1}{2}(e+e^{-1})$$

Solution

$$f(x) = 0$$ $$ ; x < 1$$
$$= e^{x-1} - e^{-(x-1)}$$ $$; x \ge 1$$

while $$g(x) \ge 1$$
so they will intresect in the region $$x > 1$$
solve $$f(x) = g(x)$$

$$e^{x-1} - e^{-(x-1)} = \dfrac{1}{2} (e^{x-1} + e^{1-x})$$

$$\dfrac{1}{2} e^{x-1} = \dfrac{3}{2} e^{1-x}$$

$$e^{2x}=3e^2$$

$$2x = 2 + \ell n3$$

$$x = 1 + \ell n \sqrt{3}$$

$$\displaystyle A = \int_0^1 g(x) dx + \int_1^{1+\ell n\sqrt{3}} (g(x) - f(x)) dx$$

$$\displaystyle = \dfrac{1}{2} \int_0^1 (e^{x-1} + e^{1-x})dx + \int_1^{1+\ell n \sqrt{3}} \dfrac{1}{2} (e^{x-1} + e^{1-x}) - (e^{x-1} - e^{1-x})dx$$

$$\displaystyle = \dfrac{1}{2} \int_1^{1+\ell n \sqrt{3}} (e^{x-1} + e^{1-x})dx - \int_1^{1+\ell n \sqrt{3}} (e^{x-1} - e^{1-x})dx$$

$$\displaystyle = \dfrac{1}{2} [e^{x-1} - e^{1-x} ]_0^{1+\ell n \sqrt{3}} - (e^{x-1} + e^{1-x})_1^{1+ \ell n \sqrt{3}}$$

$$= \displaystyle = \dfrac{1}{3} [ \sqrt{3} - \dfrac{1}{\sqrt{3}} - \dfrac{1}{e} + e] - [\sqrt{3} + \dfrac{1}{\sqrt{3}} - 1 - 1]$$

$$= 2 + \dfrac{1}{2} \left(e-\dfrac{1}{e}\right) - \dfrac{\sqrt{3}}{2} - \dfrac{3}{2\sqrt{3}}$$

$$= (2-\sqrt{3}) + \dfrac{e-\frac{1}{e}}{2}$$