Application of Integrals Test

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Application of Integrals Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

The area of the region bounded by the curve $$y=2x-x^2$$ and the line $$y=x$$ is ________ square units.

A
$$\dfrac{1}{6}$$

B
$$\dfrac{1}{2}$$

C
$$\dfrac{1}{3}$$

D
$$\dfrac{7}{6}$$

Solution

We note that the region bounded by these curves is in the region $$x \in [0,1 ]$$. In this region, the curve $$y = x$$ lies below the curve $$y = 2x-x^2$$
So, to calculate the area of said region, we evaluatie the following integral:
$$\int_0^1 2x-x^2 - x dx $$
$$= \int_0^1 x-x^2 dx $$
$$= [\frac{x^2}{2} - \frac{x^3}{3}]_{x=0}^{x=1}$$
$$= \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.$$
Question 2
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The area bounded by the curve $$ y =x^2 +2x +1 $$ and tangent at $$ ( 1 , 4) $$ and y -axis and

A
$$ \frac {2}{3} $$ sq units

B
$$ \frac {1}{3} $$ sq units

C
2 sq units

D
None of these

Solution
Question 3
1 / -0

If $$ A_n $$ is the area bounded by $$ y = ( 1 -x^2)^n $$ and coordinates axes , $$ n \epsilon N $$, then

A
$$ A_n = A_{n-1} $$

B
$$ A_n < A_{n-1} $$

C
$$ A_n > A_{n-1} $$

D
$$ A_n =2 A_{n-1} $$

Solution
Question 4
1 / -0

The area enclosed between the curves $$y=log_{e}(x+e)\, , \,x=log_{e}\left ( \dfrac{1}{y} \right ) $$ and the $$x-axis$$ is

A
$$2$$ sq.units

B
$$1$$ sq.units

C
$$4$$ sq.units

D
None of these

Solution

$$ y=log_{e}(x+e)\, , \, x=log_{e}\left ( \dfrac{1}{y} \right ) \Rightarrow y=e^{-x} $$

for $$y=log_{e}(x+e)$$ shift the graph os $$log_{e}x\, , e$$ units left hand side.

Required area = $$ \displaystyle \int_{1-e}^{0}log_e(x+e)dx\,+ \displaystyle \int_{0}^{\infty }e^{-x}dx $$

$$ = |xlog_{e}(x+e)|_{1-e}^{0}-\displaystyle \int_{1-e}^{0}\dfrac{x}{x+e}dx-|e^{x}|_{0}^{\infty } $$

$$ =\displaystyle \int_{0}^{1-e}\left ( 1-\dfrac{e}{x+e} \right )dx-e^{-\infty}+e^{0} $$

$$ = |x-elog(x+e)|_{0}^{1-e}-0+1 $$

$$=1-e +e\, loge+1 = 2sq.units $$
Question 5
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If $$\displaystyle \left ( \alpha ^2,\alpha - 2 \right )$$ be a point interior to the region of the parabola $$\displaystyle y^2 = 2x$$ bounded by the chord joining the points $$\displaystyle \left ( 2,2 \right ) and \left ( 8,-4 \right )$$ then $$\alpha$$ belongs to the interval

A
$$\displaystyle -2 + 2\sqrt {2} < \alpha <2$$

B
$$\displaystyle \alpha > -2 + 2\sqrt {2}$$

C
$$\displaystyle \alpha > -2 - 2\sqrt {2}$$

D
None of these

Solution
Question 6
1 / -0

The area of the region enclosed by the curves $$y=x\log x$$ and $$y=2x-2x^2$$ is

A
$$ \dfrac{7}{12}\,sq.units $$

B
$$ \dfrac{1}{12}\,sq.units $$

C
$$ \dfrac{5}{12}\,sq.units $$

D
None of these

Solution

Curve taking : $$y = x \log_ex$$
Clearly, $$x > 0$$,
For $$0 < x < 1, x \log_ex < 0$$, and for $$x > 1, x \log_ex > 0$$
Also $$x \log_ex = 0 \Rightarrow x = 1$$.
Further, $$\dfrac{dy}{dx} = 0 \Rightarrow 1 + \log_e x = 0 \Rightarrow x = 1/e$$, which is a point of minima.
Required area
$$= \displaystyle \int_0^1 (2x - 2x^2)dx - \int_0^1 x \log x \ dx$$

$$= \left[x^2 - \dfrac{2x^3}{3}\right]_0^1 - \left[\dfrac{x^2}{2} \log x - \dfrac{x^2}{4}\right]_0^1$$

$$= \left( 1 - \dfrac{2}{3} \right) - \left[ 0 - \dfrac{1}{4} - \dfrac{1}{2} \underset{x \to 0}{\lim} x^2 \log x\right] = \dfrac{1}{3} + \dfrac{1}{4} = \dfrac{7}{12}$$.
Question 7
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Area of the region bounded by the curve $$ y =e^x, y=e^{-x} $$ and the straight line x= 1 given by

A
$$ e-e^{-1} +2 $$

B
$$ e-e^{-1} - 2 $$

C
$$ e+ e^{-1} -2 $$

D
None of these

Solution
Question 8
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The area bounded by the curve $$ y = (x) $$ the x-axis and the ordinate $$x= 1$$ and $$x = b$$ is $$(b- 1)$$ $$cos ( 3b + 4)$$, then $$f(x)$$ is given by

A
$$(x -1 ) sin (3x +4)$$

B
$$(x-1) sin (3x-4)$$

C
$$-3(x -1) sin ( 3x+ 4) + cos (3x+ 4)$$

D
None of these

Solution
Question 9
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The area of the closed figure bounded by $$y=\dfrac{x^{2}}{2}-2x+2$$ and the tangents to it at $$(1,\dfrac{1}{2}) $$ and $$(4,2)$$ is

A
$$ \dfrac{9}{8} \, sq.units $$

B
$$ \dfrac{3}{8} \, sq.units $$

C
$$ \dfrac{3}{2} \, sq.units $$

D
$$ \dfrac{9}{4} \, sq.units $$

Solution

$$ y= \dfrac{x^{2}}{2} -2x +2 = \dfrac{(x-2)^2}{2}, $$

$$ \dfrac{dy}{dx} = x-2 \, , x-2 \,\, , \left ( \dfrac{dy}{dx} \right )_{x=1}=-1\,\, , \left ( \dfrac{dy}{dx} \right )_{x=4}=2 $$

$$ \Rightarrow$$ Tangent at $$ (1, \dfrac{1}{2}$$ is $$ y- \dfrac{1}{2}=-1(x-1) \, or \, 2x+2y-3=0 $$

Tangent at $$(4,2)$$ is $$ y-2=2(x-4) \, or\, 2x-y-6=0 $$

Hence, A= $$ \displaystyle \int_{1}^{\frac{5}{2}} \left ( \dfrac{x^{2}}{2}-2x+2-\dfrac{3-2x}{2} \right )dx+\displaystyle \int_{\frac{5}{2}}^{4}\left ( \dfrac{x^2}{2}-2x+2-(2x-6) \right )dx $$

$$ =\displaystyle \int_{1}^{4}\left ( \dfrac{x^{2}}{2}-2x+2 \right )dx- \displaystyle \int_{1}^{\frac{5}{2}}\left ( \dfrac{3-2x}{2} \right )dx- \displaystyle \int_{\frac{5}{2}}^{4}(2x-6)dx $$

$$ = \left ( \dfrac{x^{3}}{6}-x^{2}+2x \right )_{1}^{4}-\dfrac{1}{2}(3x-x^{2})_{1}^{\frac{5}{2}}-(x^{2}-6x)_{\frac{5}{2}}^{4} $$

$$ = =\left ( \dfrac{63}{6}-15+6 \right )-\dfrac{1}{2}\left ( 3 \times \dfrac{3}{2}-\left ( \dfrac{25}{4}-1 \right ) \right )-\left ( \left ( 16-\dfrac{25}{4} \right )-6\left ( 4-\dfrac{5}{2} \right ) \right ) $$

$$ = \dfrac{3}{2}- -\dfrac{1}{2}\left ( \dfrac{9}{2}-\dfrac{21}{4} \right )-\left ( \dfrac{39}{4}-6\left ( \dfrac{3}{2} \right ) \right ) $$

$$ \dfrac{9}{8} \, sq.units $$
Question 10
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The area of the region in 1st quadrant bounded by the $$y-axis, \, y=\dfrac{x}{4}, \, y=1+\sqrt{x} \, and\, y= \dfrac{2}{\sqrt{x}} $$

A
$$ \dfrac{2}{3}\, sq.units $$

B
$$ \dfrac{8}{3}\, sq.units $$

C
$$ \dfrac{11}{3}\, sq.units $$

D
$$ \dfrac{13}{6}\, sq.units $$

Solution

$$ A_{1}= \displaystyle \int_{0}^{1}\left ( 1+ \sqrt{x}-\dfrac{1}{x} \right )dx $$

$$\left [x + \dfrac{2x^{\frac{3}{2}}}{3}-\dfrac{x^{2}}{8} \right ]_{0}^{1}= 1+ \dfrac{2}{3}-\dfrac{1}{8}=\dfrac{37}{24} $$

$$ A_{2}= \displaystyle \int_{1}^{4}\left ( \dfrac{2}{\sqrt{x}}-\dfrac{x}{4} \right )dx $$

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

$$ = \left [ 8-2-4+\dfrac{1}{8} \right ]= \dfrac{17}{8} $$

$$ \Rightarrow A=A_{1}+A_{2} \, = \dfrac{88}{24}=\dfrac{11}{3} \,sq.units. $$