Application of Integration MCQ Questions & Answers in Calculus | Maths

$$xy \leqslant 8,\,\,1 \leqslant y \leqslant {x^2}$$
Intersection points of $$xy= 8$$   and $$y= 1$$   is $$\left( {8,\,1} \right);xy = 8$$    and $$y = {x^2}$$   is $$\left( {2,\,4} \right)$$  and $$y = {x^2}$$   and $$y =1$$   is $$\left( {1,\,1} \right)$$

$$\eqalign{ & {\text{Required area}}\, = \int\limits_1^2 {{x^2}dx} + \int\limits_2^8 {\frac{8}{x}dx} - \int\limits_1^8 {1\,dx} \cr & = \left( {\frac{{{x^3}}}{3}} \right)_1^2 + \left( {8\ln \,x} \right)_2^8 - \left( x \right)_1^8 \cr & = \frac{8}{3} - \frac{1}{3} + 8\,\ln \,8 - 8\,\ln \,2 - \left( {8 - 1} \right) \cr & = \frac{7}{3} + 24\,\ln \,2 - 8\,\ln \,2 - 7 \cr & = 16\,\ln \,2 - \frac{{14}}{3} \cr & \therefore {\text{ Correct option is (B)}} \cr} $$

Clearly, the curve $$y = x{\left( {3 - x} \right)^2},$$    has maximum at $$x = 1$$  and minimum at $$x = 3$$

$$\therefore $$  Required area
$$\eqalign{ & = \int_1^3 {x{{\left( {3 - x} \right)}^2}dx} \cr & = \int_1^3 {\left( {{x^3} - 6{x^2} + 9x} \right)\,dx} \cr & = \left[ {\frac{{{x^4}}}{4} - 2{x^3} + \frac{{9{x^2}}}{2}} \right]_1^3 \cr & = 4{\text{ sq}}{\text{. units}} \cr} $$

$$\eqalign{ & {I_1} = \int_{1 - k}^k {xf\left\{ {x\left( {1 - x} \right)} \right\}dx} \cr & {\text{Put }}x = 1 - z \cr & {I_1} = \int_k^{1 - k} {\left( {1 - z} \right)f\left\{ {\left( {1 - z} \right)z} \right\}d\left( {1 - z} \right)} \cr & \,\,\,\,\, = \int_{1 - k}^k {f\left\{ {z\left( {1 - z} \right)} \right\}dz} - \int_{1 - k}^k {zf\left\{ {z\left( {1 - z} \right)} \right\}dz} \cr & \,\,\,\,\, = {I_2} - {I_1} \cr & \therefore 2{I_1} = {I_2} \cr} $$

$$\eqalign{ & I = \int_{ - 1}^1 {x\,dx} - \int_{ - 1}^1 {\left[ {2x} \right]dx} \cr & \,\,\,\,\, = \left[ {\frac{{{x^2}}}{2}} \right]_{ - 1}^1 - \left\{ {\int_{ - 1}^{ - \frac{1}{2}} {\left[ {2x} \right]dx} + \int_{ - \frac{1}{2}}^0 {\left[ {2x} \right]dx} + \int_0^{\frac{1}{2}} {\left[ {2x} \right]dx} + \int_{\frac{1}{2}}^1 {\left[ {2x} \right]dx} } \right\} \cr & \,\,\,\,\, = 0 - \left\{ {\int_{ - 1}^{ - \frac{1}{2}} { - 2\,dx} + \int_{ - \frac{1}{2}}^0 { - 1\,dx} + \int_0^{\frac{1}{2}} {0\,dx} + \int_{\frac{1}{2}}^1 {1\,dx} } \right\} \cr & \,\,\,\,\, = 2\left[ x \right]_{ - 1}^{ - \frac{1}{2}} - \left[ x \right]_{ - \frac{1}{2}}^0 - \left[ x \right]_{ - \frac{1}{2}}^1 \cr & \,\,\,\,\, = 2\left( { - \frac{1}{2} + 1} \right) + \left( {0 + \frac{1}{2}} \right) - \left( {1 - \frac{1}{2}} \right) \cr & \,\,\,\,\, = 1 \cr} $$

The given curves are
$$\eqalign{ & y = \sqrt {\frac{{1 + \sin \,x}}{{\cos \,x}}} = \sqrt {\frac{{1 + \tan \frac{x}{2}}}{{1 - \tan \frac{x}{2}}}} .....(1) \cr & {\text{and }}y = \sqrt {\frac{{1 - \sin \,x}}{{\cos \,x}}} = \sqrt {\frac{{1 - \tan \frac{x}{2}}}{{1 + \tan \frac{x}{2}}}} .....(2) \cr} $$
$$\therefore $$ The area bounded by the above curves, by the lines $$x=0$$   and $$x = \frac{\pi }{4}$$   is given by
$$\eqalign{ & A = \int\limits_0^{\frac{\pi }{4}} {\left( {\sqrt {\frac{{1 + \tan \frac{x}{2}}}{{1 - \tan \frac{x}{2}}}} - \sqrt {\frac{{1 - \tan \frac{x}{2}}}{{1 + \tan \frac{x}{2}}}} } \right)} dx \cr & = \int\limits_0^{\frac{\pi }{4}} {\frac{{2\,\tan \,\frac{x}{2}}}{{\sqrt {1 - {{\tan }^2}\frac{x}{2}} }}dx} \cr & {\text{Let}}\,\tan \frac{x}{2} = t \Rightarrow \frac{1}{2}{\sec ^2}\frac{x}{2}dx = dt \Rightarrow dx = \frac{2}{{1 + {t^2}}}dt \cr} $$
Also when $$x \to 0,\,t \to 0$$    and when $$x \to \frac{\pi }{4},\,t \to \tan \frac{\pi }{8}$$
$$\eqalign{ & \therefore A = \int\limits_0^{\tan \frac{\pi }{8}} {\frac{{4t}}{{\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt} \cr & = \int\limits_0^{\sqrt 2 - 1} {\frac{{4t}}{{\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt} \cr} $$

$${\text{Area}} = \sqrt 2 \times 2\sqrt 2 $$

$$\frac{{dy}}{{dx}} = \sqrt {\sin \,x} \,\,\,\,\,\,\,\,\,\,\therefore {\left. {\frac{{dy}}{{dx}}} \right)_{x = \frac{\pi }{2}}} = 1$$

Required area $$=$$ area of the shaded region
$$ = 4$$  (area of the shaded region in first quadrant
$$\eqalign{ & = 4\int_0^{\frac{1}{{\sqrt 2 }}} {\left( {{y_1} - {y_2}} \right)} dx \cr & = 4\int_0^{\frac{1}{{\sqrt 2 }}} {\left( {\sqrt {1 - {x^2}} - x} \right)dx} \cr & = 4\left[ {\frac{1}{2} \times \sqrt {1 - {x^2}} + \frac{1}{2}{{\sin }^{ - 1}}x - \frac{{{x^2}}}{2}} \right]_0^{\frac{1}{{\sqrt 2 }}} \cr & = \frac{\pi }{2}{\text{ sq}}{\text{. unit}} \cr} $$

$$\eqalign{ & y = \int_x^{{x^2}} {\sqrt {5 - {t^2}} dt} - \int_0^x {\sqrt {5 - {t^2}} dt} \cr & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x, \cr & \frac{{dy}}{{dx}} = \sqrt {5 - {{\left( {{x^2}} \right)}^2}} .2x - \sqrt {5 - {x^2}} \cr & {\text{Put }}x = \sqrt 2 \cr} $$

Clearly, $$f\left( x \right) + f\left( { - x} \right)$$    is an even function while $$g\left( x \right) - g\left( { - x} \right)$$    is an odd function.
$$\therefore $$ the integrand is an odd function. So, $$I=0.$$