Application of Integration MCQ Questions & Answers in Calculus | Maths

Use the property $$\left| {\int_a^b {f\left( x \right)dx} } \right| \leqslant \int_a^b {\left| {f\left( x \right)} \right|dx.} $$

$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{x}{{8 + {x^3}}} \cr & {\text{Then }}f'\left( x \right) = \frac{{1.\left( {8 + {x^3}} \right) - x\left( {3{x^2}} \right)}}{{{{\left( {8 + {x^3}} \right)}^2}}} = \frac{{8 - 2{x^3}}}{{{{\left( {8 + {x^3}} \right)}^2}}} > 0{\text{ for }}x\, \in \left[ {0,\,1} \right] \cr & \therefore f\left( x \right)\,{\text{is m}}{\text{.i}}{\text{. in }}\left[ {0,\,1} \right].{\text{ So, }}f\left( 0 \right) \leqslant f\left( x \right) \leqslant f\left( 1 \right) \cr & \therefore 0 \leqslant f\left( x \right) \leqslant \frac{1}{{8 + 1}} \cr & \therefore \int_0^1 {0\,dx < } \int_0^1 {f\left( x \right)} dx < \frac{1}{9}\int_0^1 {1\,dx} \cr & \therefore 0 < \int_0^1 {f\left( x \right)} dx < \frac{1}{9} \cr} $$

$$\eqalign{ & I = \int_0^\infty {\left[ {\frac{2}{{{e^x}}}} \right]dx} \cr & {\text{Substitute }}{e^x} = t \Rightarrow x = \log \,t \cr & dx = \frac{1}{t}dt \cr & {\text{Then}} \cr & I = \int_1^\infty {\left[ {\frac{2}{t}} \right]\frac{{dt}}{t}} \cr & = \int_1^2 {\left[ { - \frac{2}{t}} \right]\frac{{dt}}{t}} + \int_2^\infty {\left[ {\frac{2}{t}} \right]\frac{{dt}}{t}} \cr & = \int_1^2 {\frac{{dt}}{t}} + \int_2^\infty {0.} \frac{{dt}}{t} \cr & = \left[ {\log \,t} \right]_1^2 \cr & = \log \,2 - \log \,1 \cr & = \log \,2 - 0 \cr & = \log \,2 \cr & = {\log _e}2 \cr & \therefore \,I = {\log _e}2 \cr} $$

$$y = a{x^2}{\text{ and }}x = a{y^2}$$
Points of intersection are $$O\,\left( {0,\,0} \right)$$   and $$A\left( {\frac{1}{a},\,\frac{1}{a}} \right)$$
$$\eqalign{ & \therefore {\text{Area}} = \int\limits_0^{\frac{1}{a}} {\left( {\sqrt {\frac{x}{a}} - a{x^2}} \right)dx} \cr & = \frac{2}{{3{a^2}}} - \frac{1}{{3{a^2}}} \cr & = \frac{1}{{3{a^2}}} \cr & = 1 \cr & \Rightarrow a = \pm \frac{1}{{\sqrt 3 }} \cr} $$

We have $$\int\limits_0^2 {f\left( x \right)dx} = \frac{3}{4}\,;$$
$$\eqalign{ & {\text{Now, }}\int\limits_0^2 {x\,f'\left( x \right)dx} \cr & = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)dx} \cr & = \left[ {x\,f\left( x \right)} \right]_0^2 - \frac{3}{4} \cr & = 2f\left( 2 \right) - \frac{3}{4} \cr & = 0 - \frac{3}{4}\,\,\,\,\,\,\,\left( {\because \,f\left( 2 \right) = 0} \right) \cr & = - \frac{3}{4} \cr} $$

Given equations of curves are $$y = {e^x}{\text{ and }}y = {e^{ - x}}$$
$$ \Rightarrow {e^x} = \frac{1}{{{e^x}}}\,\, \Rightarrow {e^{2x}} = {e^0}\,\, \Rightarrow x = 0$$
Also, equation of straight line gives $$x = 1$$
$$\therefore $$  Required area
$$\eqalign{ & = \int\limits_0^1 {\left( {{e^x} - {e^{ - x}}} \right)dx} \cr & = \left[ {{e^x} + {e^{ - x}}} \right]_0^1 \cr & = e + {e^{ - 1}} - {e^0} + {e^{ - 0}} \cr & = \left( {e + \frac{1}{e} - 2} \right){\text{ sq}}{\text{. unit}} \cr} $$

Area $$=$$ area of the semicircle of radius 2

$$y = {\log _e}\left( {x + e} \right),\,x = {\log _e}\left( {\frac{1}{y}} \right),{\text{ or }}y = {e^{ - x}}$$
For $$y = {\log _e}\left( {x + e} \right),$$    shift the graph of $$y = {\log _e}x,\,\,e$$    units to the left hand side.
Required area
$$\eqalign{ & = \int\limits_{1 - e}^0 {{{\log }_e}\left( {x + e} \right)} dx + \int\limits_0^\infty {{e^{ - x}}dx} \cr & = \left| {x\,{{\log }_e}\left( {x + e} \right)} \right|_{1 - e}^0 - \int\limits_{1 - e}^0 {\frac{x}{{x + e}}} dx - \left| {{e^{ - x}}} \right|_0^\infty \cr & = \int\limits_0^{1 - e} {\left( {1 - \frac{e}{{x + e}}} \right)dx - {e^{ - \infty }}} + {e^0} \cr & = \left| {x - e\,\log \left( {x + e} \right)} \right|_0^{1 - e} - 0 + 1 \cr & = 1 - e + e\,\log \,e + 1 \cr & = 2{\text{ sq}}{\text{. units}} \cr} $$

$$\eqalign{ & {\text{Put }}a - x = z \cr & {\text{Then }}I = \int_{\frac{{3a}}{4}}^{\frac{a}{4}} {\frac{{\sqrt {a - z} }}{{\sqrt z + \sqrt {a - z} }}\left( { - dz} \right)} = \int_{\frac{a}{4}}^{\frac{{3a}}{4}} {\frac{{\sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}dx} \cr & \therefore I + I = \int_{\frac{a}{4}}^{\frac{{3a}}{4}} {\frac{{\sqrt x }}{{\sqrt {a - x} + \sqrt x }}dx} + \int_{\frac{a}{4}}^{\frac{{3a}}{4}} {\frac{{\sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}dx} \cr & = \int_{\frac{a}{4}}^{\frac{{3a}}{4}} {dx} \cr & = \frac{{3a}}{4} - \frac{a}{4} \cr & = \frac{a}{2} \cr} $$

$$\eqalign{ & {\text{Area}} = 2\int_0^3 {x\,dy} \cr & = 2\int_0^3 {\sqrt {ky} } \,dy \cr & = \left[ {2\sqrt k .\frac{{{y^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^3 \cr & = \frac{{4\sqrt k }}{3}.3\sqrt 3 \cr & \therefore 12 = 4\sqrt k .\sqrt 3 \,\,\,\,\, \Rightarrow k = 3 \cr} $$