Definite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & f\left( x \right) = A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B \cr & \Rightarrow f'\left( x \right) = \frac{{A\pi }}{2}\cos \left( {\frac{{\pi x}}{2}} \right) \cr & \Rightarrow f'\left( {\frac{1}{2}} \right) = \frac{{A\pi }}{2}\cos \frac{\pi }{4} = \sqrt 2 \cr & \Rightarrow A = \frac{4}{\pi }\,\,{\text{and }}\int_0^1 {f\left( x \right)dx = \frac{{2A}}{\pi }} \cr & \Rightarrow \int_0^1 {\left[ {A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B} \right]dx = \frac{{2A}}{\pi }} \cr & \Rightarrow \left| { - \frac{{2A}}{\pi }\cos \left( {\frac{{\pi x}}{2}} \right) + Bx} \right|_0^1 = \frac{{2A}}{\pi } \cr & \Rightarrow B + \frac{{2A}}{\pi } = \frac{{2A}}{\pi }\,\,\,\,\, \Rightarrow B = 0 \cr} $$

$$\eqalign{ & I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 + \cos \,x}}} .....{\text{(i)}} \cr & I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 - \cos \,x}}} .....{\text{(ii)}} \cr & {\text{Using }}\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx \cr} $$
Adding (i) and (ii), we get
$$\eqalign{ & 2I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{2}{{{{\sin }^2}x}}} dx \cr & I = \int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {{\text{cose}}{{\text{c}}^2}x} \,dx \cr & I = - \left( {\cot \,x} \right)_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} = - \left[ {\cot \frac{{3\pi }}{4} - \cot \frac{\pi }{4}} \right] = 2 \cr} $$

\[\begin{array}{l} {\rm{If }}f\left( x \right) = \left\{ \begin{array}{l} {e^{\cos \,x}}\sin \,x,\,{\rm{for }}\left| x \right| \le 2\\ 2,\,\,\,\,\,\,\,\,\,\,{\rm{otherwise}} \end{array} \right\}\\ = \left\{ \begin{array}{l} {e^{\cos \,x}}\sin \,x,\,{\rm{for }} - 2 \le x \le 2\\ 2,\,\,\,\,\,\,\,\,\,\,{\rm{otherwise}} \end{array} \right\} \end{array}\]
$$\eqalign{ & \int\limits_{ - 2}^3 {f\left( x \right)} dx = \int\limits_{ - 2}^2 {f\left( x \right)} dx + \int\limits_2^3 {f\left( x \right)} dx \cr & = \int\limits_{ - 2}^2 {{e^{\cos \,x}}\sin \,x\,dx} + \int\limits_2^3 {2\,dx} = 0 + 2\left[ x \right]_2^3 \cr & \left[ {\because {e^{\cos \,x}}\sin \,x\,{\text{ is an odd function}}{\text{.}}} \right] \cr & = 2\left[ {3 - 2} \right] = 2 \cr & \therefore \int\limits_{ - 2}^3 {f\left( x \right)} dx = 2 \cr} $$

We have $$f\left( x \right) = \int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}} dt$$
Then $$f'\left( x \right) = {e^{ - {{\left( {{x^2} + 1} \right)}^2}}}.2x - {e^{ - {x^4}}}.2x$$
[Using Leibnitz theorem, ]
$$\eqalign{ & \frac{d}{{dx}}\int\limits_{\phi \left( x \right)}^{\psi \left( x \right)} {f\left( t \right)dt} \cr & = f\left[ {\psi \left( x \right)} \right].\psi '\left( x \right) - f\left[ {\phi \left( x \right)} \right].\phi '\left( x \right) \cr & = 2x\left[ {{e^{ - {{\left( {{x^2} + 1} \right)}^2}}} - {e^{ - {x^4}}}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {\left( {{x^2} + 1} \right)^2} > {x^4} \cr & \Rightarrow {e^{ + {{\left( {{x^2} + 1} \right)}^2}}} > {e^{{x^4}}} \Rightarrow {e^{ - {{\left( {{x^2} + 1} \right)}^2}}} < {e^{ - {x^4}}} \cr & \therefore {e^{ - {{\left( {{x^2} + 1} \right)}^2}}} - {e^{ - {x^4}}} < 0 \cr & \therefore f'\left( x \right) > 0,\,\forall \,x < 0 \cr} $$
$$\therefore f\left( x \right)$$   increases when $$x<0$$

$$\eqalign{ & {\text{Let }}I = \int\limits_0^1 {\frac{{dx}}{{{e^x} + e}}} = \int\limits_0^1 {\frac{{{e^x}dx}}{{{e^x}\left( {{e^x} + e} \right)}}} \cr & {\text{Put }}{e^x} = t \Rightarrow {e^x}dx = dt \cr & I = \int\limits_1^e {\frac{{dt}}{{t\left( {t + e} \right)}}} \cr & = \frac{1}{e}\int\limits_1^e {\left( {\frac{1}{t} - \frac{1}{{t + e}}} \right)} \cr & = \frac{1}{e}\int\limits_1^e {\frac{1}{t}dt} - \frac{1}{e}\int\limits_1^e {\frac{1}{{t + e}}dt} \cr & = \frac{1}{e}\left[ {\log \,t} \right]_1^e - \frac{1}{e}\left[ {\log \left( {t + e} \right)} \right]_1^e \cr & = \frac{1}{e}\left[ {\log \,t - \log \left( {t + e} \right)} \right]_1^e \cr & = \frac{1}{e}\left[ {\log \left( {\frac{t}{{t + e}}} \right)} \right]_1^e \cr & = \frac{1}{e}\left[ {\log \left( {\frac{e}{{2e}}} \right) - \log \left( {\frac{1}{{1 + e}}} \right)} \right] \cr & = \frac{1}{e}\log \left[ {\frac{{\frac{1}{2}}}{{\frac{1}{{\left( {1 + e} \right)}}}}} \right] \cr & = \frac{1}{e}\log \left( {\frac{{1 + e}}{2}} \right) \cr} $$

$$\eqalign{ & {\text{Let }}I = \int_0^1 {x{e^{{x^2}}}} dx \cr & {\text{Let }}{x^2} = t \cr & \Rightarrow 2x\,dx = dt \cr & \Rightarrow x\,dx = \frac{{dt}}{2} \cr & {\text{when }}x = 0,\,t = 0{\text{ then }}x = 1,\,t = 1 \cr & \Rightarrow I = \frac{1}{2}\int\limits_0^1 {{e^t}dt} \cr & \Rightarrow I = \frac{1}{2}\left[ {{e^t}} \right]_0^1 \cr & \Rightarrow I = \frac{1}{2}\left[ {{e^{{x^2}}}} \right]_0^1 \cr & \Rightarrow I = \frac{1}{2}\left[ {e - {e^0}} \right] \cr & \Rightarrow I = \frac{{e - 1}}{2} \cr} $$

$$\eqalign{ & I = \int\limits_0^\pi {{{\left| {\cos \,x} \right|}^3}dx} = 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^3}\,x\,dx} \cr & = \frac{2}{4}\int\limits_0^{\frac{\pi }{2}} {\left( {3\,\cos \,x + \cos \,3x} \right)dx} \cr & \left[ {\lambda \,\cos \,3\theta = 4\,{{\cos }^3}\theta - 3\,\cos \,\theta } \right] \cr & = \frac{1}{2}\left[ {3\,\sin \,x + \frac{{\sin \,3x}}{3}} \right]_0^{\frac{\pi }{2}} \cr & = \frac{1}{2}\left( {3 - \frac{1}{3}} \right) \cr & = \frac{4}{3} \cr} $$

$$\eqalign{ & {e^{ - x}}f\left( x \right) = 2 + \int_0^x {\sqrt {1 + {t^4}} } \,dt\,\forall \,x \in \left( { - 1,\,1} \right) \cr & {\text{At }}x = 0,\,\,f\left( 0 \right) = 2 \cr} $$
Now on differentiating, we get
$$\eqalign{ & - {e^{ - x}}f\left( x \right) + {e^{ - x}}f'\left( x \right) = 0\sqrt {1 + {x^4}} \cr & \Rightarrow - f\left( 0 \right) + f'\left( 0 \right) = 1 \cr & \Rightarrow f'\left( 0 \right) = 3 \cr & {\text{Now }}{f^{ - 1}}\left( {f\left( x \right)} \right) = x \cr & \Rightarrow \left[ {\left( {{f^{ - 1}}} \right)'\left( {f\left( x \right)} \right)} \right]f'\left( x \right) = 1 \cr & \Rightarrow \left( {{f^{ - 1}}} \right)'\left( {f\left( 0 \right)} \right)f'\left( 0 \right) = 1 \cr & \Rightarrow \left( {{f^{ - 1}}} \right)'\left( 2 \right) = \frac{1}{3} \cr} $$

$$\eqalign{ & {\text{Let }}I = \int\limits_0^\pi {\sqrt {1 + 4\,{{\sin }^2}\frac{x}{2} - 4\,\sin \frac{x}{2}} } \,dx \cr & = \int\limits_0^\pi {\left| {2\sin \,\frac{x}{2} - 1} \right|dx} \cr & = \int\limits_0^{\frac{\pi }{3}} {\left( {1 - 2\,\sin \,\frac{x}{2}} \right)dx} + \int\limits_{\frac{\pi }{3}}^\pi {\left( {2\,\sin \,\frac{x}{2} - 1} \right)} \,dx \cr & \left[ {\because \sin \frac{x}{2} = \frac{1}{2} \Rightarrow \frac{x}{2} = \frac{\pi }{6} \Rightarrow x = \frac{\pi }{3},\,\frac{x}{2} = \frac{{5\pi }}{6} \Rightarrow x = \frac{{5\pi }}{3}} \right] \cr & = \left[ {x + 4\,\cos \frac{x}{2}} \right]_0^{\frac{\pi }{3}} + \left[ { - 4\,\cos \frac{x}{2} - x} \right]_{\frac{\pi }{3}}^\pi \cr & = \frac{\pi }{3} + 4\frac{{\sqrt 3 }}{2} - 4 + \left( {0 - \pi + 4\frac{{\sqrt 3 }}{2} + \frac{\pi }{3}} \right) \cr & = 4\sqrt 3 - 4 - \frac{\pi }{3} \cr} $$

$$\eqalign{ & {\text{Let }}I = \int\limits_{ - \pi }^\pi {\frac{{{{\cos }^2}x}}{{1 + {a^x}}}dx} = \int\limits_{ - \pi }^\pi {\frac{{{{\cos }^2}\left( { - x} \right)}}{{1 + {a^{ - x}}}}dx.....(1)} \cr & \left[ {{\text{Using }}\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} } } \right] \cr & = \int\limits_{ - \pi }^\pi {\frac{{{{\cos }^2}x}}{{1 + {a^x}}}dx} .....(2) \cr} $$
Adding equations (1) and (2) we get
$$\eqalign{ & 2I = \int\limits_{ - \pi }^\pi {{{\cos }^2}x\left( {\frac{{1 + {a^x}}}{{1 + {a^x}}}} \right)dx = \int\limits_{ - \pi }^\pi {{{\cos }^2}x\,dx} } \cr & \Rightarrow 2\int\limits_0^\pi {{{\cos }^2}x\,dx} = 2 \times 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}x\,dx} = 4\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x\,dx} \cr & \Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x\,dx} = 2\int\limits_0^{\frac{\pi }{2}} {\left( {1 - {{\cos }^2}x} \right)\,dx} \cr & \Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {dx - } 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}x\,dx} \cr & \Rightarrow I + I = 2\left( {\frac{\pi }{2}} \right) = \pi \cr & \Rightarrow I = \frac{\pi }{2} \cr} $$