Definite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{k}{\left\{ {\frac{1}{n}.\frac{2}{n}.\frac{3}{n}......\frac{n}{n}} \right\}^{\frac{1}{n}}} \cr & = \frac{1}{k}{e^{\mathop {\lim }\limits_{n \to \infty } \,\frac{1}{n}\,\sum\limits_{r = 1}^n {\,\log \frac{r}{n}} }} \cr & = \frac{1}{k}{e^{\int_0^1 {\log \,x\,dx} }} \cr & = \frac{1}{k}{e^{\left[ {x\,\log \,x} \right]_0^1 - \int_0^1 {x.\frac{1}{x}dx} }} \cr & = \frac{1}{k}{e^{ - 1}} \cr & \left( {\because \mathop {\lim }\limits_{x \to 0} \,x\log \,x = \mathop {\lim }\limits_{x \to 0} \frac{{\log \,x}}{{\frac{1}{x}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{x}}}{{ - \frac{1}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to 0} \left( { - x} \right) = 0} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \sin \,x.\int_{ - 1}^1 {\frac{{dt}}{{1 + {t^2}}}} \cr & = \sin \,x.\left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1 \cr & = \sin \,x\left\{ {\frac{\pi }{4} - \left( { - \frac{\pi }{4}} \right)} \right\} \cr & = \frac{\pi }{2}\sin \,x \cr & \therefore f'\left( x \right) = \frac{\pi }{2}\cos \,x \cr & \therefore f'\left( {\frac{\pi }{3}} \right) = \frac{\pi }{2}.\cos \,\frac{\pi }{3} = \frac{\pi }{4} \cr} $$

Integrating by parts.
$$\eqalign{ & \int {f\left( x \right)g''\left( x \right)dx - \int {f''\left( x \right)g\left( x \right)} } dx \cr & = f\left( x \right)g'\left( x \right) - \int {f'\left( x \right)g'\left( x \right)dx - f'\left( x \right)g\left( x \right) + } \int {f'\left( x \right)g'\left( x \right)dx} \cr & = f\left( x \right)g'\left( x \right) - f'\left( x \right)g\left( x \right) \cr} $$
Hence, $$\int_0^1 {\left[ {f\left( x \right)g''\left( x \right) - f''\left( x \right)g\left( x \right)} \right]} dx$$
$$\eqalign{ & = f\left( 1 \right)g'\left( 1 \right) - f'\left( 1 \right)g\left( 1 \right) - f\left( 0 \right)g'\left( 0 \right) + f'\left( 0 \right)g\left( 0 \right) \cr & = f\left( 1 \right)g'\left( 1 \right) - f'\left( 1 \right)g\left( 1 \right) \cr} $$

$$\eqalign{ & \int\limits_0^1 {{e^t}\log \left( {1 + t} \right)} dt \cr & = \left[ {{e^t}.\frac{1}{{1 + t}}} \right]_0^1 - \int\limits_0^1 {\frac{{{e^t}}}{{t + 1}}dt} \cr & = \frac{e}{2} - 1 - A \cr} $$

We have $$l\left( {m,\,n} \right) = \int\limits_0^1 {{t^m}{{\left( {1 + t} \right)}^n}} dt$$
Integrating by parts considering $${\left( {1 + t} \right)^n}$$   as first function, we get
$$\eqalign{ & l\left( {m,\,n} \right) = \left[ {\frac{{{t^{m + 1}}}}{{m + 1}}{{\left( {1 + t} \right)}^n}} \right]_0^1 - \frac{n}{{m + 1}}\int\limits_0^1 {{t^{m + 1}}{{\left( {1 + t} \right)}^{n - 1}}dt} \cr & l\left( {m,\,n} \right) = \frac{{{2^n}}}{{m + 1}} - \frac{n}{{m + 1}}l\left( {m + 1,\,n - 1} \right) \cr} $$

$$\eqalign{ & {\text{Let }}I = \int_1^2 {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)} dx \cr & = \int_1^2 {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)} dx\,\,\,\,\,\,\,\left[ {{\text{where}}\,f\left( x \right) = \frac{1}{x}} \right] \cr & = {e^x}\left. {f\left( x \right)} \right|_1^2 \cr & \therefore \,I = \left. {\frac{{{e^x}}}{x}} \right|_1^2 = \frac{{{e^2}}}{2} - e = e\left( {\frac{e}{2} - 1} \right) \cr} $$

$$I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{x^2}\cos \,x}}{{1 + {e^x}}}\,dx} .....(1)$$
Applying $$\int_a^b {f\left( x \right)dx = \int_a^b {f\left( {a + b - x} \right)dx,} } $$       we get
$$\eqalign{ & I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{e^x}{x^2}\cos \,x}}{{1 + {e^x}}}\,dx} .....(2) \cr & {\text{Adding }}(1){\text{ and }}(2),{\text{ we get}} \cr & 2I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{x^2}\,\cos \,x\,dx} = 2\int\limits_0^{\frac{\pi }{2}} {{x^2}\cos \,x\,dx} \cr & I = \left[ {{x^2}\,\sin \,x + 2x\,\cos \,x - 2\,\sin \,x} \right]_0^{\frac{\pi }{2}}\, = \frac{{{\pi ^2}}}{4} - 2 \cr} $$

$$\eqalign{ & {\text{Given, }}f\left( x \right) = a + bx + c{x^2} \cr & \therefore \,\int_0^1 {f\left( x \right)dx} \cr & = \int_0^1 {\left( {a + bx + c{x^2}} \right)dx} \cr & = \left[ {ax + \frac{{b{x^2}}}{2} + \frac{{c{x^3}}}{3}} \right]_0^1 \cr & = a + \frac{b}{2} + \frac{c}{3}......\left( {\text{i}} \right) \cr & {\text{Here, }}f\left( 0 \right) = a,\,f\left( {\frac{1}{2}} \right) = a + \frac{b}{2} + \frac{c}{4}{\text{ and }}f\left( 1 \right) = a + b + c \cr & {\text{Now, }}\frac{{f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f\left( 1 \right)}}{6} \cr & = \frac{{a + 4\left( {a + \frac{b}{2} + \frac{c}{4}} \right) + a + b + c}}{6} \cr & = \frac{{a + 4\left( {\frac{{4a + 2b + c}}{4}} \right) + a + b + c}}{6} \cr & = \frac{{a + 4a + 2b + c + a + b + c}}{6} \cr & = \frac{{6a + 3b + 2c}}{6} \cr & = a + \frac{b}{2} + \frac{c}{3} \cr & \therefore {\text{ From equations }}\left( {\text{i}} \right)\,{\text{and }}\left( {{\text{ii}}} \right),{\text{ we get}} \cr & \int_0^1 {f\left( x \right)dx} = \frac{{f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f\left( 1 \right)}}{6} \cr} $$

$$\eqalign{ & {\text{Let }}{I_1} = \int_0^{{{\sin }^2}x} {{{\sin }^{ - 1}}\sqrt t } \,dt \cr & {\text{Put }}t = {\sin ^2}u \Rightarrow dt = 2\,\sin \,u\,\cos \,u\,du \Rightarrow dt = \sin \,2u\,du \cr & \therefore \,{I_1} = \int_0^x {u\,\sin \,2u\,du} \cr & {\text{Let }}{I_2} = \int_0^{{{\cos }^2}x} {{{\cos }^{ - 1}}\sqrt t } \,dt \cr & {\text{Put }}t = {\cos ^2}v \Rightarrow dt = - 2\,\cos \,v\,\sin \,v\,dv \Rightarrow dt = - \sin \,2v\,dv \cr & \therefore \,{I_2} = \int_{\frac{\pi }{2}}^x {v\left( { - \sin \,2v} \right)\,dv} \cr & = - \int_{\frac{\pi }{2}}^x {v\,\sin \,2v\,dv} \cr & = - \int_{\frac{\pi }{2}}^x {u\,\sin \,2u\,du\,\,\,\,\,\,\,\,\left[ {{\text{change of variable}}} \right]} \cr & \therefore \,I = {I_1} + {I_2} \cr & = \int_0^x {u\,\sin \,2u\,du} - \int_{\frac{\pi }{2}}^x {u\,\sin \,2u\,du} \cr & = \int_0^{\frac{\pi }{2}} {u\,\sin \,2u\,du} + \int_{\frac{\pi }{2}}^x {u\,\sin \,2u\,du} - \int_{\frac{\pi }{2}}^x {u\,\sin \,2u\,du} \cr & = \int_0^{\frac{\pi }{2}} {u\,\sin \,2u\,du} \cr & = \frac{\pi }{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Integrate by parts}}} \right] \cr} $$

$$\eqalign{ & {\text{Limit}} = \int_0^1 {\sin \frac{{\pi x}}{2}dx} \cr & = \left[ {\frac{{ - \cos \frac{{\pi x}}{2}}}{{\frac{\pi }{2}}}} \right]_0^1 \cr & = \frac{2}{\pi }\left\{ { - \cos \frac{\pi }{2} + \cos \,0} \right\} \cr & = \frac{2}{\pi } \cr} $$