Definite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & f\left( x \right) = a{x^2} + bx + c \cr & f\left( 0 \right) = 2 \Rightarrow c = 2 \cr & f'\left( x \right) = 2ax + b \cr & \therefore \,f'\left( 0 \right) - 3 \Rightarrow b = - 3 \cr & f''\left( x \right) = 2a \cr & \therefore \,f''\left( 0 \right) = 4 \Rightarrow a = 2 \cr & f\left( x \right) = 2{x^2} - 3x + 2 \cr & \int_{ - 1}^1 {f\left( x \right)dx} = \int_{ - 1}^1 {\left( {2{x^2} - 3x + 2} \right)dx} \cr & = \left[ {\frac{{2{x^3}}}{3} - \frac{{3{x^2}}}{2} + 2x} \right]_{ - 1}^1 \cr & = \frac{2}{3} - \frac{3}{2} + 2 + \frac{2}{3} + \frac{3}{2} + 2 \cr & = \frac{{16}}{3} \cr} $$

$$\eqalign{ & f\left( {p,\,q} \right) = \int_0^{\frac{\pi }{2}} {{{\cos }^p}x\,\cos \,qx\,dx} \cr & \Rightarrow f\left( {p,\,q} \right) = \left[ {{{\cos }^p}x.\frac{{\sin \,qx}}{q}} \right]_0^{\frac{\pi }{2}} + \int_0^{\frac{\pi }{2}} {\frac{p}{q}{{\cos }^{p - 1}}x\,\sin \,x\,\sin \,qx\,dx} \cr & \Rightarrow f\left( {p,\,q} \right) = 0 + \frac{p}{q}\int_0^{\frac{\pi }{2}} {{{\cos }^{p - 1}}} x\left[ {\cos \left( {q - 1} \right)x - \cos \,qx\,\cos \,x} \right]dx \cr} $$
\[\left[ \begin{gathered} \because \,\cos \left( {q - 1} \right)x = \cos \,qx\,\cos \,x + \sin \,qx\,\sin \,x \hfill \\ \therefore \,\cos \left( {q - 1} \right)x - \cos \,qx\,\cos \,x = \sin \,qx\,\sin \,x \hfill \\ \end{gathered} \right]\]
$$\eqalign{ & \Rightarrow f\left( {p,\,q} \right) = \frac{p}{q}f\left( {p - 1,\,q - 1} \right) - \frac{p}{q}f\left( {p,\,q} \right) \cr & \Rightarrow \left( {1 + \frac{p}{q}} \right)f\left( {p,\,q} \right) = \frac{p}{q}f\left( {p - 1,\,q - 1} \right) \cr & \Rightarrow f\left( {p,\,q} \right) = \frac{p}{{p + q}}f\left( {p - 1,\,q - 1} \right) \cr} $$

$$\eqalign{ & {\text{Let }}{I_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx} \cr & {\text{Consider,}} \cr & {I_n} + {I_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}x\,dx} + \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x\,dx} \cr & = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}x\left( {{{\tan }^2}x + 1} \right)dx} \cr & = \int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}x\,{{\tan }^{n - 2}}x\,dx} \cr & {\text{Put }}\tan \,x = t \cr & {\sec ^2}x\,dx = dt \cr & {\text{when }}x = 0{\text{ then }}t = 0{\text{ and when }}x = \frac{\pi }{4},\,t = 1 \cr & \therefore \,{I_n} + {I_{n - 2}} = \int\limits_0^1 {{t^{n - 2}}dt} \cr & = \left. {\frac{{{t^{n - 2 + 1}}}}{{n - 2 + 1}}} \right|_0^1 \cr & = \left. {\frac{{{t^{n - 1}}}}{{n - 1}}} \right|_0^1 \cr & = \frac{1}{{n - 1}}\left[ {1 - 0} \right] \cr & = \frac{1}{{n - 1}} \cr} $$

$$\eqalign{ & {\text{Let }}I = \int\limits_0^1 {\frac{{dx}}{{\left( {{x^2} + 16} \right)\left( {{x^2} + 25} \right)}}} \cr & = \frac{1}{9}\int\limits_0^1 {\left( {\frac{1}{{{x^2} + 16}} - \frac{1}{{{x^2} + 25}}} \right)} dx \cr & = \frac{1}{9}\left( {\frac{1}{4}{{\tan }^{ - 1}}\frac{x}{4} - \frac{1}{5}{{\tan }^{ - 1}}\frac{x}{5}} \right)_0^1 \cr & = \frac{1}{9}\left[ {\frac{1}{4}{{\tan }^{ - 1}}\frac{1}{4} - \frac{1}{5}{{\tan }^{ - 1}}\frac{1}{5}} \right] \cr} $$

$$\eqalign{ & {\text{Let }}I = \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\cot \,x} \right)dx} \cr & = \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\cos \,x} \right)dx} - \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} \cr & = \int\limits_0^{\frac{\pi }{2}} {\sin \,\left[ {2\left( {\frac{\pi }{2} + x} \right)} \right]\ell n\,\cos \left( {\frac{\pi }{2} + x} \right)dx} - \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} \cr & = \int\limits_0^{\frac{\pi }{2}} {\sin \left( {\pi + 2x} \right)\ell n\left( {\sin \,x} \right)dx} - \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} \cr & = \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} - \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} \cr & = 0 \cr} $$

$$\eqalign{ & {\text{Let }}I = \int\limits_0^4 {f\left( x \right)\phi \left( x \right)dx} \cr & \Rightarrow I = \int\limits_0^4 {f\left( {4 - x} \right)\phi \left( {4 - x} \right)dx} \cr & \Rightarrow I = \int\limits_0^4 {f\left( x \right).\left( {3 - \phi \left( x \right)} \right)dx} \cr & \left[ {\because \,f\left( x \right) = f\left( {4 - x} \right){\text{ and }}\phi \left( x \right) + \phi \left( {4 - x} \right) = 3} \right] \cr & \Rightarrow I = 3\int\limits_0^4 {f\left( x \right)dx - I} \cr & \Rightarrow 2I = 3.2 \cr & \therefore I = 3 \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{{\frac{r}{n}}}{{\sqrt {1 + {{\left( {\frac{r}{n}} \right)}^2}} }}} \cr & = \int_\alpha ^\beta {\frac{x}{{\sqrt {1 + {x^2}} }}dx} \cr & = \int_0^2 {\frac{{x\,dx}}{{\sqrt {1 + {x^2}} }}} \cr & = \frac{1}{2}\int_0^2 {\frac{{d\left( {1 + {x^2}} \right)}}{{\sqrt {1 + {x^2}} }}} \cr & = \left[ {\frac{1}{2}.2\sqrt {1 + {x^2}} } \right]_0^2 \cr & = \sqrt 5 - 1 \cr} $$

$$\eqalign{ & {\text{Use }}\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} \cr & \int_0^\pi {\frac{{\sin \,2mx}}{{\sin \,x}}dx} \cr & = \int_0^\pi {\frac{{\sin \left( {2m\pi - 2mx} \right)}}{{\sin \left( {\pi - x} \right)}}dx} \cr & = \int_0^\pi {\frac{{ - \sin \,2mx}}{{\sin \,x}}dx} \cr & = - I \cr & \Rightarrow 2I = 0 \cr & \Rightarrow I = 0 \cr} $$

$$\eqalign{ & I = \frac{1}{2}\int_{\sqrt {\ell n2} }^{\sqrt {\ell n3} } {\frac{{2x\,\sin \,{x^2}}}{{\sin \,{x^2} + \sin \left( {\ell n6 - {x^2}} \right)}}dx} \cr & {\text{Let }}\,{x^2} = t\,\,\, \Rightarrow 2x\,dx = dt \cr & {\text{Also when}}\,x = \sqrt {\ell n2} ,\,\,t = \ell n2 \cr & {\text{when }}x = \sqrt {\ell n3} ,\,\,t = \ell n3 \cr & \therefore I = \frac{1}{2}\int_{\ell n2}^{\ell n3} {\frac{{\sin \,t\,dt}}{{\sin \,t + \sin \left( {\ell n6 - t} \right)\,}}.....(1)} \cr & {\text{Using }}\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( {a + b - x} \right)dx} \cr & {\text{We get, }}I = \frac{1}{2}\int_{\ell n2}^{\ell n3} {\frac{{\sin \,\left( {\ell n6 - t} \right)}}{{\sin \,t + \sin \left( {\ell n6 - t} \right)\,}}dt.....(2)} \cr} $$
Adding values of $$I$$ in equation (1) and (2)
$$\eqalign{ & 2I = \frac{1}{2}\int_{\ell n2}^{\ell n3} {1\,dt} = \frac{1}{2}\left( {\ell n3 - \ell n2} \right) = \frac{1}{2}\ell n\frac{3}{2} \cr & \Rightarrow I = \frac{1}{4}\ell n\frac{3}{2} \cr} $$

$$\eqalign{ & I = \int\limits_{ - 2}^0 {\left[ {{x^3} + 3{x^2} + 3x + 3 + \left( {x + 1} \right)\,\cos \left( {x + 1} \right)} \right]dx} \cr & = \left[ {\frac{{{x^4}}}{4} + {x^3} + \frac{{3{x^2}}}{2} + 3x + \left( {x + 1} \right)\sin \left( {x + 1} \right) + \cos \left( {x + 1} \right)} \right] \cr & = \left( {\sin \,1 + \cos \,1} \right) - \left( {4 - 8 + 6 - 6 + \sin \,1 + \cos \,1} \right) \cr & = 4 \cr} $$