Definite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} .....(1) \cr & = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,\left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\cot \,\left( {\frac{\pi }{2} - x} \right)} + \sqrt {\tan \,\left( {\frac{\pi }{2} - x} \right)} }}dx} \cr & I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {tan\,x} }}{{\sqrt {\tan \,x} + \sqrt {cot\,x} }} = dx.....(2)} \cr & {\text{Adding (1) and (2) we get}} \cr & 2I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }} = dx = \int_0^{\frac{\pi }{2}} {1.dx} } \cr & = \left( x \right)_0^{\frac{\pi }{2}} = \frac{\pi }{2} \cr & \therefore I = \frac{\pi }{4} \cr} $$

$$\eqalign{ & g\left( x \right) = \int_0^x {{{\left( {\frac{{1 + \cos \,2t}}{2}} \right)}^2}dt} \cr & = \frac{1}{4}\int_0^x {\left( {1 + 2\cos \,2t + {{\cos }^2}2t} \right)} dt \cr & = \frac{1}{4}\left[ {t + \sin \,2t} \right]_0^x + \frac{1}{8}\int_0^x {\left( {1 + \cos \,4t} \right)dt} \cr & = \frac{1}{4}\left( {x + \sin \,2x} \right) + \frac{1}{8}\left[ {t + \frac{{\sin \,4t}}{4}} \right]_0^x \cr & = \frac{1}{4}\left( {x + \sin \,2x} \right) + \frac{1}{8}\left( {x + \frac{{\sin \,4x}}{4}} \right) \cr & = \frac{3}{8}x + \frac{1}{4}\sin \,2x + \frac{1}{{32}}\sin \,4x \cr & \therefore \,g\left( {x + \pi } \right) = \frac{3}{8}\left( {x + \pi } \right) + \frac{1}{4}\sin \,2x + \frac{1}{{32}}\sin \,4x \cr & = \frac{3}{8}\pi + g\left( x \right) \cr & = g\left( \pi \right) + g\left( x \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{{e^x}}}{{1 + {e^x}}} \cr & \Rightarrow f\left( { - x} \right) = \frac{{{e^{ - x}}}}{{1 + {e^{ - x}}}} = \frac{1}{{{e^x} + 1}} \cr & \therefore f\left( x \right) + f\left( { - x} \right) = 1\,\forall \,x \cr & {\text{Now}}\,\,{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {x\,g\left\{ {x\left( {1 - x} \right)} \right\}dx} \cr & = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)\,g\left\{ {x\left( {1 - x} \right)} \right\}dx} \cr & \left[ {{\text{Using }}\int\limits_a^b {f\left( x \right)} dx\,a = \int\limits_a^b {f\left( {a + b - x} \right)dx} } \right] \cr & = {I_2} - {I_1}\,\,\, \Rightarrow 2{I_1} = {I_2} \cr & \therefore \frac{{{I_2}}}{{{I_1}}} = 2 \cr} $$

$$\eqalign{ & I = \int\limits_0^\pi {x\,f\left( {\sin \,x} \right)dx} = \int\limits_0^\pi {\left( {\pi - x} \right)\,f\left( {\sin \,x} \right)dx} \cr & = \pi \int\limits_0^\pi {\,f\left( {\sin \,x} \right)dx} - I\,\, \Rightarrow 2I = \pi \int\limits_0^\pi {\,f\left( {\sin \,x} \right)dx} \cr & I = \frac{\pi }{2}\int\limits_0^\pi {\,f\left( {\sin \,x} \right)dx} = \pi \int\limits_0^{\frac{\pi }{2}} {\,f\left( {\sin \,x} \right)dx} \cr & = \pi \int\limits_0^{\frac{\pi }{2}} {\,f\left( {\cos \,x} \right)dx} \cr} $$

$$\eqalign{ & {\text{Let}} \cr & I = \int_\pi ^{2\pi } {\left[ {2\,\sin \,x} \right]dx} \cr & \pi \leqslant x < \frac{{7\pi }}{6}\, \Rightarrow - 1 \leqslant 2\,\sin \,x < 0 \cr & \Rightarrow \left[ {2\,\sin \,x} \right] = - 1 \cr & \frac{{7\pi }}{6} \leqslant x < \frac{{11\pi }}{6}\, \Rightarrow - 2 \leqslant 2\,\sin \,x < - 1 \cr & \Rightarrow \left[ {2\,\sin \,x} \right] = - 1 \cr & \therefore \,I = \int\limits_\pi ^{\frac{{7\pi }}{6}} { - 1\,dx + } \int\limits_{\frac{{7\pi }}{6}}^{\frac{{11\pi }}{6}} { - 2\,dx + } \int\limits_{\frac{{11\pi }}{6}}^{2\pi } { - 1\,dx} \cr & = \left( { - \frac{{7\pi }}{6} + \pi } \right) + 2\left( { - \frac{{11\pi }}{6} + \frac{{7\pi }}{6}} \right) + \left( { - 2\pi + \frac{{11\pi }}{6}} \right) \cr & = - \frac{\pi }{6} - \frac{{8\pi }}{6} - \frac{\pi }{6} \cr & = - \frac{{10\pi }}{6} \cr & = \frac{{ - 5\pi }}{3} \cr} $$

$$\eqalign{ & {\text{Let }}I = \int\limits_0^1 {\sqrt {\frac{{1 - x}}{{1 + x}}} dx} \cr & = \int\limits_0^1 {\frac{{1 - x}}{{\sqrt {1 - {x^2}} }}dx} = \left. {{{\sin }^{ - 1}}x} \right|_0^1\left( { - \frac{1}{2}} \right)\int\limits_0^1 {\frac{{2x}}{{\sqrt {1 - {x^2}} }}dx} \cr & = \frac{\pi }{2} + \frac{1}{2}\left[ {\left. {2\sqrt {1 - {x^2}} } \right|_0^1} \right] \cr & = \frac{\pi }{2} + \left( {0 - 1} \right) \cr & = \frac{\pi }{2} - 1 \cr} $$

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

$$\eqalign{ & {\text{Given that }}\,g\left( x \right) = \int\limits_0^x {{{\cos }^4}t\,dt} \cr & \therefore g\left( {x + \pi } \right) = \int\limits_0^{x + \pi } {{{\cos }^4}t\,dt} \cr & = \int\limits_0^\pi {{{\cos }^4}t\,dt} + \int\limits_\pi ^{x + \pi } {{{\cos }^4}t\,dt} \cr & g\left( {x + \pi } \right) = g\left( \pi \right) + I,{\text{ where }}I = \int\limits_\pi ^{x + \pi } {{{\cos }^4}t\,dt} \cr & {\text{Put }}t = \pi + y,\,dt = dy \cr & {\text{Also as }}t \to \pi ,\,y \to 0 \cr & {\text{as }}t \to x + \pi ,\,y \to x \cr & \therefore I = \int\limits_0^x {{{\cos }^4}\left( {\pi + y} \right)\,dy} \cr & = \int\limits_0^x {{{\cos }^4}y\,dy} = \int\limits_0^x {{{\cos }^4}t\,dt} = g\left( x \right) \cr & \therefore g\left( {x + \pi } \right) = g\left( \pi \right) + g\left( x \right) \cr} $$

$$\eqalign{ & {\text{Given : }}{u_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^n}\theta \,d\theta } \cr & \Rightarrow {u_n} = \int\limits_0^{\frac{\pi }{4}} {{{\tan }^2}\theta \,{{\tan }^{n - 2}}\theta \,d\theta } \cr & \Rightarrow {u_n} = \int\limits_0^{\frac{\pi }{4}} {\left( {{{\sec }^2}\theta - 1} \right)\,{{\tan }^{n - 2}}\theta \,d\theta } \cr & \Rightarrow {u_n} = \int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}\theta \,{{\tan }^{n - 2}}\theta \,d\theta } - \int\limits_0^{\frac{\pi }{4}} {{{\tan }^{n - 2}}\theta \,d\theta } \cr & \Rightarrow {u_n} = \int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}\theta \,{{\tan }^{n - 2}}\theta \,d\theta } - {u_{n - 2}} \cr & \Rightarrow {u_n} + {u_{n - 2}} = \int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}\theta \,{{\tan }^{n - 2}}\theta \,d\theta } \cr & \Rightarrow {u_n} + {u_{n - 2}} = \left. {\frac{{{{\tan }^{n - 1}}\theta }}{{n - 1}}} \right|_0^{\frac{\pi }{4}} \cr & \Rightarrow {u_n} + {u_{n - 2}} = \frac{1}{{n - 1}} \cr} $$

$$\eqalign{ & {\text{Let}} \cr & I = \int_0^{\frac{\pi }{2}} {\frac{{dx}}{{1 + {{\tan }^3}x}}} \cr & = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}x}}{{{{\sin }^3}x + {{\cos }^3}x}}dx.....(1)} \cr & I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}\left( {\frac{\pi }{2} - x} \right)}}{{{{\sin }^3}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^3}\left( {\frac{\pi }{2} - x} \right)}}dx} \cr & = \int_0^{\frac{\pi }{2}} {\frac{{si{n^3}x}}{{co{s^3}x + {{\sin }^3}x}}dx.....(2)} \cr & {\text{Adding (1) and (2) we get}} \cr & {\text{2}}I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^3}x + {{\sin }^3}x}}{{{{\sin }^3}x + {{\cos }^3}x}}dx} \cr & = \int_0^{\frac{\pi }{2}} {1.dx} = \frac{\pi }{2} \cr & \therefore I = \frac{\pi }{4} \cr} $$