Definite Integration MCQ Questions & Answers in Calculus | Maths

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

In the range $$\frac{\pi }{2}$$ to $$\frac{{3\pi }}{2},$$  we have to find the value of $${\left[ {2\,\sin \,x} \right]}$$
\[\left[ {2\,\sin \,x} \right] = \left\{ \begin{array}{l} 2\,\,\,\,\,\,\,\,\,\,{\rm{if }}x = \frac{\pi }{2}\\ 1\,\,\,\,\,\,\,\,\,\,\,{\rm{if }}\frac{\pi }{2} < x \le \frac{{5\pi }}{6}\\ 0\,\,\,\,\,\,\,\,\,\,{\rm{if }}\frac{{5\pi }}{6} < x \le \pi \\ - 1\,\,\,\,\,\,\,{\rm{if }}\pi < x \le \frac{{7\pi }}{6}\\ - 2\,\,\,\,\,\,\,{\rm{if }}\frac{{7\pi }}{6} < x \le \frac{{3\pi }}{2} \end{array} \right.\]
$$\eqalign{ & {\text{Thus}} \cr & I = \int_{\frac{\pi }{2}}^{\frac{{5\pi }}{6}} {1.dx} + \int_{\frac{{5\pi }}{6}}^\pi {0.dx} + \int_\pi ^{\frac{{7\pi }}{6}} {\left( { - 1} \right).dx} + \int_{\frac{{7\pi }}{6}}^{\frac{{3\pi }}{2}} {\left( { - 2} \right).dx} \cr & {\text{or }}I = \left[ {\frac{{5\pi }}{6} - \frac{\pi }{2}} \right] + 0 - 1\left[ {\frac{{7\pi }}{6} - \pi } \right] - 2\left[ {\frac{{3\pi }}{2} - \frac{{7\pi }}{6}} \right] \cr & = \frac{{2\pi }}{6} - \frac{\pi }{6} - \frac{{4\pi }}{6} \cr & = \frac{{ - 3\pi }}{6} \cr & = \frac{{ - \pi }}{2} \cr} $$

$$\eqalign{ & {\text{Consider }}{I_1} = \int\limits_0^\pi {x\,f\left[ {{{\sin }^3}x + {{\cos }^2}x} \right]dx} \cr & \Rightarrow {I_1} = \int\limits_0^\pi {\left( {\pi - x} \right)f\left[ {{{\sin }^3}\left( {\pi - x} \right) + {{\cos }^2}\left( {\pi - x} \right)} \right]dx} \cr & \Rightarrow {I_1} = \int\limits_0^\pi {\left( {\pi - x} \right)f\left[ {{{\sin }^3}x + {{\cos }^2}x} \right]dx} \cr & \Rightarrow {I_1} = \int\limits_0^\pi {\pi f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx - } \int\limits_0^\pi {\pi f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx} \cr & \Rightarrow {I_1} = \int\limits_0^\pi {\pi f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx - } {I_1} \cr & \Rightarrow 2{I_1} = \int\limits_0^{\frac{\pi }{2}} {\pi \,f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx} \cr & \Rightarrow 2{I_1} = 2\pi \int\limits_0^{\frac{\pi }{2}} {\,f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx} \cr & \Rightarrow {I_1} = \pi \int\limits_0^{\frac{\pi }{2}} {\,f\left( {{{\sin }^3}x + {{\cos }^2}x} \right)dx} \cr & {I_1} = {I_2}\,\,\,\,\,\left( {{\text{By definition of }}{I_2}} \right) \cr} $$

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

$$\eqalign{ & \int_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx \cr & = \int_0^1 {\left( {1 + 1 - \frac{{{x^2}}}{{1!}} + \frac{{{x^4}}}{{2!}} - \frac{{{x^6}}}{{3!}} + .....\infty } \right)} dx \cr & = \left[ {2x - \frac{{{x^3}}}{{3.1!}} + \frac{{{x^5}}}{{5.2!}} - \frac{{{x^7}}}{{7.3!}} + .....\infty } \right]_0^1 \cr & = \left[ {2 - \frac{1}{{3.1!}} + \frac{1}{{5.2!}} - \frac{1}{{7.3!}} + .....\infty } \right] \cr} $$

$$\eqalign{ & \int_{ - \pi }^\pi {\frac{{2x\left( {1 + \sin \,x} \right)}}{{1 + {{\cos }^2}\,x}}dx} \cr & = \int_{ - \pi }^\pi {\frac{{2x\,dx}}{{1 + {{\cos }^2}\,x}} + 2\int_{ - \pi }^\pi {\frac{{x\,\sin \,x}}{{1 + {{\cos }^2}\,x}}dx} } \cr & = 0 + 4\int_0^\pi {\frac{{x\,\sin \,x\,dx}}{{1 + {{\cos }^2}\,x}}\,;\,\,\,\,\,\,\,\,\,\left[ {\because \int\limits_{ - a}^a {f\left( x \right)dx = 0} } \right]} \cr} $$
if $$f\left( x \right)$$  is odd $$ = 2\int\limits_0^a {f\left( x \right)dx} $$    if $$f\left( x \right)$$  is even.
$$\eqalign{ & I = 4\int_0^\pi {\frac{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)}}{{1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx \cr & I = 4\int_0^\pi {\frac{{\left( {\pi - x} \right)\sin \,x}}{{1 + {{\cos }^2}x}}} dx \cr & \Rightarrow I = 4\pi \int_0^\pi {\frac{{\sin \,x\,dx}}{{1 + {{\cos }^2}x}}} - 4\int {\frac{{x\,\sin \,x\,dx}}{{1 + {{\cos }^2}x}}} \cr & \Rightarrow 2I = 4\pi \int_0^\pi {\frac{{\sin \,x}}{{1 + {{\cos }^2}x}}dx} \cr & {\text{Put cos }}x = t\,\, \Rightarrow - \sin \,x\,dx = dt \cr & \therefore I = - 2\pi \int\limits_1^{ - 1} {\frac{1}{{1 + {t^2}}}dt} = 2\pi \int\limits_{ - 1}^1 {\frac{1}{{1 + {t^2}}}dt} \cr & = 2\pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1 \cr & = 2\pi \left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right] \cr & = 2\pi \left[ {\frac{\pi }{4} - \left( {\frac{{ - \pi }}{4}} \right)} \right] \cr & = 2\pi .\frac{\pi }{2} \cr & = {\pi ^2} \cr} $$

$$\eqalign{ & \int\limits_0^2 {\left[ {{x^2}} \right]dx} = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]dx} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right] + } \int\limits_{\sqrt 3 }^2 {\left[ {{x^2}} \right]dx} \cr & = \int\limits_0^1 {0\,dx + } \int\limits_1^{\sqrt 2 } {1\,dx + } \int\limits_{\sqrt 2 }^{\sqrt 3 } {2\,dx} + \int\limits_{\sqrt 3 }^2 {3\,dx} \cr & = \left[ x \right]_1^{\sqrt 2 } + \left[ {2x} \right]_{\sqrt 2 }^{\sqrt 3 } + \left[ {3x} \right]_{\sqrt 3 }^2 \cr & = \sqrt 2 - 1 + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 \cr & = 5 - \sqrt 3 - \sqrt 2 \cr} $$

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

We have,
$$\eqalign{ & f'\left( x \right) - 2f\left( x \right) < 0 \cr & \Rightarrow {e^{ - 2x}}f'\left( x \right) - 2{e^{ - 2x}}f\left( x \right) < 0 \cr & \Rightarrow \frac{d}{{dx}}\left( {{e^{ - 2x}}f\left( x \right)} \right) < 0 \cr & \Rightarrow {e^{ - 2x}}f\left( x \right)\,{\text{is strictly decreasing function on }}\left[ {\frac{1}{2},\,1} \right] \cr & \therefore {e^{ - 2x}}f\left( x \right) < {e^{ - 1}}f\left( {\frac{1}{2}} \right){\text{ or }}f\left( x \right) < {e^{2x - 1}} \cr} $$
Also given that $$f\left( x \right)$$  is positive function so $$f\left( x \right) > 0$$
$$\eqalign{ & \therefore 0 < f\left( x \right) < {e^{2x - 1}} \cr & \Rightarrow 0 < \int\limits_{\frac{1}{2}}^1 {f\left( x \right)dx} < \int\limits_{\frac{1}{2}}^1 {{e^{2x - 1}}dx} \cr & \Rightarrow 0 < \int\limits_{\frac{1}{2}}^1 {f\left( x \right)dx} < \left[ {\frac{{{e^{2x - 1}}}}{2}} \right]_{\frac{1}{2}}^1 \cr & \Rightarrow \int\limits_{\frac{1}{2}}^1 {f\left( x \right)dx} \, \in \left( {0,\,\frac{{e - 1}}{2}} \right) \cr} $$

$$\eqalign{ & \int\limits_0^{2\pi } {\log \left( {\frac{{a + b\,\sec \,x}}{{a - b\,\sec \,x}}} \right)dx} \cr & = 2\int\limits_0^\pi {\log \left( {\frac{{a + b\,\sec \,x}}{{a - b\,\sec \,x}}} \right)dx} \cr & = 2\int\limits_0^\pi {\log \left( {a + b\,\sec \,x} \right)dx} - 2\int\limits_0^\pi {\log \left( {a - b\,\sec \left( {\pi - x} \right)} \right)dx} \cr & = 2\int\limits_0^\pi {\log \left( {a + b\,\sec \,x} \right)dx} - 2\int\limits_0^\pi {\log \left( {a + b\,\sec \,x} \right)dx} \cr & = 0 \cr} $$