Definite Integration MCQ Questions & Answers in Calculus | Maths

$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}.\frac{1}{{\sqrt {\frac{r}{n}} }} = \int_0^1 {\frac{1}{{\sqrt x }}dx = \left[ {2\sqrt x } \right]_0^1 = 2} } $$

$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{{{{\sin }^5}x\,{{\cos }^3}x}}{{{x^4}}} \cr & f\left( { - x} \right) = \frac{{{{\sin }^5}\left( { - x} \right)\,{{\cos }^3}\left( { - x} \right)}}{{{{\left( { - x} \right)}^4}}} \cr & f\left( { - x} \right) = \frac{{ - {{\sin }^5}x\,{{\cos }^3}x}}{{{x^4}}} \cr & f\left( { - x} \right) = - f\left( x \right) \cr} $$
$$ \Rightarrow f\left( x \right)$$   is an odd function.
Hence, $$\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{{{\sin }^5}x\,{{\cos }^3}x}}{{{x^4}}}dx} = 0$$

$$\eqalign{ & {\text{Let }}I = \int_0^\pi {\left[ {\cot \,x} \right]dx} .....(1) \cr & = \int_0^\pi {\left[ {\cot \,\left( {\pi - x} \right)} \right]dx} \cr & = \int_0^\pi {\left[ { - \cot \,x} \right]dx.....(2)} \cr} $$
Adding two values of 1 in equations (1) & (2), We get
$$\eqalign{ & 2I = \int_0^\pi {\left( {\left[ {\cot \,x} \right] + \left[ { - \cot \,x} \right]} \right)dx} \,\,\, = \int_0^\pi {\left[ { - 1} \right]dx} \cr & \left[ {\because \left[ x \right] + \left[ { - x} \right] = 1,\,{\text{if}}\,x \notin z\,{\text{and}}\left[ x \right] + \left[ { - x} \right] = 0{\text{ if }}x \in z} \right] \cr & = \left[ { - x} \right]_0^\pi = - \pi \,\,\, \Rightarrow I = - \frac{\pi }{2} \cr} $$

$$\eqalign{ & {\text{We have, }}\int\limits_0^9 {\left[ {\sqrt x + 2} \right]dx} \cr & = \int\limits_0^1 {2dx} + \int\limits_1^4 {3dx} + \int\limits_4^9 {4dx} \cr & = 2 + \left( {12 - 3} \right) + \left( {36 - 16} \right) \cr & = 2 + 9 + 20 \cr & = 31 \cr} $$

$$\eqalign{ & \int_{\sqrt 2 }^x {\frac{{dt}}{{t\sqrt {{t^2} - 1} }} = \frac{\pi }{2}} \cr & \therefore \left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = \frac{\pi }{2}\,\,\,\,\,\,\left[ {\because \int {\frac{{dx}}{{x\sqrt {{x^2} - 1} }} = {{\sec }^{ - 1}}x} } \right] \cr & \Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = \frac{\pi }{2} \cr & \Rightarrow {\sec ^{ - 1}}x - \frac{\pi }{4} = \frac{\pi }{2} \cr & \Rightarrow {\sec ^{ - 1}}x = \frac{\pi }{2} + \frac{\pi }{4} \cr & \Rightarrow {\sec ^{ - 1}}x = \frac{{3\pi }}{4} \cr & \Rightarrow x = \sec \frac{{3\pi }}{4} \cr & \Rightarrow x = - \sqrt 2 \cr} $$

Given function $$f\left( x \right) = A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B$$
Differentiating w.r.t. $$x$$
$$\eqalign{ & f'\left( x \right) = A\,\cos \left( {\frac{{\pi x}}{2}} \right).\frac{\pi }{2} \cr & f'\left( {\frac{1}{2}} \right) = \sqrt 2 = A\left( {\cos \frac{\pi }{4}} \right)\frac{\pi }{2} = A.\frac{1}{{\sqrt 2 }}.\frac{\pi }{2} \cr & \Rightarrow A = \frac{{\left( {\sqrt 2 \times \sqrt 2 } \right) \times 2}}{\pi } = \frac{4}{\pi } \cr & {\text{Now, }}\int_0^1 {f\left( x \right)dx = \frac{{2A}}{\pi }} \cr & \Rightarrow \int_0^1 {\left\{ {A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B} \right\}dx = \frac{{2 \times 4}}{{{\pi ^2}}}} \cr & \Rightarrow \left[ { - A\,\cos \frac{{\pi x}}{2}.\frac{2}{\pi } + Bx} \right]_0^1 = \frac{8}{{{\pi ^2}}} \cr & \Rightarrow - \frac{4}{\pi }.\frac{2}{\pi }\cos \frac{\pi }{2} + B + \frac{4}{\pi }.\frac{2}{\pi }\cos \,0 = \frac{8}{{{\pi ^2}}} \cr & \Rightarrow B + \frac{8}{{{\pi ^2}}} = \frac{8}{{{\pi ^2}}} \cr & \Rightarrow B = 0 \cr} $$

We have,
$$\eqalign{ & I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left[ {f\left( x \right) + f\left( { - x} \right)} \right]} \left[ {g\left( x \right) - g\left( { - x} \right)} \right]dx \cr & {\text{Let }}F\left( x \right) = \left( {f\left( x \right) + f\left( { - x} \right)} \right)\left( {g\left( x \right) - g\left( { - x} \right)} \right) \cr & {\text{then }}F\left( { - x} \right) = \left( {f\left( { - x} \right) + f\left( x \right)} \right)\left( {g\left( { - x} \right) - g\left( x \right)} \right) \cr & = - \left[ {f\left( x \right) + f\left( { - x} \right)} \right]\left[ {g\left( x \right) - g\left( { - x} \right)} \right] \cr & = - F\left( x \right) \cr} $$
$$\therefore F\left( x \right)$$  is an odd function,
$$\therefore $$ We get $$I=0$$

$$\eqalign{ & {\text{Let }}\phi \left( x \right) = \left\{ {f\left( x \right) + f\left( { - x} \right)} \right\}\left\{ {g\left( x \right) - g\left( { - x} \right)} \right\} \cr & \phi \left( { - x} \right) = \left\{ {f\left( { - x} \right) + f\left( x \right)} \right\}\left\{ {g\left( { - x} \right) - g\left( x \right)} \right\} \cr & \phi \left( { - x} \right) = - \left\{ {f\left( x \right) + f\left( { - x} \right)} \right\}\left\{ {g\left( x \right) - g\left( { - x} \right)} \right\} \cr & \phi \left( { - x} \right) = - \phi \left( x \right) \cr & \therefore \,\varphi \left( x \right)\,{\text{is an odd function}} \cr & \Rightarrow \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\phi \left( x \right)} dx = 0 \cr} $$

Given integral is $$I = \int_0^1 {\left( {x - 1} \right){e^{ - x}}dx} $$
Integrating by parts taking $$\left( {x - 1} \right)$$  as first function
We get,
$$\eqalign{ & I = \left[ {\left( {x - 1} \right)\left\{ { - {e^{ - x}}} \right\}} \right]_0^1 - \int_0^1 {1.\left( { - {e^{ - x}}} \right)dx} \cr & = - \left( {1 - 1} \right)\frac{1}{e} + \left( { - 1} \right){e^0} + \left[ { - {e^{ - x}}} \right]_0^1 \cr & = - 1 - \frac{1}{e} + 1 \cr & = - \frac{1}{e} \cr} $$

Given $$f'\left( x \right) = f\left( x \right) \Rightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = 1$$
Integrating $$\log f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$
$$\eqalign{ & f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x} \cr & \therefore \int\limits_0^1 {f\left( x \right)\,g\left( x \right)} dx = \int\limits_0^1 {{e^x}\left( {{x^2} - {e^x}} \right)} dx \cr & = \int\limits_0^1 {{x^2}{e^x}dx} - \int\limits_0^1 {{e^{2x}}dx} \cr & = \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - \frac{1}{2}\left[ {{e^{2x}}} \right]_0^1 \cr & = e - \left[ {\frac{{{e^2}}}{2} - \frac{1}{2}} \right] - 2\left[ {e - e + 1} \right] \cr & = e - \frac{{{e^2}}}{2} - \frac{3}{2} \cr} $$