Definite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {I_1} = \int\limits_0^{\frac{\pi }{2}} {\cos \left( {\sin \,x} \right)dx} \cr & {I_2} = \int\limits_0^{\frac{\pi }{2}} {\sin \left( {\cos \,x} \right)dx} \cr & {I_3} = \int\limits_0^{\frac{\pi }{2}} {\cos \,x\,dx} \cr} $$

$$\eqalign{ & {\text{Let }}{f_1}\left( x \right) = \cos \left( {\sin \,x} \right),\,{f_2}\left( x \right) = \sin \left( {\cos \,x} \right),\,{f_3}\left( x \right) = \cos \,x \cr & {\text{If }}x > 0,{\text{ then }}\sin x < x \cr & \Rightarrow {\text{ for }}0 < x < \frac{\pi }{2}{\text{,}}\,{\text{sin}}\left( {\cos \,x} \right) < \cos \,x \cr & {\text{Also, }}0 < x < \frac{\pi }{2}{\text{ then }}\sin \,x < x \cr & \Rightarrow \cos \left( {\sin \,x} \right) > \cos \,x \cr & \therefore \,\cos \left( {\sin \,x} \right) > \cos \,x > \sin \left( {\cos \,x} \right){\text{ if }}0 < x < \frac{\pi }{2} \cr & \therefore \,{I_1} > {I_3} > {I_2} \cr} $$

Given that $$f$$ is a non negative function defined on $$\left[ {0,\,1} \right]$$  and $$\int\limits_0^x {\sqrt {1 - {{\left( {f'\left( t \right)} \right)}^2}} } dt = \int\limits_0^x {f\left( t \right)dt,\,\,\,\,0 \leqslant x \leqslant 1} $$
Differentiating both sides with respect to $$x,$$ we get
$$\eqalign{ & \sqrt {1 - {{\left[ {f'\left( x \right)} \right]}^2}} = f\left( x \right) \cr & \Rightarrow 1 - {\left[ {f'\left( x \right)} \right]^2} = {\left[ {f\left( x \right)} \right]^2} \cr & \Rightarrow {\left[ {f'\left( x \right)} \right]^2} = 1 - {\left[ {f\left( x \right)} \right]^2} \cr & \Rightarrow \frac{d}{{dx}}f\left( x \right) = \pm \sqrt {1 - {{\left[ {f\left( x \right)} \right]}^2}} \cr & \Rightarrow \pm \frac{{d\,f\left( x \right)}}{{\sqrt {1 - {{\left[ {f\left( x \right)} \right]}^2}} }} = dx \cr} $$
Integrating both sides with respect to $$x,$$ we get
$$ \pm \frac{{d\,f\left( x \right)}}{{\sqrt {1 - {{\left[ {f\left( x \right)} \right]}^2}} }} = \int {dx} \,\,\, \Rightarrow \pm {\sin ^{ - 1}}f\left( x \right) = x + C$$
$$\because $$ Given that $$f\left( 0 \right) = 0\,\,\, \Rightarrow C = 0$$
Hence $$f\left( x \right) = \pm \sin \,x$$
But as $$f\left( x \right)$$  is a non negative function on $$\left[ {0,\,1} \right]$$
$$\therefore f\left( x \right) = \sin \,x$$
Now $$\sin \,x < x,\,\forall \,x > 0$$
$$\therefore \,f\left( {\frac{1}{2}} \right) < \frac{1}{2}{\text{ and }}f\left( {\frac{1}{3}} \right) < \frac{1}{3}$$

$$\eqalign{ & I = \int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{1}{2}{{\sec }^2}\frac{x}{2}dx} = \left[ {\tan \frac{x}{2}} \right]_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} \cr & = \tan \frac{{3\pi }}{8} - \tan \frac{\pi }{8} \cr & = \tan \left( {\frac{\pi }{2} - \frac{\pi }{8}} \right) - \tan \frac{\pi }{8} \cr & = \cot \frac{\pi }{8} - \tan \frac{\pi }{8} \cr & = \frac{{{{\cos }^2}\frac{\pi }{8} - {{\sin }^2}\frac{\pi }{8}}}{{\sin \frac{\pi }{8}.\cos \frac{\pi }{8}}} \cr & = \frac{{\cos \frac{\pi }{4}}}{{\frac{1}{2}\sin \frac{\pi }{4}}} \cr & = 2 \cr} $$

$$\eqalign{ & {\text{Let }}I = \int\limits_0^1 {\frac{{{{\tan }^{ - 1}}}}{{1 + {x^2}}}dx} \cr & {\text{Put }}{\tan ^{ - 1}}x = t \cr & \frac{1}{{1 + {x^2}}}dx = dt \cr & x = 0,\, \Rightarrow t = 0 \cr & x = 1,\, \Rightarrow t = \frac{\pi }{4} \cr & \therefore \,I = \int\limits_0^{\frac{\pi }{4}} {t\,dt} = \left. {\frac{{{t^2}}}{2}} \right|_0^{\frac{\pi }{4}} = \frac{{{\pi ^2}}}{{32}} \cr} $$

Since $$f\left( x \right)$$  is an even function therefore $$\int\limits_0^\pi {f\left( x \right)} dx = 2\int\limits_0^{\frac{\pi }{2}} {f\left( x \right)} dx$$
Hence, $$\int\limits_0^\pi {f\left( {\cos \,x} \right)} dx = 2\int\limits_0^{\frac{\pi }{2}} {f\left( {\cos \,x} \right)} dx$$

Value of the integral $$\int\limits_2^9 {f\left( x \right)dx} $$
$$ = \int\limits_{ - 3}^9 {f\left( x \right)dx} - \int\limits_{ - 3}^2 {f\left( x \right)dx} ......\left( {\text{i}} \right)$$
Given, $$\int\limits_{ - 3}^9 {f\left( x \right)dx} = - \frac{5}{6}{\text{ and }}\int\limits_{ - 3}^2 {f\left( x \right)dx} = \frac{7}{3}$$
Putting these values in equation $$\left( {\text{i}} \right)$$
$$\int\limits_2^9 {f\left( x \right)dx} = - \frac{5}{6} - \frac{7}{3} = - \frac{{19}}{6}$$

Given that $$f\left( x \right) = \ln \left( {x - \sqrt {1 + {x^2}} } \right)$$
$$\int {f''\left( x \right)dx} = f'\left( x \right) + c$$       where $$c$$ is a constant
$$\eqalign{ & = \frac{1}{{\left( {x - \sqrt {1 + {x^2}} } \right)}}.\left( {1 - \frac{{2x}}{{2\sqrt {1 + {x^2}} }}} \right) + c \cr & = \frac{{ - \left( {x - \sqrt {1 + {x^2}} } \right)}}{{\left( {\sqrt {1 + {x^2}} } \right)\left( {x - \sqrt {1 + {x^2}} } \right)}} + c \cr & = - \frac{1}{{\sqrt {1 + {x^2}} }} + c \cr} $$

Given that $$T>0$$  is a fixed real number. $$f$$ is continuous $$\forall \,x \in R$$   such that $$f\left( {x + T} \right) = f\left( x \right)$$
$$ \Rightarrow f$$  is a periodic function of period $$T$$
Also given $$I = \int_0^T {f\left( x \right)dx} $$
Then, let $${I_1} = \int_3^{3 + 3T} {f\left( {2x} \right)dx} $$
$$\eqalign{ & {\text{Put }}2x = z\,\, \Rightarrow dx = \frac{{dz}}{2} \cr & {\text{Also as }}x \to 3,\,z \to 6;\, \cr & {\text{As }}x \to 3 + 3T,\,z \to 6 + 6T \cr & {I_1} = \frac{1}{2}\int_6^{6 + 6T} {f\left( z \right)dz} \cr & = \frac{1}{2}\left[ {\int_6^T {f\left( z \right)dz} + \sum\limits_{n = 1}^5 {\int_{nT}^{\left( {n + 1} \right)T} {f\left( z \right)dz} } + \int_{6T}^{6T + 6} {f\left( z \right)dz} } \right] \cr & {\text{Now, }}\int_{nT}^{\left( {n + 1} \right)T} {f\left( z \right)dz} = \int_0^T {f\left( {nT + u} \right)du,} \cr & {\text{where }}z = nT + u \cr & = \int_0^T {f\left( u \right)du = 1\,\,\,\,\,\,\,\,\,\,\left[ {\because f\left( {nT + u} \right) = f\left( u \right)} \right]} \cr} $$
Similarly, we can show that
$$\eqalign{ & \int_{6T}^{6T + 6} {f\left( z \right)dz} = \int_0^6 {f\left( z \right)dz} \cr & \therefore {I_1} = \frac{1}{2}\left[ {\int\limits_6^T {f\left( z \right)dz} + 5I + \int\limits_0^6 {f\left( z \right)dz} } \right] \cr & = \frac{1}{2}\left[ {\int\limits_6^T {f\left( z \right)dz} + 5I} \right] \cr & = \frac{1}{2}\left( {6I} \right) \cr & = 3I \cr} $$

$${R_1} = \int_a^b {{{\left( {x - 1} \right)}^2}dx} = \left[ {\frac{{{{\left( {x - 1} \right)}^3}}}{3}} \right]_0^b = \frac{{{{\left( {b - 1} \right)}^3} + 1}}{3}$$

$$\eqalign{ & {R_2} = \int_b^1 {{{\left( {x - 1} \right)}^2}dx} = \left[ {\frac{{{{\left( {x - 1} \right)}^3}}}{3}} \right]_b^1 = - \frac{{{{\left( {b - 1} \right)}^3}}}{3} \cr & {\text{As }}{R_1} - {R_2} = \frac{1}{4}\,\, \Rightarrow \frac{{2{{\left( {b - 1} \right)}^3}}}{3} + \frac{1}{3} = \frac{1}{4} \cr & {\text{or }}{\left( {b - 1} \right)^3} = - \frac{1}{8} \cr & {\text{or }}b - 1 = \frac{{ - 1}}{2} \cr & {\text{or }}b = \frac{1}{2} \cr} $$

Let $$a=k+ h$$   where $$k$$ is an integer such that $$\left[ a \right] = k$$   and $$0 \leqslant h < 1$$
$$\eqalign{ & \therefore \int\limits_1^a {\left[ x \right]f'\left( x \right)dx} = \int\limits_1^2 {1\,f'\left( x \right)dx} + \int\limits_2^3 {2\,f'\left( x \right)dx} + .....\int\limits_{k - 1}^k {\left( {k - 1} \right)dx} + \int\limits_k^{k + h} {k\,f'\left( x \right)dx} \cr & = \left\{ {f\left( 2 \right) - f\left( 1 \right)} \right\} + 2\left\{ {f\left( 3 \right) - f\left( 2 \right)} \right\} + 3\left\{ {f\left( 4 \right) - f\left( 3 \right)} \right\} + ..... + \left( {k - 1} \right)\left\{ {f\left( k \right) - f\left( {k - 1} \right)} \right\} + k\left\{ {f\left( {k + h} \right) - f\left( k \right)} \right\} \cr & = - f\left( 1 \right) - f\left( 2 \right) - f\left( 3 \right)..... - f\left( k \right) - k\,f\left( {k + h} \right) \cr & = \left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right) + f\left( 3 \right) + .....f\left( {\left[ a \right]} \right)} \right\} \cr} $$