Application of Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \int_0^{2a} {f\left( x \right)dx} = \int_0^a {f\left( x \right)dx} + \int_a^{2a} {f\left( x \right)dx} \cr & {\text{Putting }}x = 2a - z{\text{ in the second integrand,}} \cr & \int_0^{2a} {f\left( x \right)dx} = \int_0^a {f\left( x \right)dx} + \int_a^0 {f\left( {2a - z} \right)\left( { - dz} \right)} \cr & = \int_0^a {f\left( x \right)dx} + \int_0^a {f\left( {2a - z} \right)dz} \cr & = \int_0^a {f\left( x \right)dx} + \int_0^a {f\left( z \right)dz} \cr & = \lambda + \lambda \cr & = 2\lambda \cr} $$

$${\text{Area}} = \int_{ - \infty }^0 {{2^x}dx} = \left[ {\frac{{{2^x}}}{{{{\log }_e}2}}} \right]_{ - \infty }^0 = \frac{1}{{{{\log }_e}e}} = {\log _2}e$$

$$\eqalign{ & \int_{ - 2}^3 {f\left( x \right)dx} - \int_2^3 {f\left( x \right)} dx = 0 \cr & {\text{or }}\int_{ - 2}^2 {f\left( x \right)} dx = 0 \cr & {\text{or }}\int_{ - 2}^0 {f\left( x \right)} dx + \int_0^2 {f\left( x \right)} dx = 0 \cr} $$
$${\text{or }}\int_0^2 {\left\{ {f\left( x \right) + f\left( { - x} \right)} \right\}dx = 0} ,$$       which may imply $$f\left( { - x} \right) = - f\left( x \right)$$    in $$\left[ { - 2,\,2} \right].$$
Nothing can be said for the whole of $$R.$$

The required area
$$\eqalign{ & = \int_0^4 {\sqrt {25 - {x^2}} } dx - \int_0^2 {\frac{{4 - {x^2}}}{4}} dx - \int_2^4 {\frac{{{x^2} - 4}}{4}} dx \cr & = \left[ {\frac{x}{2}\sqrt {25 - {x^2}} + \frac{{25}}{2}{{\sin }^{ - 1}}\frac{x}{5}} \right]_0^4 - \frac{1}{4}\left[ {4x - \frac{{{x^3}}}{3}} \right]_0^2 - \frac{1}{4}\left[ {\frac{{{x^3}}}{3} - 4x} \right]_2^4 \cr & = 2 + \frac{{25}}{2}{\sin ^{ - 1}}\frac{4}{5} \cr} $$

Point of intersection of $${y^2} = 4ax$$   and $$y = ax$$  are $$\left( {0,\,0} \right)$$  and $$\left( {\frac{4}{a},\,4} \right)$$

$$\eqalign{ & {\text{Given,}}\,\int\limits_0^4 {\left[ {\frac{y}{a} - \frac{{{y^2}}}{{4a}}} \right]} dy = \frac{1}{3} \cr & \Rightarrow \frac{8}{a} - \frac{1}{{12a}} \times 64 = \frac{1}{3} \cr & \Rightarrow \frac{8}{{3a}} = \frac{1}{3} \cr & \Rightarrow a = 8 \cr} $$
So, the parabola is $${y^2} = 32x$$

Area enclosed by $$y = 4x$$   is $$\int\limits_0^8 {\left[ {\frac{y}{4} - \frac{{{y^2}}}{{32}}} \right]} dy = \left[ {\frac{{{y^2}}}{8} - \frac{{{y^3}}}{{96}}} \right]_0^8 = \frac{8}{3}$$

$$\left| x \right| \geqslant - \left[ x \right]$$    for all positive $$x,$$ and $$\left| x \right| \leqslant - \left[ x \right]$$    for all negative $$x.$$
$$\eqalign{ & \therefore \int_{ - 2}^2 {f\left( x \right)dx} = \int_{ - 2}^0 {\left( {x - \left[ x \right]} \right)dx} + \int_0^2 {\left( {x + \left| x \right|} \right)dx} \cr & = \int_{ - 2}^0 {x\,dx} - \int_{ - 2}^0 {\left[ x \right]dx} + \int_0^2 {2x\,dx} \cr & = \left[ {\frac{{{x^2}}}{2}} \right]_{ - 2}^0 - \left\{ {\int_{ - 2}^{ - 1} {\left[ x \right]dx} + \int_{ - 1}^0 {\left[ x \right]dx} } \right\} + \left[ {{x^2}} \right]_0^2 \cr & = 0 - 2 - \int_{ - 2}^{ - 1} { - 2\,dx} - \int_{ - 1}^0 { - 1\,dx + 4} \cr & = - 2 + 2\left[ x \right]_{ - 2}^{ - 1} + \left[ x \right]_{ - 1}^0 + 4 \cr & = - 2 + 2\left( { - 1 + 2} \right) + \left( {0 + 1} \right) + 4 \cr & = 5 \cr} $$

Since, $$f\left( x \right) = f\left( {a + x} \right),$$    therefore $$a$$ is the period of $$f\left( x \right)$$
$$\eqalign{ & {\text{Again, }}\int_{ma}^{na} {f\left( x \right)dx = \left( {n - m} \right)} \int_0^a {f\left( x \right)dx} \cr & {\text{Here, }}\int_a^{na} {f\left( x \right)dx = \left( {n - 1} \right)} \int_0^a {f\left( x \right)dx} \cr & = \left( {n - 1} \right)p \cr & {\text{Hence, answer is option B}}{\text{.}} \cr} $$

$$\eqalign{ & \phi \left( t \right) = tf\left( t \right) = t.\frac{{{e^t} + 1}}{{{e^t} - 1}} \cr & {\text{So, }}\phi \left( { - t} \right) = - t.\frac{{{e^{ - t}} + 1}}{{{e^{ - t}} - 1}} = t.\frac{{{e^t} + 1}}{{{e^t} - 1}} \cr & \therefore \phi \left( t \right) = \phi \left( { - t} \right){\text{.}}\,{\text{Hence, }}\phi \left( t \right){\text{ is an even function}} \cr & \therefore \int_{ - 1}^1 {\phi \left( t \right)dt} = 2\int_0^1 {\phi \left( t \right)dt} = 2\lambda \cr} $$

$$\eqalign{ & \int_0^a {f\left( x \right)} dx = \lambda \cr & f\left( a \right) - f\left( 0 \right) = \lambda \to 1 \cr & \int_0^a {f\left( {2a - x} \right)dx} = \mu \cr & f\left( a \right) - f\left( {2a} \right) = \mu \to 2 \cr & {\text{from 1 and 2}} \cr & \,\,\,\,f\left( a \right) - f\left( 0 \right) = \lambda \cr & \,\,\,\,f\left( a \right) - f\left( {2a} \right) = \mu \cr & + \,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \cr & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \cr & f\left( {2a} \right) - f\left( 0 \right) = \lambda - \mu \cr & \int_0^{2a} {f\left( x \right)} dx = \lambda - \mu \cr} $$

$$\eqalign{ & {\text{Here }}0 < x < 1{\text{ and }}{e^{{x^2}}} > 0.\,\,{\text{So, }}0 < x{e^{{x^2}}} < {e^{{x^2}}} \cr & \therefore \int_0^1 {0\,dx} < \int_0^1 {x{e^{{x^2}}}dx} < \int_0^1 {{e^{{x^2}}}} dx \cr & \therefore 0 < \lambda \int_0^1 {{e^{{x^2}}}dx} < \int_0^1 {{e^{{x^2}}}dx} \,\,\,\,\,\, \Rightarrow 0 < \lambda < 1 \cr} $$