Application of Integration MCQ Questions & Answers in Calculus | Maths

Required area $$ = 2\int\limits_{y = 0}^4 {\frac{y}{2}dy = } \left. {\frac{{{y^2}}}{2}} \right|_0^4 = \frac{{16}}{2} = 8\,{\text{sq}}{\text{. units}}$$
or, Area $$ = 2 \times 4 = 8\,{\text{sq}}{\text{. units}}$$

Required area
$$\eqalign{ & = \int\limits_{y = 1}^4 {\sqrt {\frac{y}{9}} dy} \cr & = \frac{1}{3}\int\limits_{y = 1}^4 {{y^{\frac{1}{2}}}dy} \cr & = \frac{1}{3} \times \frac{2}{3}\left| {\left( {{y^{\frac{3}{2}}}} \right)} \right|_1^4 \cr & = \frac{2}{9}\left[ {{{\left( {{4^{\frac{1}{2}}}} \right)}^3} - {{\left( {{1^{\frac{1}{2}}}} \right)}^3}} \right] \cr & = \frac{{14}}{9}{\text{ sq}}{\text{. units}} \cr} $$

Given curves are $$y = {x^2} + 3$$   and $$y = 2x + 3$$   points of intersection are $$\left( {0,\,3} \right)$$  and $$\left( {2,\,7} \right).$$
$$\therefore $$  Required area
$$ = \left| {\int\limits_0^2 {\left( {{x^2} - 2x} \right)dx} } \right| = \left| {\frac{{{x^3}}}{3} - \frac{{2{x^2}}}{2}} \right|_0^2 = \left| {\frac{8}{3} - 4} \right| = \frac{4}{3}{\text{ sq}}{\text{. unit}}$$

First we draw each curve as separate graph

Clearly the bounded area is as shown in the following figure.

Required area
$$\eqalign{ & = 4\int\limits_0^1 {\left( { - \ell n\,x} \right)} dx \cr & = - 4\left[ {x\,\ell n\,x - x} \right]_0^1 \cr & = 4{\text{ sq}}{\text{. units}} \cr} $$

$$\frac{{\mu \sin \,x}}{{1 + \cos \,x}}$$   is an odd function. So, $$\int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\frac{{\mu \sin \,x}}{{1 + \cos \,x}}dx = 0.} $$
$$\therefore $$ the equation is $$\lambda \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left| {\sin \,x} \right|dx + \nu \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {dx = 0} } $$
Or $$2\lambda \int_0^{\frac{\pi }{4}} {\left| {\sin \,x} \right|dx + \nu .\frac{\pi }{2} = 0,} $$       which is a relation between $$\lambda ,\,\nu .$$

$$\eqalign{ & {\text{For }}c < 1,\,\int_c^1 {\left( {8{x^2} - {x^5}} \right)} dx = \frac{{16}}{3} \cr & \Rightarrow \frac{8}{3} - \frac{1}{6} - \frac{{8{c^3}}}{3} + \frac{{{c^6}}}{6} = \frac{{16}}{3} \cr & \Rightarrow {c^3}\left[ { - \frac{8}{3} + \frac{{{c^3}}}{6}} \right] = \frac{{17}}{6} \cr} $$
Again, for $$c \geqslant 1,$$  none of the values of $$c$$ satisfy the required condition that
$$\int_1^c {\left( {8{x^2} - {x^5}} \right)} dx = \frac{{16}}{3} \Rightarrow c + 2 = 1$$

$$\eqalign{ & \int_0^x {f\left( t \right)dt} + \int_1^x {tf\left( t \right)dt = x} \cr & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x \cr & f\left( x \right) + xf\left( x \right) = 1 \cr & \therefore f\left( x \right) = \frac{1}{{1 + x}} \cr & {\text{So, }}f\left( 1 \right) = \frac{1}{2} \cr} $$

$${\text{Area}} = \int_q^p {c{e^x}dx} = \left[ {c{e^x}} \right]_q^p = c\left( {{e^p} - {e^q}} \right) = f\left( p \right) - f\left( q \right)$$

$$\eqalign{ & A = \int\limits_{ - 1}^0 {\left\{ {\left( {3 + x} \right) - \left( { - x + 1} \right)} \right\}dx + } \int\limits_0^1 {\left\{ {\left( {3 - x} \right) - \left( { - x + 1} \right)} \right\}dx + } \int\limits_1^2 {\left\{ {\left( {3 - x} \right) - \left( {x - 1} \right)} \right\}dx} \cr & = \int\limits_{ - 1}^0 {\left( {2 + 2x} \right)dx + } \int\limits_0^1 {2\,dx + } \int\limits_1^2 {\left( {4 - 2x} \right)dx} \cr & = \left[ {2x - {x^2}} \right]_{ - 1}^0 + \left[ {2x} \right]_0^1 + \left[ {4x - {x^2}} \right]_1^2 \cr & = 0 - \left( { - 2 + 1} \right) + \left( {2 - 0} \right) + \left( {8 - 4} \right) - \left( {4 - 1} \right) \cr & = 1 + 2 + 4 - 3 \cr & = 4\,{\text{sq}}{\text{.}}\,{\text{units}} \cr} $$

Required area

$$\eqalign{ & = {\text{Area of ABCD}} - {\text{ar (ABOCD)}} \cr & = \int\limits_{ - \frac{1}{2}}^1 {\frac{{y + 1}}{4}dy} - \int\limits_{ - \frac{1}{2}}^1 {\frac{{{y^2}}}{2}dy} \cr & = \frac{1}{4}\left[ {\frac{{{y^2}}}{2} + y} \right]_{ - \frac{1}{2}}^1 - \frac{1}{2}\left[ {\frac{{{y^3}}}{2}} \right]_{ - \frac{1}{2}}^1 \cr & = \frac{1}{4}\left[ {\frac{3}{2} + \frac{3}{8}} \right] - \frac{9}{{48}} \cr & = \frac{{15}}{{32}} - \frac{9}{{48}} \cr & = \frac{{27}}{{96}} \cr & = \frac{9}{{32}} \cr} $$