Application of Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & I = \int_0^{\frac{\pi }{2}} {\left[ {\cos \,x} \right]dx} + \int_{\frac{\pi }{2}}^\pi {\left[ {\cos \,x} \right]dx} \cr & \,\,\,\,\, = \int_0^{\frac{\pi }{2}} {0\,dx} + \int_{\frac{\pi }{2}}^\pi {\left( { - 1} \right)dx} \cr & \,\,\,\,\, = - \left( {\pi - \frac{\pi }{2}} \right) \cr & \,\,\,\,\, = - \frac{\pi }{2} \cr} $$

$$\eqalign{ & I = \int_0^3 {\left| x \right|\,\left| {x - 1} \right|\,\left| {x - 2} \right|dx} \cr & = \int_0^1 {\left( {{x^3} - 3{x^2} + 2x} \right)} dx + \int_1^2 { - \left( {{x^3} - 3{x^2} + 2x} \right)} dx + \int_2^3 {\left( {{x^3} - 3{x^2} + 2x} \right)dx} \cr & = \left[ {\frac{{{x^4}}}{4} - {x^3} + {x^2}} \right]_0^1 - \left[ {\frac{{{x^4}}}{4} - {x^3} + {x^2}} \right]_1^2 + \left[ {\frac{{{x^4}}}{4} - {x^3} + {x^2}} \right]_2^3 \cr & = \frac{1}{4} - \left\{ {\left( {4 - 8 + 4} \right) - \frac{1}{4}} \right\} + \left\{ {\left( {\frac{{81}}{4} - 27 + 9} \right) - \left( {4 - 8 + 4} \right)} \right\} \cr & = \frac{{11}}{4} \cr} $$

$$\eqalign{ & I = \int_\alpha ^0 {x\left| x \right|dx} + \int_0^\beta {x\left| x \right|dx} \cr & \,\,\,\,\, = \int_\alpha ^0 { - {x^2}dx + \int_0^\beta {{x^2}dx} } \cr & \,\,\,\,\, = \left[ { - \frac{{{x^3}}}{3}} \right]_\alpha ^0 + \left[ {\frac{{{x^3}}}{3}} \right]_0^\beta \cr & \,\,\,\,\, = \frac{{{\alpha ^3}}}{3} + \frac{{{\beta ^3}}}{3} \cr} $$

$$y = a{x^2}\,\& \,x = a{y^2}$$
Points of intersection are $$O\left( {0,\,0} \right)\,\& \,A\left( {\frac{1}{a},\,\frac{1}{a}} \right)$$

$$\eqalign{ & \therefore \,{\text{Area}} = \int\limits_0^{\frac{1}{a}} {\left( {\sqrt {\frac{x}{a}} - a{x^2}} \right)} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\frac{2}{{3{a^2}}} - \frac{1}{{3{a^2}}} = \frac{1}{{3{a^2}}} = 1 \cr & \Rightarrow a = \frac{1}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & {\text{Here}},\,\,\,18{x^2} - 9\pi x + {\pi ^2} = 0 \cr & \Rightarrow \left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0 \cr & \Rightarrow \alpha = \frac{\pi }{6},\,\,\beta = \frac{\pi }{3} \cr & {\text{Also, }}\,gof\left( x \right) = \cos \,x \cr & \therefore {\text{Required area :}} \cr & = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\cos \,x\,dx} = \frac{{\sqrt 3 - 1}}{2} \cr} $$

$$\eqalign{ & y = \int_0^{\frac{4}{a}} {\left( {a.x - \sqrt {4a.x} } \right)dx} \cr & \frac{1}{3} = \int_0^{\frac{4}{a}} {ax\,dx - \int_0^{\frac{4}{a}} {\sqrt {4ax} \,dx} } \cr & \frac{1}{3} = \left[ {\frac{{a{x^2}}}{2}} \right]_0^{\frac{4}{a}} - 2\left[ {\frac{{{{\left( {4ax} \right)}^{\frac{3}{2}}}}}{3}} \right]_0^{\frac{4}{a}} \cr & \frac{1}{3} = \frac{{\frac{{16a}}{{{a^2}}}}}{2} - \frac{2}{3}\left[ {4a{{\left( {\frac{4}{a}} \right)}^{\frac{3}{2}}}} \right],\,a = 8 \cr} $$
Putting the value of $$a$$ in $${x^2} + 2x - a = 0,$$    we get its roots i.e., $$ – 4$$ and $$2.$$

We, given $$\frac{{dy}}{{dx}} = 2x + 1 \Rightarrow y = {x^2} + x + k$$

Since, the curve passes through the point $$\left( {1,\,2} \right)$$
$$\eqalign{ & \therefore \,2 = 1 + 1 + k \Rightarrow k = 0 \cr & \therefore \,{\text{The curve is }}y = {x^2} + x \cr} $$
So, the required area
$$\eqalign{ & = \int_0^1 {\left( {{x^2} + x} \right)dx} \cr & = \left[ {\frac{{{x^3}}}{3} + \frac{{{x^2}}}{2}} \right]_0^1 \cr & = \frac{1}{3} + \frac{1}{2} \cr & = \frac{5}{6}{\text{ sq}}{\text{. unit}} \cr} $$

Equation of given curve is $${y^2} = 12x$$
At $$y = 6,\,36 = 12x \Rightarrow x = 3$$
$$\therefore $$  Required area $$ = \int_0^3 {\left( {{y_1} - {y_2}} \right)dx} $$    where $${y_1}$$ represents line and $${y_2}$$ represents the curve.
$$\eqalign{ & = \int_0^3 {\left( {6 - \sqrt {12x} } \right)dx} \cr & = \left[ {6x} \right]_0^3 - \sqrt {12} \left[ {\frac{{2{x^{\frac{3}{2}}}}}{3}} \right]_0^3 \cr & = \left[ {6 \times 3} \right] - \frac{{\sqrt {12} \times 2 \times \sqrt {27} }}{3} \cr & = 18 - 12 \cr & = 6{\text{ sq}}{\text{. units}} \cr} $$

We have, $$\int\limits_0^a {f\left( x \right)dx} = \frac{{{a^2}}}{2} + \frac{a}{2}\sin \,a + \frac{\pi }{2}\cos \,a$$
Differentiating w.r.t. $$a,$$ we get
$$\eqalign{ & f\left( a \right) = a + \frac{1}{2}\left( {\sin \,a + a\,\cos \,a} \right) - \frac{\pi }{2}\sin \,a \cr & {\text{Put }}a = \frac{\pi }{2}\,; \cr & f\left( {\frac{\pi }{2}} \right) = \frac{\pi }{2} + \frac{1}{2} - \frac{\pi }{2} = \frac{1}{2} \cr} $$

The area enclosed between the curves $${y^2} = x$$   and $$y = \left| x \right|$$
From the figure, area lies between $${y^2} = x$$   and $$y=x$$

$$\eqalign{ & \therefore {\text{Required area}} = \int_0^1 {\left( {{y_2} - {y_1}} \right)dx} \cr & = \int_0^1 {\left( {\sqrt x - x} \right)dx} \cr & = \left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \frac{{{x^2}}}{2}} \right]_0^1 \cr & \therefore {\text{Required area}} = \frac{2}{3}\left[ {{x^{\frac{3}{2}}}} \right]_0^1 - \frac{1}{2}\left[ {{x^2}} \right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6} \cr} $$