Application of Integration MCQ Questions & Answers in Calculus | Maths

$$I = \int_0^{100\pi } {\sqrt 2 } \left| {\cos \,x} \right|dx.$$     But $$\left| {\cos \,x} \right|$$   is a periodic function of period $$\pi .$$
$$\eqalign{ & \therefore I = 100\int_0^\pi {\sqrt 2 \left| {\cos \,x} \right|dx} \cr & = 100\sqrt 2 \left\{ {\int_0^{\frac{\pi }{2}} {\left| {\cos \,x} \right|dx + } \int_{\frac{\pi }{2}}^\pi {\left| {\cos \,x} \right|dx} } \right\} \cr & = 100\sqrt 2 \left\{ {\int_0^{\frac{\pi }{2}} {\cos \,x\,dx + } \int_{\frac{\pi }{2}}^\pi { - \cos \,x\,dx} } \right\} \cr & = 100\sqrt 2 \left\{ {\left[ {\sin \,x} \right]_0^{\frac{\pi }{2}} + \left[ { - \sin \,x} \right]_{\frac{\pi }{2}}^x} \right\} \cr & = 100\sqrt 2 \left\{ {1 - 0 - \left( {0 - 1} \right)} \right\} \cr & = 200\sqrt 2 \cr} $$

Area of shaded region :
$$\eqalign{ & = \int\limits_0^1 {\left( {1 + \sqrt x } \right)dx} + \int\limits_1^2 {\left( {3 - x} \right)dx} - \int\limits_0^2 {\frac{{{x^2}}}{4}dx} \cr & = \left. x \right]_0^1 + \left. {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^1 + \left. {3x} \right]_1^2 - \left. {\frac{{{x^2}}}{2}} \right]_1^2 - \left. {\frac{{{x^3}}}{{12}}} \right]_0^2 \cr & = \frac{5}{2}\,{\text{sq}}{\text{. units}} \cr} $$

$$\eqalign{ & y = \cos \left| x \right| \cr & {\text{Area}} = 2\int_0^1 {\cos \,x\,dx} \cr & = 2\left[ {\sin \,x} \right]_0^1 \cr & = 2\,\sin \,1 \cr} $$

$$\eqalign{ & {\text{Here }}2n\pi < x < 2n\pi + \pi \cr & \therefore \int_0^x {\left[ {\sin \,x} \right]dx = } \int_0^{2n\pi } {\left[ {\sin \,x} \right]dx} + \int_{2n\pi }^x {\left[ {\sin \,x} \right]dx} \cr & \int_0^{2n\pi } {\left[ {\sin \,x} \right]} dx = n\int_0^{2\pi } {\left[ {\sin \,x} \right]dx} \,\,\,\,\,\left( {\because \,\left[ {\sin \,\overline {x + 2\pi } } \right] = \left[ {\sin \,x} \right]} \right) \cr & = n\left\{ {\int_0^{\frac{\pi }{2}} {\left[ {\sin \,x} \right]dx} + \int_{\frac{\pi }{2}}^\pi {\left[ {\sin \,x} \right]dx} + \int_\pi ^{\frac{{3\pi }}{2}} {\left[ {\sin \,x} \right]dx} + \int_{\frac{{3\pi }}{2}}^{2\pi } {\left[ {\sin \,x} \right]dx} } \right\} \cr & = n\left\{ {\int_0^{\frac{\pi }{2}} {0\,dx} + \int {0\,dx} + \int_\pi ^{\frac{{3\pi }}{2}} { - 1\,dx} + \int_{\frac{{3\pi }}{2}}^{2\pi } { - 1\,dx} } \right\} \cr & = n\left\{ { - \left( {\frac{{3\pi }}{2} - \pi } \right) - \left( {2\pi - \frac{{3\pi }}{2}} \right)} \right\} \cr & = - n\pi \cr & \int_{2n\pi }^x {\left[ {\sin \,x} \right]dx = } \int_{2n\pi }^x {0\,dx = 0,{\text{ if }}\,2n\pi } < x < 2n\pi + \frac{\pi }{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int_{2n\pi + \frac{\pi }{2}}^x {0\,dx = 0,{\text{ if }}\,2n\pi + \frac{\pi }{2} \leqslant x < 2n\pi + \pi } \cr & \therefore I = - n\pi + 0 = - n\pi \cr} $$

Let the equation of curve $${y^2}\left( {2a - x} \right) = {x^3}......\left( {\text{i}} \right)$$
and equation of line $$x = 2a......\left( {{\text{ii}}} \right)$$
The given curve is symmetrical about $$x$$-axis and passes through origin.
From $$\left( {\text{i}} \right)$$ we have $${y^2} = \frac{{{x^3}}}{{2a - x}}$$
But $$\frac{{{x^3}}}{{2a - x}} < 0$$   for $$x > 2a$$   and $$x < 0$$
So, curve does not lie in the portion $$x > 2a$$   and $$x < 0,$$  therefore curve lies in $$0 \leqslant x \leqslant 2a.$$
$$\therefore $$  Area bounded by the curve and line $$ = \int\limits_0^{2a} {y\,dx} = \int\limits_0^{2a} {\frac{{{x^{\frac{3}{2}}}}}{{\sqrt {2a} - x}}} dx$$
Put $$x = 2a\,{\sin ^2}\theta {\text{ and }}dx = 4a\,\sin \,\theta \,\cos \,\theta \,d\theta $$
$$\eqalign{ & \therefore \,I = \int\limits_0^{\frac{\pi }{2}} {8{a^2}{{\sin }^4}\theta \,d\theta } \cr & = 8{a^2}\left[ {\frac{3}{4}.\frac{1}{2}.\frac{\pi }{2}} \right] \cr & = \frac{{3\pi {a^2}}}{2}{\text{sq}}{\text{. units}} \cr} $$

$$\eqalign{ & {\text{Here }}f\left( x \right) = \frac{1}{{{{\sin }^3}x + \sin \,x}} \cr & \therefore f\left( { - x} \right) = - f\left( x \right). \cr & {\text{So, }}f\left( x \right){\text{is odd}}{\text{. Hence, }}I = 0. \cr} $$

Points of intersection of the two curves are $$\left( {0,\,0} \right),\,\left( {2,\,2} \right)$$   and $$\left( {2,\, - 2} \right)$$
Area $$=$$ Area $$\left( {OAB} \right) - $$   area under parabola $$\left( {0{\text{ to }}2} \right)$$
$$\eqalign{ & = \frac{{\pi \times {{\left( 2 \right)}^2}}}{4} - \int\limits_0^2 {\sqrt 2 } \sqrt x \,dx \cr & = \pi - \frac{8}{3} \cr} $$

Area above $$x$$-axis $$=$$ Area below $$x$$-axis
$$\eqalign{ & \therefore \,{\text{Required area :}} \cr & = 2\left[ {\int\limits_0^{\frac{\pi }{4}} {\left( {\cos \,x - \sin \,x} \right)dx} + \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\left( {\sin \,x - \cos \,x} \right)dx} + \int\limits_{\frac{\pi }{2}}^\pi {\sin \,x\,dx}} \right] \cr & = 4\sqrt 2 - 2 \cr} $$

$$\eqalign{ & I = \int_0^{\frac{\pi }{2}} {\frac{{f\left( x \right)dx}}{{f\left( x \right) + f\left( {\frac{\pi }{2} - x} \right)}}} \cr & \,\,\,\,\,\, = \int_0^{\frac{\pi }{2}} {\frac{{f\left( {\frac{\pi }{2} - x} \right)}}{{f\left( {\frac{\pi }{2} - x} \right) + f\left( x \right)}}dx} \cr & \left( {{\text{using }}\int_0^a {f\left( x \right)dx = } \int_0^a {f\left( {a - x} \right)dx} } \right) \cr & \therefore I + I = \int_0^{\frac{\pi }{2}} {\frac{{f\left( x \right) + f\left( {\frac{\pi }{2} - x} \right)}}{{f\left( x \right) + f\left( {\frac{\pi }{2} - x} \right)}}} dx\, = \int_0^{\frac{\pi }{2}} {dx} \, = \left[ x \right]_0^{\frac{\pi }{2}}\, = \frac{\pi }{2} \cr & \therefore I = \frac{\pi }{4} \cr} $$

Given $$\int_1^b {f\left( x \right)dx} = \left( {b - 1} \right)\sin \left( {3b + 4} \right)$$
Differentiating both sides w.r.t. $$b,$$ we get
$$\eqalign{ & \Rightarrow f\left( b \right) = 3\left( {b - 1} \right)\cos \left( {3b + 4} \right) + \sin \left( {3b + 4} \right) \cr & \Rightarrow f\left( x \right) = \sin \left( {3x + 4} \right) + 3\left( {x - 1} \right)\cos \left( {3x + 4} \right) \cr} $$