Application of Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Here, }}4 = \mathop {\lim }\limits_{x \to 1} \frac{1}{{x - 1}}\int_2^{f\left( x \right)} {2t\,dt} \cr & = \mathop {\lim }\limits_{x \to 1} \frac{{2f\left( x \right).f'\left( x \right)}}{1}\,\,\,\left( {{\text{using L'Hospital's}}\,{\text{rule}}} \right) \cr & \therefore 4 = 2f\left( 1 \right).f'\left( 1 \right) \cr & \therefore f'\left( 1 \right) = 1 \cr} $$

$$\eqalign{ & f\left( x \right) = \int_0^{{x^2}} {\frac{{dt}}{{1 + {t^3}}}} - \int_0^x {\frac{{dt}}{{1 + {t^3}}}} \cr & \therefore f'\left( x \right) = \frac{d}{{d{x^2}}}\left\{ {\int_0^{{x^2}} {\frac{{dt}}{{1 + {t^3}}}} } \right\}\frac{{d\left( {{x^2}} \right)}}{{dx}} - \frac{1}{{1 + {x^3}}} \cr & \therefore f'\left( x \right) = \frac{1}{{1 + {x^6}}}.2x - \frac{1}{{1 + {x^3}}} \cr & \therefore f'\left( 2 \right) = \frac{{2 \times 2}}{{1 + {2^6}}} - \frac{1}{{1 + {2^3}}} = \frac{4}{{65}} - \frac{1}{9} = \frac{{ - 29}}{{585}} \cr} $$

Area of $$\Delta AOB = \frac{1}{2} \times 2a \times {a^2} = {a^3}{\text{units}}$$

Area of region $$AOB = 2\int\limits_0^{{a^2}} {x\,dy} = 2\int\limits_0^{{a^2}} {\sqrt y \,dy} = 2\left[ {\frac{{{y^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^{{a^2}} = \frac{4}{3}{a^3}{\text{units}}$$
$$\therefore $$   ratio of areas $$ = \frac{{{a^3}}}{{\frac{4}{3}{a^3}}} = \frac{3}{4}$$

$$\eqalign{ & {\text{We have, }}0 \leqslant \frac{{{x^2}}}{{64}} < 1,{\text{ if}}\, - 8 < x < 8 \cr & \Rightarrow 2 \leqslant \frac{{{x^2}}}{{64}} + 2 < 3,{\text{ if }}\,\left| x \right| < 8 \cr & \Rightarrow y = \left[ {\frac{{{x^2}}}{{64}} + 2} \right] = 2,{\text{ if }}\,\left| x \right| < 8 \cr} $$
The graphs of the given curves is as shown in figure.

$$\eqalign{ & {\text{Required area}} = {\text{area of the shaded region}} \cr & = \int_0^2 {x\,dy} \cr & = \int_0^2 {\left( {y + 1} \right)dy} \cr & = \frac{1}{2}\left[ {\left( {y + 1} \right)} \right]_0^2 \cr & = \frac{9}{2} - \frac{1}{2} \cr & = 4{\text{ sq}}{\text{. units}} \cr} $$

$$\eqalign{ & {\text{Put }}x = \frac{\pi }{2} - z \cr & {\text{Then}}\,\,I = \int_{\frac{{3\pi }}{{10}}}^{\frac{\pi }{5}} {\frac{{\cos \left( {\frac{\pi }{2} - z} \right)}}{{\cos \left( {\frac{\pi }{2} - z} \right) + \sin \left( {\frac{\pi }{2} - z} \right)}}} d\left( {\frac{\pi }{2} - z} \right) \cr & \therefore I = - \int_{\frac{{3\pi }}{{10}}}^{\frac{\pi }{5}} {\frac{{\sin \,z}}{{\sin \,z + \cos \,z}}dz} = \int_{\frac{\pi }{5}}^{\frac{{3\pi }}{{10}}} {\frac{{\sin \,x}}{{\sin \,x + \cos \,x}}dx} \cr & \therefore I + I = \int_{\frac{\pi }{5}}^{\frac{{3\pi }}{{10}}} {\frac{{\cos \,x}}{{\cos \,x + \sin \,x}}dx} + \int_{\frac{\pi }{5}}^{\frac{{3\pi }}{{10}}} {\frac{{\sin \,x}}{{\sin \,x + \cos \,x}}dx} \cr & = \int_{\frac{\pi }{5}}^{\frac{{3\pi }}{{10}}} {dx} \cr & = \frac{{3\pi }}{{10}} - \frac{\pi }{5} \cr & = \frac{\pi }{{10}} \cr} $$

$$\eqalign{ & \int_0^a {f\left( x \right).g\left( x \right)dx = } \int_0^a {f\left( {a - x} \right).g\left( {a - x} \right)dx} \,\,\,\,\left( {{\text{by property}}} \right) \cr & = \int_0^a {f\left( x \right).\left\{ {2 - g\left( x \right)} \right\}dx} \,\,\,\,\,\,\left( {{\text{from the question}}} \right) \cr & = 2\int_0^a {f\left( x \right)dx - } \int_0^a {f\left( x \right).g\left( x \right)dx} \cr & \therefore \,\,2\int_0^a {f\left( x \right).g\left( x \right)dx} = 2\int_0^a {f\left( x \right)dx} \cr} $$

$$\eqalign{ & I = \int_0^a {f\left( x \right)dx} + \int_0^a {f\left( { - x} \right)dx} \cr & \,\,\,\,\, = \int_0^a {f\left( x \right)dx + } \int_0^{ - a} {f\left( z \right)d\left( { - z} \right)} \cr & \,\,\,\,\, = \int_0^a {f\left( x \right)dx + } \int_{ - a}^0 {f\left( z \right)dz} \cr & \,\,\,\,\, = \int_0^a {f\left( x \right)dx} + \int_{ - a}^0 {f\left( x \right)dx} \cr & \,\,\,\,\, = \int_{ - a}^a {f\left( x \right)dx} \cr} $$

$$y = 0,$$  when $$t = 0$$  and then $$x = a$$
So desire area
$$\eqalign{ & A = \int\limits_0^a {y\,dx} \cr & \Rightarrow A = \int\limits_{\frac{\pi }{2}}^0 {b\,{{\sin }^3}t\left( { - 3a\,{{\cos }^2}t\,\sin \,t\,dt} \right)} \cr & \Rightarrow A = 3ab\int\limits_0^{\frac{\pi }{2}} {{{\sin }^4}t\,{{\cos }^2}t\,dt} \cr & \Rightarrow A = 3ab\int\limits_0^{\frac{\pi }{2}} {{{\cos }^4}t\,{{\sin }^2}t\,dt} \cr & \therefore \,2A = 3ab\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}t\,{{\sin }^2}t\,dt} \cr & \Rightarrow 2A = 3ab.\frac{\pi }{{16}} \cr & \Rightarrow A = \frac{{3\pi ab}}{{32}} \cr} $$

Area bounded by $$y = \frac{1}{{1 + {x^2}}}$$   and $$x$$-axis is $$\int_{ - \infty }^\infty {\frac{1}{{1 + {x^2}}}} dx = \pi $$
Area bounded by two curves is $$ = \int_{ - 1}^1 {\left( {\frac{1}{{1 + {x^2}}} - \frac{{{x^2}}}{2}} \right)} dx = \frac{\pi }{2} - 1$$

Required area
$$\eqalign{ & = \int_0^{\frac{\pi }{4}} {\tan \,x\,dx} \cr & = \ln \left| {\sec \,1} \right|_0^{\frac{\pi }{4}} \cr & = \ln \sqrt 2 \cr & = \frac{{\ln \,2}}{2} \cr} $$