Application of Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & f\left( x \right) = \int_0^{\sin \,x} {\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}t} \right)\,dt} + \int_0^{\cos \,x} {\left( {\frac{\pi }{2} - {{\cos }^{ - 1}}t} \right)\,dt} \cr & = \frac{\pi }{2}\left[ t \right]_0^{\sin \,x} - \int_0^{\sin \,x} {{{\sin }^{ - 1}}} t\,dt + \frac{\pi }{2}\left[ t \right]_0^{\cos \,x} - \int_0^{\cos \,x} {{{\cos }^{ - 1}}t\,dt} \cr & = \frac{\pi }{2}\left( {\sin \,x + \cos \,x} \right) - \int_0^{\sin \,x} {\sin {\,^{ - 1}}t\,dt} - \int_0^{\cos \,x} {{{\cos }^{ - 1}}t\,dt} \cr & \int_0^{\sin \,x} {{{\sin }^{ - 1}}t\,dt} = \int_0^x {\theta \cos \,\theta \,d\theta \,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{putting }}t = \sin \,\theta } \right)} \cr & = \left[ {\theta \sin \,\theta } \right]_0^x - \int_0^x {\sin \,\theta \,d\theta } \cr & = x\sin \,x + \left[ {\cos \,\theta } \right]_0^x \cr & = x\sin \,x + \cos \,x - 1 \cr & \int_0^{\cos \,x} {{{\cos }^{ - 1}}t\,dt} = \int_0^x { - \phi \sin \,\phi \,d\phi \,\,\,\,\,\,\,\,\,\left( {{\text{putting }}t = \cos \,\phi } \right)} \cr & = \left[ {\phi \cos \,\phi } \right]_0^x - \int_0^x {\cos \phi \,d\phi } \cr & = x\cos \,x - \left[ {\sin \,\phi } \right]_0^x \cr & = x\cos \,x - \sin \,x \cr & \therefore f\left( x \right) = \frac{\pi }{2}\left( {\sin \,x + \cos \,x} \right) - x\sin \,x - \cos \,x + 1 - x\cos \,x + \sin \,x \cr & \therefore f\left( {\frac{\pi }{4}} \right) = \frac{\pi }{2}.\sqrt 2 - \frac{\pi }{4}.\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} + 1 - \frac{\pi }{4}.\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = 1 + \frac{\pi }{{2\sqrt 2 }} \cr} $$

Area of $$OABL = \int_0^2 {y\,dx} $$
$$\eqalign{ & = \int_0^2 {\left( {a + bx + c{x^2}} \right)} dx \cr & = 2a + 2b + \frac{8}{3}c \cr & = \frac{1}{3}\left[ {6a + 6b + 8c} \right]......\left( {\text{i}} \right) \cr & {\text{But, }}f\left( x \right) = a + bx + c{x^2}\,; \cr & f\left( 0 \right) = a, \cr & f\left( 1 \right) = a + b + c \cr & f\left( 2 \right) = a + 2b + 4c \cr & \Rightarrow \frac{1}{3}\left\{ {f\left( 0 \right) + 4f\left( 1 \right) + f\left( 2 \right)} \right\} \cr & = \frac{1}{3}\left\{ {a + 4\left( {a + b + c} \right) + \left( {a + 2b + 4c} \right)} \right\} \cr & = \frac{1}{3}\left\{ {6a + 6b + 8c} \right\} \cr} $$

The rough graph of $$y = \sin \,x + \cos \,x$$    and $$y = \left| {\cos \,x - \sin \,x} \right|$$    suggest the required area is
$$ = \int_0^{\frac{\pi }{2}} {\left[ {\left( {\sin \,x + \cos \,x} \right) - \left| {\cos \,x - \sin \,x} \right|} \right]} dx$$

$$\eqalign{ & = \int_0^{\frac{\pi }{4}} {2\,\sin \,x\,dx} + \int_{\frac{\pi }{4}}^{\frac{\pi }{2}} {2\,\cos \,x\,dx} \cr & = 2\left[ {\left( { - \cos \,x} \right)_0^{\frac{\pi }{4}} + \left( {\sin \,x} \right)_{\frac{\pi }{4}}^{\frac{\pi }{2}}} \right] \cr & = 2\sqrt 2 \left( {\sqrt 2 - 1} \right) \cr} $$

$$\eqalign{ & {a_{n + 1}} - {a_n} = \int_0^{\frac{\pi }{2}} {\left\{ {{{\cos }^{n + 1}}x.\cos \left( {n + 1} \right)x - {{\cos }^n}x.\cos \,nx} \right\}dx} \cr & = \int_0^{\frac{\pi }{2}} {{{\cos }^n}x.\left\{ {\cos \,x.\cos \left( {n + 1} \right)x - \cos \,nx} \right\}} dx \cr & = \int_0^{\frac{\pi }{2}} {{{\cos }^n}x.\left\{ { - \sin \,x.\sin \left( {n + 1} \right)x} \right\}dx} \cr & = \left[ {\sin \left( {n + 1} \right)x.\frac{{{{\cos }^{n + 1}}x}}{{n + 1}}} \right]_0^{\frac{\pi }{2}} - \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{n + 1}}x}}{{n + 1}}.\left( {n + 1} \right)\cos \left( {n + 1} \right)x\,dx} \cr & = - {a_{n + 1}} \cr & \therefore 2{a_{n + 1}} = {a_n} \cr} $$

$$\eqalign{ & \left| {1 - {x^2}} \right|{\text{ is an even function}}{\text{.}} \cr & \therefore I = 2\int_0^2 {\left| {1 - {x^2}} \right|dx} \cr & \therefore I = 2\int_0^1 {\left( {1 - {x^2}} \right)dx} + 2\int_1^2 {\left( {{x^2} - 1} \right)dx} \cr & = 2.\left[ {x - \frac{{{x^3}}}{3}} \right]_0^1 + 2.\left[ {\frac{{{x^3}}}{3} - x} \right]_1^2 \cr & = 2\left\{ {\left( {1 - \frac{1}{3}} \right) - 0} \right\} + 2\left\{ {\left( {\frac{8}{3} - 2} \right) - \left( {\frac{1}{3} - 1} \right)} \right\} \cr & = \frac{4}{3} + \frac{8}{3} \cr & \,\,\,\,\,\,\,\,\,\, = 4 \cr} $$

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

$$\eqalign{ & A = \int_0^1 {\left( {\sqrt {4x} - \frac{{{x^2}}}{4}} \right)dx} = \frac{5}{4} \cr & B = \int_1^4 {\left( {\sqrt {4x} - \frac{{{x^2}}}{4}} \right)dx} = \frac{{49}}{{12}} \cr} $$

Given equation of the curves are for $$y = \sqrt x $$   and $$x = 2y + 3$$   in the first quadrant.
On solving both the equation for $$y$$, we get
$$\eqalign{ & y = \sqrt {2y + 3} \cr & \Rightarrow {y^2} = 2y + 3 \cr & \Rightarrow {y^2} - 2y - 3 = 0 \cr & \Rightarrow {y^2} - 3y + y - 3 = 0 \cr & \Rightarrow y\left( {y - 3} \right) + 1\left( {y - 3} \right) = 0 \cr & \Rightarrow \left( {y + 1} \right)\left( {y - 3} \right) = 0 \cr & \Rightarrow y = - 1,\,\,3 \cr} $$
∴ Required area of shaded region,
$$\eqalign{ & A = \int_0^3 {\left( {2y + 3 - {y^2}} \right)dy} = \left[ {\frac{{2{y^2}}}{2} + 3y - \frac{{{y^3}}}{3}} \right]_0^3 \cr & = \left[ {\frac{{18}}{2} + 9 - 9 - 0} \right] \cr & = 9\,{\text{sq}}{\text{. units}} \cr} $$

Area of $$ABCD = 2 \times 1 = 2\,{\text{sq}}{\text{. units}}$$
Area of $$BAODC = \int_{ - 1}^1 {{x^2}dx} = \left. {\frac{{{x^3}}}{3}} \right|_{ - 1}^1 = \frac{2}{3}\,{\text{sq}}{\text{. units}}$$
Required area $$ = 2 - \frac{2}{3} = \frac{4}{3}{\text{ sq}}{\text{. units}}$$

$$\eqalign{ & I = \int_{ - 1}^0 {\max \left\{ {2 - x,\,2,\,1 + x} \right\}} dx + \int_0^1 {\max \left\{ {2 - x,\,2,\,1 + x} \right\}} dx \cr & \,\,\,\,\, = \int_{ - 1}^0 {\left( {2 - x} \right)dx} + \int_0^1 {2\,dx} \cr & \,\,\,\,\, = \left[ {2x - \frac{{{x^2}}}{2}} \right]_{ - 1}^0 + 2\left[ x \right]_0^1 \cr & \,\,\,\,\, = - \left( { - 2 - \frac{1}{2}} \right) + 2 \cr & \,\,\,\,\, = \frac{9}{2} \cr} $$