Application of Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Here, }}\int_0^a {f\left( x \right)dx} = \frac{{{a^2}}}{2} + \frac{a}{2}\sin \,a + \frac{\pi }{2}\cos \,a.\, \cr & {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}a, \cr & f\left( a \right) = a + \frac{1}{2}\left( {\sin \,a + a\cos \,a} \right) - \frac{\pi }{2}\sin a \cr & {\text{So, }}f\left( {\frac{\pi }{2}} \right) = \frac{\pi }{2} + \frac{1}{2}\left( {1 + 0} \right) - \frac{\pi }{2} = \frac{1}{2} \cr} $$

Given, equation of the circle is $${x^2} + {y^2} = 9$$

$$\therefore $$  Area of the smaller segment cut off from the circle $${x^2} + {y^2} = 9$$   by $$x = 1,$$  is given by
$$\eqalign{ & A = 2\int_1^3 {\sqrt {9 - {x^2}} } dx \cr & \,\,\,\,\, = 2.\frac{1}{2}\left[ {x\sqrt {9 - {x^2}} + 9\,{{\sin }^{ - 1}}\frac{x}{3}} \right]_1^3 \cr & \,\,\,\,\, = \left[ {3.\sqrt {9 - 9} + 9\,{{\sin }^{ - 1}}\left( {\frac{3}{3}} \right) - 1.\sqrt {9 - 1} - 9\,{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)} \right] \cr & \,\,\,\,\, = \left( {9\,{{\sec }^{ - 1}}3 - \sqrt 8 } \right){\text{sq}}{\text{.unit}} \cr} $$

The area bounded by the curve $$y = 1 + \frac{8}{{{x^2}}},\,\,x$$   -axis and the ordinates $$x = 2,\,x = 4$$    is
$$\eqalign{ & = \int_2^4 {y\,dx} \cr & = \int_2^4 {\left( {1 + \frac{8}{{{x^2}}}} \right)dx} \cr & = \left[ {x - \frac{8}{x}} \right]_2^4 \cr & = 4 \cr} $$

Since, $$x = a$$  divides this area into two equal parts.
$$\therefore $$  Required area $$ = 2\int_2^a {y\,dx} $$
$$\eqalign{ & \therefore \,4 = 2\int_2^a {\left( {1 + \frac{8}{{{x^2}}}} \right)dx} \cr & \Rightarrow 2 = \left[ {x - \frac{8}{x}} \right]_2^a \cr & \Rightarrow 2 = \left( {a - \frac{8}{a}} \right) - \left( {2 - 4} \right) \cr & \Rightarrow {a^2} = 8 \cr & \therefore \,a = 2\sqrt 2 \cr} $$

Area of required region $$AOBC :$$
$$\eqalign{ & = \int\limits_0^1 {x\,dx} + \int\limits_1^e {\frac{1}{x}dx} \cr & = \frac{1}{2} + 1 \cr & = \frac{3}{2}\,\,{\text{sq}}{\text{. units}} \cr} $$

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
$$f\left( a \right) = a + \frac{1}{2}\left( {\sin \,a + a\,\cos \,a} \right) - \frac{\pi }{2}\sin \,a$$
Put $$a = \frac{\pi }{2}\,;\,f\left( {\frac{\pi }{2}} \right) = \frac{\pi }{2} + \frac{1}{2} - \frac{\pi }{2} = \frac{1}{2}$$

Given curves are $$y = {e^x}{\text{ and }}y = {e^{ - x}}$$
Now, $${e^x} = {e^{ - x}} \Rightarrow x = 0$$
$$\therefore $$  Area
$$\eqalign{ & = A = \int\limits_0^1 {\left( {{e^x} - {e^{ - x}}} \right)dx} \cr & = \left( {{e^x} + {e^{ - x}}} \right)_0^1 \cr & = \left[ {\left( {e + {e^{ - 1}}} \right) - \left( {{e^0} + {e^{ - 0}}} \right)} \right] \cr & = e + \frac{1}{e} - 2 \cr} $$