Application of Integration MCQ Questions & Answers in Calculus | Maths

The required area
$$\eqalign{ & {\text{ = }}\int_2^{\frac{5}{2}} {x\,dx - } \int_2^{\frac{5}{2}} {\left[ {2 - \sqrt {1 - {{\left( {x - 3} \right)}^2}} } \right]dx} \cr & = \left[ {\frac{{{x^2}}}{2}} \right]_2^{\frac{5}{2}} - \left[ {2x} \right]_2^{\frac{5}{2}} + \left[ {\frac{{x - 3}}{2}\sqrt {1 - {{\left( {x - 3} \right)}^2}} + \frac{1}{2}{{\sin }^{ - 1}}\left( {x - 3} \right)} \right]_2^{\frac{5}{2}} \cr & = \frac{9}{8} - 1 + \left( { - \frac{{\sqrt 3 }}{8} + \frac{\pi }{6}} \right) \cr & = \frac{\pi }{6} - \frac{{\sqrt 3 - 1}}{8}{\text{ sq}}{\text{. unit}} \cr} $$

The given parabola is $${\left( {y - 2} \right)^2} = x - 1$$
Vertex $$\left( {1,\,2} \right)$$  and it meets $$x$$-axis at $$\left( {5,\,0} \right)$$
Also it gives $${y^2} - 4y - x + 5 = 0$$
So, that equation of tangent to the parabola at $$\left( {2,\,3} \right)$$  is
$$\eqalign{ & y.3 - 2\left( {y + 3} \right) - \frac{1}{2}\left( {x + 2} \right) + 5 = 0 \cr & {\text{or }}x - 2y + 4 = 0\,\,{\text{which meets }}x{\text{ - axis at }}\left( { - 4,\,0} \right) \cr} $$

In the figure shaded area is the required area.
Let us draw $$PD$$  perpendicular to $$y$$-axis.
Then required area
$$\eqalign{ & = Ar\left( {\Delta BOA} \right) + Ar\left( {OCPD} \right) - Ar\left( {\Delta APD} \right) \cr & = \frac{1}{2} \times 4 \times 2 + \int_0^3 {x\,dy - \frac{1}{2} \times 2 \times 1} \cr & = 3 + \int_0^3 {{{\left( {y - 2} \right)}^2} + 1\,dy} \cr & = 3 + \left[ {\frac{{{{\left( {y - 2} \right)}^3}}}{3} + y} \right]_0^3 \cr & = 3 + \left[ {\frac{1}{3} + 3 + \frac{8}{3}} \right] \cr & = 3 + 6 \cr & = 9\,{\text{sq}}{\text{. units}} \cr} $$

$$\eqalign{ & \left| x \right|\,{\text{for }}x \geqslant 0 \cr & = x{\text{ and }}\left| {x - 1} \right|{\text{ for }}x \leqslant 1 = - \left( {x - 1} \right), \cr & {\text{So, }}\int_0^1 {\left( {\left| x \right| + \left| {x - 11} \right|} \right)} = {\text{required area}} \cr & a = \int_0^1 {x\,dx} - \int_0^1 {\left( {x - 1} \right)dx} \cr & \,\,\,\,\, = \left[ {\frac{{{x^2}}}{2}} \right]_0^1 - \left[ {\frac{{{x^2}}}{2} - x} \right]_0^1 \cr & \,\,\,\,\, = \frac{1}{2} - \left( {\frac{1}{2} - 1} \right) \cr & \,\,\,\,\, = 1{\text{ sq}}{\text{. unit}} \cr} $$

Area $$ABE$$  (under parabola) $$ = \int_{ - 4}^{ - 3} {\sqrt { - x - 3} dx} = \frac{2}{3}$$
Area $$BCD$$  (under parabola) $$ = \int_{ - 3}^1 {\sqrt {x + 3} dx} = \frac{{16}}{3}$$
Area of trapezium $$ACDE = \frac{1}{2}\left( {1 + 2} \right)5 = \frac{{15}}{2}$$
Required area $$ = \frac{{15}}{2} - \frac{{16}}{3} - \frac{2}{3} = \frac{3}{2}$$

$$\eqalign{ & f'\left( x \right) = x\left( {{x^2} - 3x + 2} \right) = x\left( {x - 1} \right)\left( {x - 2} \right) \cr & {\text{The sign scheme for}}\,{\text{ }}f'\left( x \right){\text{ is as below}}{\text{.}} \cr} $$

$$\eqalign{ & \therefore f'\left( x \right) \leqslant 0{\text{ in }}1 \leqslant x \leqslant 2{\text{ and }}f'\left( x \right) \geqslant 0{\text{ in }}2 \leqslant x \leqslant 3 \cr & \therefore f\left( x \right)\,{\text{is m}}{\text{.d}}{\text{. in }}\left[ {1,\,2} \right]{\text{ and m}}{\text{.i}}{\text{. in }}\left[ {2,\,3} \right] \cr & \therefore \min f\left( x \right) = f\left( 2 \right) = \int_1^2 {x\left( {{x^2} - 3x + 2} \right)dx} \cr & = \left[ {\frac{{{x^4}}}{4} - {x^3} + {x^2}} \right]_1^2 \cr & = - \frac{1}{4} \cr & \max f\left( x \right) = {\text{the greatest among}}\left\{ {f\left( 1 \right),\,f\left( 3 \right)} \right\} \cr & f\left( 1 \right) = \int_1^1 {x\left( {{x^2} - 3x + 2} \right)dx} = 0 \cr & f\left( 3 \right) = \int_1^3 {x\left( {{x^2} - 3x + 2} \right)dx} = \left[ {\frac{{{x^4}}}{4} - {x^3} + {x^2}} \right]_1^3 = 2 \cr & \therefore \max f\left( x \right) = 2.\,\,{\text{So, the range }} = \left[ { - \frac{1}{4},\,2} \right] \cr} $$

$$\eqalign{ & I = \int_0^\pi {{{\sin }^6}x.{{\cos }^5}x\,dx} = \int_0^\pi {{{\sin }^6}\left( {\pi - x} \right).{{\cos }^5}\left( {\pi - x} \right)\,dx} = - I \cr & \therefore I = 0 \cr} $$

Given equation of circle is $${x^2} + {y^2} = 2 \Rightarrow y = \sqrt {2 - {x^2}} $$

Required area $$ = 4 \times $$  Area of shaded portion
$$\eqalign{ & = 4\int_0^{\sqrt 2 } {\sqrt {2 - {x^2}} } dx \cr & = 4\left[ {\frac{x}{2}\sqrt {2 - {x^2}} + \frac{2}{2}{{\sin }^{ - 1}}\frac{x}{{\sqrt 2 }}} \right]_0^{\sqrt 2 } \cr & = 4\left[ {{{\sin }^{ - 1}}\frac{{\sqrt 2 }}{{\sqrt 2 }}} \right] \cr & = 4\,{\sin ^{ - 1}}1 \cr & = 4 \times \frac{\pi }{2} \cr & = 2\pi {\text{ sq}}{\text{. units}} \cr} $$

The graph of the curve $$y = {\log _e}\left( {x + e} \right)$$     is as shown in the figure

$$\eqalign{ & {\text{Required area}}\,{\text{:}} \cr & A = \int\limits_{1 - e}^0 {y\,dx} = \int\limits_{1 - e}^0 {{{\log }_e}\left( {x + e} \right)dx} \cr & {\text{Put}}\,x + e = t\,\, \Rightarrow dx = dt \cr & {\text{Also}}\,x = 1 - e,\,\,t = 1 \cr & {\text{At}}\,x = 0,\,t = e \cr & \therefore A = \int\limits_1^e {{{\log }_e}t\,dt} = \left[ {t\,{{\log }_e}t - t} \right]_1^e \cr & e - e - 0 + 1 = 1 \cr} $$
Hence the required area is 1 square unit.

$$\eqalign{ & {\text{Let, }}I = \int_0^\pi {\frac{{dx}}{{1 + {3^{\cos \,x}}}}} .....\left( 1 \right) \cr & {\text{Using above formula,}} \cr & I = \int_0^\pi {\frac{{dx}}{{1 + {3^{\cos \,\left( {\pi - x} \right)}}}}} = \int_0^\pi {\frac{{{3^{\cos \,x}}dx}}{{1 + {3^{\cos \,x}}}}} ......\left( 2 \right) \cr & {\text{Adding equation }}\left( 1 \right){\text{ and }}\left( 2 \right){\text{ we get,}} \cr & I = \frac{1}{2}\int_0^\pi {dx} = \frac{\pi }{2} \cr} $$

Let $$f\left( x \right) = {\log _e}\left( {x + \sqrt {1 + {x^2}} } \right).$$      Clearly, $$f\left( { - x} \right) = - f\left( { - x} \right).$$
So, $$f\left( x \right)$$  is an odd function.
$$\therefore \,I = 0$$