Application of Integration MCQ Questions & Answers in Calculus | Maths

The given curves are
$$y = {\left( {x + 1} \right)^2}.....(1)$$
upward parabola with vertex at $$\left( { - 1,\,0} \right)$$  meeting $$y$$-axis at $$\left( {0,\,1} \right)$$
$$y = {\left( {x - 1} \right)^2}.....(2)$$
upward parabola with vertex at $$\left( {1,\,0} \right)$$  meeting $$y$$-axis at $$\left( {1,\,0} \right)$$
$$y = \frac{1}{4}.....(3)$$
a line parallel to $$x$$-axis meeting (1) at $$\left( { - \frac{1}{2},\,\frac{1}{4}} \right),\,\left( { - \frac{3}{2},\,\frac{1}{4}} \right)$$
and meeting (2) at $$\left( {\frac{3}{2},\,\frac{1}{4}} \right),\left( {\frac{1}{2},\,\frac{1}{4}} \right)$$
The graph is as shown

The required area is the shaded portion given by ar $$\left( {BPCQB} \right) = 2Ar\left( {PQCP} \right)$$       (by symmetry)
$$\eqalign{ & = 2\left[ {\int\limits_0^{\frac{1}{2}} {\left( {{{\left( {x - 1} \right)}^2} - \frac{1}{4}} \right)dx} } \right] \cr & = 2\left[ {\left( {\frac{{{{\left( {x - 1} \right)}^3}}}{3} - \frac{x}{4}} \right)_0^{\frac{1}{2}}} \right] \cr & = 2\left[ {\left( { - \frac{1}{{24}} - \frac{1}{8}} \right) - \left( { - \frac{1}{3}} \right)} \right] \cr & = 2\left[ {\frac{{ - 1 - 3 + 8}}{{24}}} \right] \cr & = \frac{1}{3}{\text{ sq}}{\text{. units}} \cr} $$

$$\eqalign{ & {\text{Let }}\phi \left( x \right) = \int_a^x {f\left( t \right)dt.\,{\text{Then }}\phi } \left( { - x} \right) = \int_a^{ - x} {f\left( t \right)dt} \cr & \therefore \phi \left( { - x} \right) = \int_a^x {f\left( t \right)dt} + \int_x^{ - x} {f\left( t \right)dt} \cr & = \phi \left( x \right) + \int_x^0 {f\left( t \right)dt} + \int_0^{ - x} {f\left( t \right)dt} \cr & = \phi \left( x \right) - \int_0^x {f\left( t \right)dt} + \int_0^x { - f\left( { - z} \right)dz,{\text{ using }}t} = - z \cr & = \phi \left( x \right) - \int_0^x {\left\{ {f\left( t \right) + f\left( { - t} \right)} \right\}dt} \cr & = \phi \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because f\left( t \right) + f\left( { - t} \right) = 0,{\text{ from the question}}} \right) \cr & \therefore \,\,\,\phi \left( x \right){\text{ is even}}{\text{.}} \cr} $$

The given lines are
$$\eqalign{ & y = x - 1;\,\,\,y = - x - 1; \cr & y = x + 1\,{\text{and}}\,y = - x + 1 \cr} $$
which are two pairs of parallel lines and distance between the lines of each pair is $$\sqrt 2 .$$ Also non parallel lines are perpendicular. Thus lines represents a square of side $$\sqrt 2 .$$
Hence, area $$ = {\left( {\sqrt 2 } \right)^2} = 2$$   sq. units.

Given $$\int\limits_1^b {f\left( x \right)dx} = \sqrt {{b^2} + 1} - \sqrt 2 $$
Differentiate with respect to $$b$$
$$f\left( b \right) = \frac{b}{{\sqrt {{b^2} + 1} }} \Rightarrow f\left( x \right) = \frac{x}{{\sqrt {{x^2} + 1} }}$$

Given curves are $${x^2} + {y^2} = 1$$   and $${y^2} = 1 - x.$$
Intersecting points are $$x=0,\,1$$
Area of shaded portion is the required area.
So, Required Area $$=$$ Area of semi-circle $$+$$ Area bounded by parabola
$$\eqalign{ & = \frac{{\pi {r^2}}}{2} + 2\int\limits_0^1 {\sqrt {1 - x} \,dx} \cr & = \frac{\pi }{2} + 2\int\limits_0^1 {\sqrt {1 - x} \,dx} \,\,\,\,\,\left( {\because {\text{radius of circle}} = 1} \right) \cr & = \frac{\pi }{2} + 2\left[ {\frac{{{{\left( {1 - x} \right)}^{\frac{3}{2}}}}}{{ - \frac{3}{2}}}} \right]_0^1 \cr & = \frac{\pi }{2} - \frac{4}{3}\left( { - 1} \right) \cr & = \frac{\pi }{2} + \frac{4}{3}{\text{ sq}}{\text{. unit}} \cr} $$

$$\eqalign{ & \int_1^3 {\left\{ {2 - f\left( x \right)} \right\}dx = 6\,\,\,\,\,\,\,\,\,} \Rightarrow 2\left( {3 - 1} \right) - 6 = \int_1^3 {f\left( x \right)dx} \cr & \therefore \int_1^3 {f\left( x \right)dx = - 2} \cr & {\text{Now, }}\int_{ - 2}^1 {f\left( x \right)dx} = \int_{ - 2}^3 {f\left( x \right)dx} - \int_1^3 {f\left( x \right)dx} = 5 - \left( { - 2} \right) = 7 \cr} $$

$$\eqalign{ & I = \int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{x}{{1 + \sin \,x}}} dx \cr & I = \int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{\left( {\pi - x} \right)}}{{1 + \sin \,x}}} dx \cr & 2I = \int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{\pi }{{1 + \sin \,x}}} dx \cr & I = \frac{\pi }{2}\int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{{dx}}{{1 + \sin \,x}}} \cr & I = \frac{\pi }{2}\int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\left( {\frac{{1 - \sin \,x}}{{{{\cos }^2}x}}} \right)dx} \cr & I = \frac{\pi }{2}\int_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\left( {{{\sec }^2}x - \tan \,x\,\sec \,x} \right)dx} \cr & I = \frac{\pi }{2}\left[ {\tan \,x - \sec \,x} \right]_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} \cr & I = \frac{\pi }{2}\left[ {\left( { - 1 + \sqrt 2 } \right) - \left( {1 - \sqrt 2 } \right)} \right] \cr & I = \pi \left[ {\left( {\sqrt 2 - 1} \right)} \right] \cr} $$
Option A is correct answer.

$$\eqalign{ & I = \int_{ - 2}^0 {\left| {x\left( {x - 1} \right)} \right|dx} + \int_0^1 {\left| {x\left( {x - 1} \right)} \right|dx} + \int_1^2 {\left| {x\left( {x - 1} \right)} \right|dx} \cr & \,\,\,\,\, = \int_{ - 2}^0 {x\left( {x - 1} \right)dx} + \int_0^1 { - x\left( {x - 1} \right)dx} + \int_1^2 {x\left( {x - 1} \right)dx} \cr & \,\,\,\,\, = \left[ {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right]_{ - 2}^0 - \left[ {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right]_0^1 + \left[ {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right]_1^2 \cr & \,\,\,\,\, = 0 - \left( { - \frac{8}{3} - 2} \right) - \left( {\frac{1}{3} - \frac{1}{2}} \right) + \left( {\frac{8}{3} - 2} \right) - \left( {\frac{1}{3} - \frac{1}{2}} \right) \cr & \,\,\,\,\, = \frac{{17}}{3} \cr} $$

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

Given curve is $$y = 4x - {x^2} - 3$$
Since, area bounded by $$x$$-axis   $$\therefore \,y = 0$$
$$\eqalign{ & \Rightarrow 4x - {x^2} - 3 = 0 \cr & \Rightarrow {x^2} - 4x + 3 = 0 \cr & \Rightarrow {x^2} - 3x - x + 3 = 0 \cr & \Rightarrow \left( {x - 3} \right)\left( {x - 1} \right) = 0 \cr & \Rightarrow x = 1,\,3 \cr} $$
$$\therefore $$  Required area
$$\eqalign{ & = \int_1^3 {\left( {4x - {x^2} - 3} \right)dx} \cr & = \left. {\frac{{4{x^2}}}{2} - \frac{{{x^3}}}{3} - 3x} \right|_1^3 \cr & = \left( {\frac{{36}}{2} - \frac{{27}}{3} - 9} \right) - \left( {\frac{4}{2} - \frac{1}{3} - 3} \right) \cr & = \left( {18 - 9 - 9} \right) - \left( {2 - \frac{{10}}{3}} \right) \cr & = 0 - \left( {\frac{{ - 4}}{3}} \right) \cr & = \frac{4}{3}{\text{ sq}}{\text{. unit}} \cr} $$