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Class 12 Chapter 8 Maths Extra Questions
Application of Integrals Class 12 Maths Extra Questions
Chapter 8 Application of Integrals
The area bounded by the curves {tex}{y^2} = 20x{/tex} and {tex}{x^2} = 16y{/tex} is equal to
- {tex}\frac{{320}}{3}{\text{ }}sq.{\text{ }}units{/tex}
- {tex}80\pi {\text{ }}sq.{\text{ }}units{/tex}
- none of these
- {tex}100\pi {\text{ }}sq.{\text{ }}units{/tex}
The area of the region bounded by the parabola ( y – 2)2 = x – 1, the tangent to the parabola at the point ( 2 , 3 ) and the x – axis is equal to
- none of these
- 6 sq. units
- 9 sq. units
- 12 sq. units
The area bounded by the curves {tex}y = \sqrt x,{/tex} 2y + 3 = xand the x – axis in the first quadrant is
- 36
- 18
- 9
- none of these
If the area cut off from a parabola by any double ordinate is k times the corresponding rectangle contained by that double ordinate and its distance from the vertex, then k is equal to
- {tex}\frac{2}{3}{/tex}
- 3
- {tex}\frac{1}{3}{/tex}
- {tex}\frac{3}{2}{/tex}
The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and {tex}x = \frac{\pi }{2}{/tex}is equal to
- {tex}2(\sqrt 2 + 1){/tex} sq. units
- {tex}2(\sqrt 2 – 1){/tex} sq. units
- {tex}\left( {4\sqrt 2 – 1\;} \right){/tex} sq. units
- {tex}\left( {4\sqrt 2 + 1\;} \right){/tex} sq. units
The area of the bounded by the lines y = 2, x = 1, x = a and the curve y = f(x), which cuts the last two lines above the first line for all {tex}a\ge1{/tex}, is equal to {tex}\frac{2}{3}\left[{(2a)^{3/2}-3a+3-2\sqrt2}\right]{/tex}. Find f(x)
Let f(x) be a continuous function such that the area bounded by the curve y=f(x), x-axis and the lines x=0 and x=a is {tex}\frac{a^2}{2}+\frac{a}{2}sin\ a+\frac{π}{2}\ cos\ a,{/tex} then find {tex} f(\frac{π}{2}){/tex}.
Find the area of the region enclosed by the curves y = x , x = e, y = {tex}\frac{1}{x}{/tex} and the positive x-axis.
Calculate the area of the region enclosed between the circles: x2 + y2 = 16 and (x + 4)2 + y2 = 16.
Using integration, find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).
Find the area of the region {tex}\left\{ {\left( {x,y} \right);{x^2} \leqslant y \leqslant x} \right\}{/tex}.
Evaluate {tex}\lim_{x\to\infty} \left({\frac{x^x}{x!}}\right)^{1/x}{/tex}.
Evaluate {tex}\lim_{x\to\infty} \left[{\frac{1}{x}+\frac{x^2}{(x+1)^3}+\frac{x^2}{(x+2)^3}+………+\frac{1}{8x}}\right]{/tex}.
Find the area of the region enclosed by the parabola x2= y and the line y=x + 2.
Using integration, find the area of the region enclosed between the two circles x2 + y2 = 4 and (x – 2)2 + y2 = 4.
Chapter 8 Application of Integrals
Solution
- (a) {tex}\frac{{320}}{3}{\text{ }}sq.{\text{ }}units{/tex}
Explanation: Eliminating y, we get: {tex}{x^4} = 256 \times 20x{/tex}
{tex} \Rightarrow x = 0,x = 8{(10)^{\frac{1}{3}}}{/tex}
Required area:
{tex}=\int\limits_0^{8{{(10)}^{\frac{1}{3}}}} {\left( {\sqrt {20x} – \frac{{{x^2}}}{{16}}} \right)dx} {/tex}
{tex} = \frac{{640}}{3} – \frac{{320}}{3} = \frac{{320}}{3}{/tex} sq units - (c) 9 sq. units
Explanation: Given parabola is: {tex}{\left( {y – 2} \right)^2} = x – 1 \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{2(y – 2)}} {/tex}
When y= 3, x= 2
{tex}\; \therefore \frac{{dy}}{{dx}} = \frac{1}{2} \\{/tex}
Therefore, tangent at ( 2, 3 ) is y – 3 = ½ ( x – 2 ). i.e. x – 2y +4 = 0 . therefore required area is: {tex}\int \limits_0^3 {{{(y – 2)}^2} + 1.dy} – \int\limits_0^3 {(2y – 4)dy} {/tex}{tex}= \left[ {\frac{{{{(y – 2)}^3}}}{3} + y} \right]_0^3 – \left[ {{y^2} – 4y} \right]_0^3 = 9{/tex} - (c) 9
Explanation: Required area: {tex}\int\limits_0^9 {\sqrt x dx} – \int\limits_3^9 {\left( {\frac{{x – 3}}{2}} \right)dx} {/tex}{tex}= \left[ {\frac{{{x^{\frac{3}{2}}}}}{{3/2}}} \right]_0^9 – \frac{1}{2}\left[ {\frac{{{x^2}}}{2} – 3x} \right]_3^9 = 9sq.units{/tex} - (a) {tex}\frac{2}{3}{/tex}
Explanation: Required area: {tex}2\int\limits_0^a {\sqrt {4ax} dx} {/tex}
{tex}= k\alpha (2\sqrt {4a\alpha } ){/tex}
{tex}= \frac{{8\sqrt a }}{3}{\alpha ^{\frac{3}{2}}}{/tex}
{tex} = 4\sqrt a k{\alpha ^{\frac{3}{2}}} \Rightarrow k = \frac{2}{3} {/tex} - (b) {tex}2(\sqrt 2 – 1){/tex}sq. units
Explanation: Required area = {tex}\int\limits_0^{\frac{\pi }{2}} {\left| {\sin x – \cos x} \right|dx} {/tex}
{tex}= \int\limits_0^{\frac{\pi }{4}} {(\cos x – \sin x)dx + } \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {(sinx – \cos x)dx} {/tex}
{tex}= \left[ {\sin x + \cos x} \right]_0^{\frac{\pi }{4}} + \left[ { – cosx – sinx} \right]_{\frac{\pi }{4}}^{\frac{\pi }{2}} {/tex}
{tex}= \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} – (0 + 1) – \left\{ {1 – \left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)} \right\} {/tex}
{tex}= \frac{4}{{\sqrt 2 }} – 2 = 2\sqrt 2 – 2 = 2(\sqrt 2 – 1) {/tex} - we are given,
{tex}\int_a^1[f(x)-2]dx=\frac{2}{3}\left[{(2a)^{3/2}-3a+3-2\sqrt2}\right]{/tex}
Differentiating w.r.t a, we get
f(a) – 2 {tex}=\frac{2}{3}\left[{\frac{3}{2}\sqrt{2a}.2-3}\right]{/tex}
f(a)= 2{tex}\sqrt{2a},a\ge1{/tex}
{tex}\therefore\ f(x)=2\sqrt{2x},x\ge1{/tex} - we have, {tex}\int_0^af(x)dx=\frac{a^2}{2}+\frac{a}{2}sin\ a+\frac{\pi}{2}cos\ a{/tex}
Differentiating w.r.t a,we get,
f(a)=a+ {tex}\frac{1}{2}(sin \ a+acos\ a)-\frac{\pi}{2}sin\ a{/tex}
put a={tex}\frac{\pi}{2}, {/tex} {tex}f\left( {\frac{\pi }{2}} \right) = \frac{\pi }{2} + \frac{1}{2} – \frac{\pi }{2} = \frac{1}{2}{/tex} - We have {tex}y=4x^2\ and\ y=\frac{1}{9}x^2{/tex}
Required area ={tex}2\int_0^2\left({3\sqrt y-\frac{\sqrt y}{2}}\right)dy{/tex}
{tex}=2\left({\frac{5y}{2}\frac{\sqrt y}{3/2}}\right)_0^2{/tex}
{tex}=2.\frac{5}{3}2\sqrt2=\frac{20\sqrt2}{3}{/tex}
x2 + y2 = 16
(x + 4)2 + y2 = 16
Intersecting at x = -2
Area{tex} = 4\int_{ – 4}^{ – 2} {\sqrt {16 – {x^2}} dx}{/tex}
{tex}=4 \left[ {\int_{-4}^{-2} \sqrt {4^2-x^2}}dx \right] {/tex} {tex}= 4\left[ {\frac{x}{2}\sqrt{1-x^2}+\frac{4^2}{2}sin^{-1}\frac{x}{4}} \right] _{-4}^{-2}{/tex} {tex} = 4\left[ {( – 2\sqrt 3 – \frac{{4\pi }}{3}) – ( – 4\pi )} \right]{/tex}
{tex}= \left( { – 8\sqrt 3 + \frac{{32\pi }}{3}} \right){/tex}
A (-1, 0) B (1, 3) C (3, 2)
Equation of AB
{tex}y – {y_1} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\left( {x – {x_1}} \right){/tex}
{tex}y – 0 = \frac{{3 – 0}}{{1 + 1}}\left( {x + 1} \right){/tex}
{tex}y = \frac{3}{2}\left( {x + 1} \right){/tex}
Similarly,
Equation of BC {tex}y = \frac{{ – 1}}{2}\left( {x – 7} \right){/tex}
Equation of AC {tex}= \frac{1}{2}\left( {x + 1} \right){/tex}
Area {tex}\Delta ABC = \int_{ – 1}^1 {\frac{3}{2}\left( {x + 1} \right)dx + \int_1^3 {\frac{1}{2}\left( {x – 7} \right)dx} } {/tex} {tex} – \int_{ – 1}^3 {\frac{1}{2}\left( {x + 1} \right)dx} {/tex}
{tex}=\frac{3}{2} \left[ {\frac{x^2}{2}+x} \right] _{-1}^1+\frac{1}{2} \left[ {7x-\frac{x^2}{2}} \right] _1^3{/tex}{tex}- \left[ {\frac{x^2}{2}+x} \right] _{-1}^3{/tex}
{tex}=\frac{3}{2} \left[ {(\frac{1}{2}+1)-(\frac{1}{2}-1)} \right] +\frac{1}{2} \left[ {(21-\frac{9}{2})-(7-\frac{1}{2}} \right] {/tex}
{tex}-\frac{1}{2} \left[ {(\frac{9}{2}+3)-(\frac{1}{2}-1)} \right] {/tex}
{tex}=\frac{3}{2}(2)+\frac{1}{2}(10)-\frac{1}{2}(8)=3+5-4{/tex}
= 4 sq. units- y = x2
y = x
{tex} \Rightarrow {/tex} x = 0, y = 0
x = 1, y = 1
Area {tex} = \int_0^1 {xdx – \int_0^1 {{x^2}dx} } {/tex}
{tex}=\int_0^1(x-x^2)dx{/tex}
{tex}= \left[ {\frac{x^2}{2}-\frac{x^3}{3}} \right]_0^1{/tex}
{tex}=\frac{1}{2}-\frac{1}{3}{/tex}
{tex} = \frac{1}{6}{/tex} sq. units - Given {tex}L=\lim_{x\to\infty} \left({\frac{x^x}{x!}}\right)^{1/x}{/tex}
Taking logarithm on both sides
{tex}log \ L = \lim_{x\to\infty} \frac{1}{x}\left({log\frac{x}{1}+log\frac{x}{2}+…..+log\frac{x}{x}}\right){/tex}
= {tex}\lim_{x\to\infty} \frac{1}{x}\sum_{r=1}^xlog\ \frac{x}{r}{/tex}
={tex}\lim_{x\to\infty}\frac{1}{x}\sum_{r=1}^xlog\ \frac{1}{(r/x)}{/tex}
{tex}=\int_0^1log\frac{1}{x}\ dx{/tex}
{tex}=-\int_0^1log\ x\ dx{/tex}
{tex}=-[xlog\ x+x]_0^1{/tex}
{tex}=-[(1log\ 1+1)-(0\log0-0)]{/tex} = 1
{tex}\therefore{/tex} {tex}Log\ L=1\ {/tex}
{tex}\Rightarrow L=e\quad {/tex}
{tex}\Rightarrow \lim_{x\to\infty} \left({\frac{x^x}{x!}}\right)^{1/x}=e{/tex} - Given, {tex}\lim_{x\to\infty} \left[{\frac{1}{x}+\frac{x^2}{(x+1)^3}+\frac{x^2}{(x+2)^3}+………+\frac{1}{8x}}\right]{/tex}
{tex}=\lim_{x\to\infty} \sum_{r=0}^x\frac{x^2}{(x+r)^3}{/tex}
{tex}=\lim_{x\to\infty} \sum_{r=0}^x\frac{1/x}{(1+r/x)^2}{/tex}
{tex}=\int_0^1\frac{dy}{(1+y)^3},{/tex} replace {tex}\ \frac{r}{x}{/tex} by y and {tex}\frac {1}{x}{/tex} by dy
{tex}=\left[{\frac{-1}{2(1+y)^2}}\right]_0^1{/tex}
{tex}=\left[\frac{-1}{2(1+1^2)}-\frac{-1}{2(1+0^2)}\right]{/tex}
{tex}=\left[\frac{-1}{2(2)}-\frac{-1}{2(1)}\right]{/tex}
{tex}=\left[\frac{-1}{4}-\frac{-1}{2}\right]{/tex} {tex}=\frac{1}{4}{/tex} - We have, x2 = y and y = x + 2
{tex} \Rightarrow {x^2} = x + 2{/tex}
{tex}\Rightarrow {x^2} – x – 2 = 0{/tex}
{tex}\Rightarrow {x^2} – 2x + x – 2 = 0{/tex}
{tex}\Rightarrow x\left( {x – 2} \right) + 1\left( {x – 2} \right) = 0{/tex}
{tex}\Rightarrow \left( {x + 1} \right)\left( {x – 2} \right) = 0{/tex}
{tex}\Rightarrow x = – 1,2{/tex}
{tex}\therefore{/tex} Required area of shaded region, {tex}= \int_{ – 1}^2 {\left( {x + 2 – {x^2}} \right)dx = \left[ {\frac{{{x^2}}}{2} + 2x – \frac{{{x^3}}}{3}} \right]} _{ – 1}^2{/tex}
{tex}= \left( {8 – 3 – \frac{1}{2}} \right) = \frac{9}{2}{/tex} - Given circles are {tex}x^2 + y^2 = 4{/tex}…(i)
{tex}(x – 2)^2 + y^2 = 4{/tex}…(ii)
Eq. (i) is a circle with centre origin and
Radius = 2.
Eq. (ii) is a circle with centre C (2, 0) and
Radius = 2.
On solving Eqs. (i) and (ii), we get
{tex}(x – 2)^2+ y^2 = x^2+ y^2{/tex}
{tex} \Rightarrow {x^2}{\text{ – }}4x + 4 + {y^2} = {x^2} + {y^2}{/tex}
{tex}\Rightarrow x = 1{/tex}
On putting x = 1 in Eq. (i), we get
{tex}y = \pm \sqrt { 3 }{/tex}
Thus, the points of intersection of the given circles are A (1, {tex}\sqrt3{/tex}) and A'(1,-{tex}\sqrt3{/tex}).
Clearly, required area= Area of the enclosed region OACA’O between circles
= 2 [ Area of the region ODCAO] =2 [Area of the region ODAO + Area of the region DCAD] {tex}= 2 \left[ \int _ { 0 } ^ { 1 } y _ { 2 } d x + \int _ { 1 } ^ { 2 } y _ { 1 } d x \right]{/tex}
{tex}= 2 \left[ \int _ { 0 } ^ { 1 } \sqrt { 4 – ( x – 2 ) ^ { 2 } } d x + \int _ { 1 } ^ { 2 } \sqrt { 4 – x ^ { 2 } } d x \right]{/tex}
{tex}= 2 \left[ \frac { 1 } { 2 } ( x – 2 ) \sqrt { 4 – ( x – 2 ) ^ { 2 } } + \frac { 1 } { 2 } \times 4 \sin ^ { – 1 } \left( \frac { x – 2 } { 2 } \right) \right] _ { 0 } ^ { 1 }{/tex}{tex}+ 2 \left[ \frac { 1 } { 2 } x \sqrt { 4 – x ^ { 2 } } + \frac { 1 } { 2 } \times 4 \sin ^ { – 1 } \frac { x } { 2 } \right] _ { 1 } ^ { 2 }{/tex}
{tex}= \left[ ( x – 2 ) \sqrt { 4 – ( x – 2 ) ^ { 2 } } + 4 \sin ^ { – 1 } \left( \frac { x – 2 } { 2 } \right) \right] _ { 0 } ^ { 1 }{/tex}{tex}+ \left[ x \sqrt { 4 – x ^ { 2 } } + 4 \sin ^ { – 1 } \frac { x } { 2 } \right] _ { 1 } ^ { 2 } {/tex}
{tex}= \left[ \left\{ – \sqrt { 3 } + 4 \sin ^ { – 1 } \left( \frac { – 1 } { 2 } \right) \right\} – 0 – 4 \sin ^ { – 1 } ( – 1 ) \right]{/tex}{tex}+ \left[ 0 + 4 \sin ^ { – 1 } 1 – \sqrt { 3 } – 4 \sin ^ { – 1 } \frac { 1 } { 2 } \right]{/tex}
{tex}= \left[ \left( – \sqrt { 3 } – 4 \times \frac { \pi } { 6 } \right) + 4 \times \frac { \pi } { 2 } \right] +{/tex}{tex}\left[ 4 \times \frac { \pi } { 2 } – \sqrt { 3 } – 4 \times \frac { \pi } { 6 } \right]{/tex}
{tex}= \left( – \sqrt { 3 } – \frac { 2 \pi } { 3 } + 2 \pi \right) + \left( 2 \pi – \sqrt { 3 } – \frac { 2 \pi } { 3 } \right){/tex}
{tex}= \frac { 8 \pi } { 3 } – 2 \sqrt { 3 }{/tex} sq units.
Chapter Wise Important Questions Class 12 Maths Part I and Part II
- Relations and Functions
- Inverse Trigonometric Functions
- Matrices
- Determinants
- Continuity and Differentiability
- Application of Derivatives
- Integrals
- Application of Integrals
- Differential Equations
- Vector Algebra
- Three Dimensional Geometry
- Linear Programming
- Probability
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