CBSE Class 12 Maths Important Questions Application of Integrals

CBSE Class 12 Maths Important Questions Application of Integrals. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in myCBSEguide website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 8 Application of Integrals Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Chapter 8 Maths Extra Questions

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Application of Integrals Class 12 Maths Extra Questions

Chapter 8 Application of Integrals

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1. The area bounded by the curves {tex}{y^2} = 20x{/tex} and {tex}{x^2} = 16y{/tex} is equal to

   1. {tex}\frac{{320}}{3}{\text{ }}sq.{\text{ }}units{/tex}
   2. {tex}80\pi {\text{ }}sq.{\text{ }}units{/tex}
   3. none of these
   4. {tex}100\pi {\text{ }}sq.{\text{ }}units{/tex}

2. 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

   1. none of these
   2. 6 sq. units
   3. 9 sq. units
   4. 12 sq. units

3. The area bounded by the curves {tex}y = \sqrt x,{/tex} 2y + 3 = xand the x – axis in the first quadrant is

   1. 36
   2. 18
   3. 9
   4. none of these

4. 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

   1. {tex}\frac{2}{3}{/tex}
   2. 3
   3. {tex}\frac{1}{3}{/tex}
   4. {tex}\frac{3}{2}{/tex}

5. 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

   1. {tex}2(\sqrt 2 + 1){/tex} sq. units
   2. {tex}2(\sqrt 2 – 1){/tex} sq. units
   3. {tex}\left( {4\sqrt 2 – 1\;} \right){/tex} sq. units
   4. {tex}\left( {4\sqrt 2 + 1\;} \right){/tex} sq. units

6. 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)

7. 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}.

8. 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.

9. Calculate the area of the region enclosed between the circles: x² + y² = 16 and (x + 4)² + y² = 16.

10. Using integration, find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).

11. Find the area of the region {tex}\left\{ {\left( {x,y} \right);{x^2} \leqslant y \leqslant x} \right\}{/tex}.

12. Evaluate {tex}\lim_{x\to\infty} \left({\frac{x^x}{x!}}\right)^{1/x}{/tex}.

13. 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}.

14. Find the area of the region enclosed by the parabola x²= y and the line y=x + 2.

15. 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

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Solution

1. (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
2. (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}
3. (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}
4. (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}
5. (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}
6. 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}
7. 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}
8. 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}
9. 
   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}
10. 
    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
11. 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
12. 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}
13. 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}
14. 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}
15. 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

1. Relations and Functions
2. Inverse Trigonometric Functions
3. Matrices
4. Determinants
5. Continuity and Differentiability
6. Application of Derivatives
7. Integrals
8. Application of Integrals
9. Differential Equations
10. Vector Algebra
11. Three Dimensional Geometry
12. Linear Programming
13. Probability