{"id":27921,"date":"2019-10-19T10:41:37","date_gmt":"2019-10-19T05:11:37","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=27921"},"modified":"2019-10-25T11:51:01","modified_gmt":"2019-10-25T06:21:01","slug":"cbse-class-12-maths-important-questions-application-of-integrals","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-class-12-maths-important-questions-application-of-integrals\/","title":{"rendered":"CBSE Class 12 Maths Important Questions Application of Integrals"},"content":{"rendered":"

CBSE Class 12 Maths Important Questions Application of Integrals. <\/strong>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\u00a0myCBSEguide<\/a>\u00a0<\/strong>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<\/a><\/strong>\u00a0and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.<\/p>\n

Class 12 Chapter 8 Maths Extra Questions<\/strong><\/p>\n

Download as PDF<\/a><\/strong><\/p>\n

Application of Integrals Class 12 Maths Extra Questions<\/h2>\n

Chapter 8 Application of Integrals<\/strong><\/p>\n

\n
    \n
  1. \n
    \n
    \n

    The area bounded by the curves {tex}{y^2} = 20x{\/tex}<\/span> and {tex}{x^2} = 16y{\/tex}<\/span> is equal to<\/p>\n<\/div>\n<\/div>\n

      \n
    1. {tex}\\frac{{320}}{3}{\\text{ }}sq.{\\text{ }}units{\/tex}<\/span><\/li>\n
    2. {tex}80\\pi {\\text{ }}sq.{\\text{ }}units{\/tex}<\/span><\/li>\n
    3. none of these<\/li>\n
    4. {tex}100\\pi {\\text{ }}sq.{\\text{ }}units{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
    5. \n
      \n
      \n

      The area of the region bounded by the parabola ( y – 2)2<\/sup> = x – 1, the tangent to the parabola at the point ( 2 , 3 ) and the x \u2013 axis is equal to<\/p>\n<\/div>\n<\/div>\n

        \n
      1. none of these<\/li>\n
      2. 6 sq. units<\/li>\n
      3. 9 sq. units<\/li>\n
      4. 12 sq. units<\/li>\n<\/ol>\n<\/li>\n
      5. \n
        \n
        \n

        The area bounded by the curves {tex}y = \\sqrt x,{\/tex}<\/span> 2y + 3 = xand the x \u2013 axis in the first quadrant is<\/p>\n<\/div>\n<\/div>\n

          \n
        1. 36<\/li>\n
        2. 18<\/li>\n
        3. 9<\/li>\n
        4. none of these<\/li>\n<\/ol>\n<\/li>\n
        5. \n
          \n
          \n

          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<\/p>\n<\/div>\n<\/div>\n

            \n
          1. {tex}\\frac{2}{3}{\/tex}<\/span><\/li>\n
          2. 3<\/li>\n
          3. {tex}\\frac{1}{3}{\/tex}<\/span><\/li>\n
          4. {tex}\\frac{3}{2}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
          5. \n
            \n
            \n

            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}<\/span>is equal to<\/p>\n<\/div>\n<\/div>\n

              \n
            1. {tex}2(\\sqrt 2 + 1){\/tex}<\/span> sq. units<\/li>\n
            2. {tex}2(\\sqrt 2 – 1){\/tex}<\/span> sq. units<\/li>\n
            3. {tex}\\left( {4\\sqrt 2 – 1\\;} \\right){\/tex}<\/span> sq. units<\/li>\n
            4. {tex}\\left( {4\\sqrt 2 + 1\\;} \\right){\/tex}<\/span> sq. units<\/li>\n<\/ol>\n<\/li>\n
            5. \n
              \n
              \n

              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}<\/span>, is equal to {tex}\\frac{2}{3}\\left[{(2a)^{3\/2}-3a+3-2\\sqrt2}\\right]{\/tex}<\/span>. Find f(x)<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            6. \n
              \n
              \n

              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{\u03c0}{2}\\ cos\\ a,{\/tex}<\/span> then find {tex} f(\\frac{\u03c0}{2}){\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            7. \n
              \n
              \n

              Find the area of the region enclosed by the curves y = x , x = e, y = {tex}\\frac{1}{x}{\/tex}<\/span> and the positive x-axis.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            8. \n
              \n
              \n

              Calculate the area of the region enclosed between the circles: x2<\/sup> + y2<\/sup> = 16 and (x + 4)2<\/sup> + y2<\/sup> = 16.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            9. \n
              \n
              \n

              Using integration, find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            10. \n
              \n
              \n

              Find the area of the region {tex}\\left\\{ {\\left( {x,y} \\right);{x^2} \\leqslant y \\leqslant x} \\right\\}{\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            11. \n
              \n
              \n

              Evaluate {tex}\\lim_{x\\to\\infty} \\left({\\frac{x^x}{x!}}\\right)^{1\/x}{\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            12. \n
              \n
              \n

              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}<\/span><\/u>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            13. \n
              \n
              \n

              Find the area of the region enclosed by the parabola x2<\/sup>= y and the line y=x + 2.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            14. \n
              \n
              \n

              Using integration, find the area of the region enclosed between the two circles x2<\/sup> + y2<\/sup> = 4 and (x – 2)2<\/sup> + y2<\/sup> = 4.<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n

              Chapter 8 Application of Integrals<\/strong><\/p>\n

              \n

              Solution<\/strong><\/p>\n

                \n
              1. (a) {tex}\\frac{{320}}{3}{\\text{ }}sq.{\\text{ }}units{\/tex}<\/span>
                \nExplanation: <\/strong>Eliminating y, we get: {tex}{x^4} = 256 \\times 20x{\/tex}<\/span>
                \n{tex} \\Rightarrow x = 0,x = 8{(10)^{\\frac{1}{3}}}{\/tex}<\/span>
                \nRequired area:
                \n{tex}=\\int\\limits_0^{8{{(10)}^{\\frac{1}{3}}}} {\\left( {\\sqrt {20x} – \\frac{{{x^2}}}{{16}}} \\right)dx} {\/tex}<\/span>
                \n{tex} = \\frac{{640}}{3} – \\frac{{320}}{3} = \\frac{{320}}{3}{\/tex}<\/span> sq units<\/li>\n
              2. (c) 9 sq. units
                \nExplanation: <\/strong>Given parabola is: {tex}{\\left( {y – 2} \\right)^2} = x – 1 \\Rightarrow \\frac{{dy}}{{dx}} = \\frac{1}{{2(y – 2)}} {\/tex}<\/span>
                \nWhen y= 3, x= 2
                \n{tex}\\; \\therefore \\frac{{dy}}{{dx}} = \\frac{1}{2} \\\\{\/tex}<\/span>
                \nTherefore, tangent at ( 2, 3 ) is y \u2013 3 = \u00bd ( x \u2013 2 ). i.e. x \u2013 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}<\/span>{tex}= \\left[ {\\frac{{{{(y – 2)}^3}}}{3} + y} \\right]_0^3 – \\left[ {{y^2} – 4y} \\right]_0^3 = 9{\/tex}<\/span><\/li>\n
              3. (c) 9
                \nExplanation: <\/strong>Required area: {tex}\\int\\limits_0^9 {\\sqrt x dx} – \\int\\limits_3^9 {\\left( {\\frac{{x – 3}}{2}} \\right)dx} {\/tex}<\/span>{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}<\/span><\/li>\n
              4. (a) {tex}\\frac{2}{3}{\/tex}<\/span>
                \nExplanation: <\/strong>Required area: {tex}2\\int\\limits_0^a {\\sqrt {4ax} dx} {\/tex}<\/span>
                \n{tex}= k\\alpha (2\\sqrt {4a\\alpha } ){\/tex}<\/span>
                \n{tex}= \\frac{{8\\sqrt a }}{3}{\\alpha ^{\\frac{3}{2}}}{\/tex}<\/span>
                \n{tex} = 4\\sqrt a k{\\alpha ^{\\frac{3}{2}}} \\Rightarrow k = \\frac{2}{3} {\/tex}<\/span><\/li>\n
              5. (b) {tex}2(\\sqrt 2 – 1){\/tex}<\/span>sq. units
                \nExplanation: <\/strong>Required area = {tex}\\int\\limits_0^{\\frac{\\pi }{2}} {\\left| {\\sin x – \\cos x} \\right|dx} {\/tex}<\/span>
                \n{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}<\/span>
                \n{tex}= \\left[ {\\sin x + \\cos x} \\right]_0^{\\frac{\\pi }{4}} + \\left[ { – cosx – sinx} \\right]_{\\frac{\\pi }{4}}^{\\frac{\\pi }{2}} {\/tex}<\/span>
                \n{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}<\/span>
                \n{tex}= \\frac{4}{{\\sqrt 2 }} – 2 = 2\\sqrt 2 – 2 = 2(\\sqrt 2 – 1) {\/tex}<\/span><\/li>\n
              6. we are given,
                \n{tex}\\int_a^1[f(x)-2]dx=\\frac{2}{3}\\left[{(2a)^{3\/2}-3a+3-2\\sqrt2}\\right]{\/tex}<\/span>
                \nDifferentiating w.r.t a, we get
                \nf(a) – 2 {tex}=\\frac{2}{3}\\left[{\\frac{3}{2}\\sqrt{2a}.2-3}\\right]{\/tex}<\/span>
                \nf(a)= 2{tex}\\sqrt{2a},a\\ge1{\/tex}<\/span>
                \n{tex}\\therefore\\ f(x)=2\\sqrt{2x},x\\ge1{\/tex}<\/span><\/li>\n
              7. we have, {tex}\\int_0^af(x)dx=\\frac{a^2}{2}+\\frac{a}{2}sin\\ a+\\frac{\\pi}{2}cos\\ a{\/tex}<\/span>
                \nDifferentiating w.r.t a,we get,
                \nf(a)=a+ {tex}\\frac{1}{2}(sin \\ a+acos\\ a)-\\frac{\\pi}{2}sin\\ a{\/tex}<\/span>
                \nput a={tex}\\frac{\\pi}{2}, {\/tex}<\/span> {tex}f\\left( {\\frac{\\pi }{2}} \\right) = \\frac{\\pi }{2} + \\frac{1}{2} – \\frac{\\pi }{2} = \\frac{1}{2}{\/tex}<\/span><\/li>\n
              8. We have {tex}y=4x^2\\ and\\ y=\\frac{1}{9}x^2{\/tex}<\/span>
                \n\"CBSE
                \nRequired area ={tex}2\\int_0^2\\left({3\\sqrt y-\\frac{\\sqrt y}{2}}\\right)dy{\/tex}<\/span>
                \n{tex}=2\\left({\\frac{5y}{2}\\frac{\\sqrt y}{3\/2}}\\right)_0^2{\/tex}<\/span>
                \n{tex}=2.\\frac{5}{3}2\\sqrt2=\\frac{20\\sqrt2}{3}{\/tex}<\/span><\/li>\n
              9. \"CBSE
                \nx2<\/sup> + y2<\/sup> = 16
                \n(x + 4)2<\/sup> + y2<\/sup> = 16
                \nIntersecting at x = -2
                \nArea{tex} = 4\\int_{ – 4}^{ – 2} {\\sqrt {16 – {x^2}} dx}{\/tex}<\/span>
                \n{tex}=4 \\left[ {\\int_{-4}^{-2} \\sqrt {4^2-x^2}}dx \\right] {\/tex}<\/span> {tex}= 4\\left[ {\\frac{x}{2}\\sqrt{1-x^2}+\\frac{4^2}{2}sin^{-1}\\frac{x}{4}} \\right] _{-4}^{-2}{\/tex}<\/span> {tex} = 4\\left[ {( – 2\\sqrt 3 – \\frac{{4\\pi }}{3}) – ( – 4\\pi )} \\right]{\/tex}<\/span>
                \n{tex}= \\left( { – 8\\sqrt 3 + \\frac{{32\\pi }}{3}} \\right){\/tex}<\/span><\/li>\n
              10. \"CBSE
                \nA (-1, 0) B (1, 3) C (3, 2)
                \nEquation of AB
                \n{tex}y – {y_1} = \\frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\\left( {x – {x_1}} \\right){\/tex}<\/span>
                \n{tex}y – 0 = \\frac{{3 – 0}}{{1 + 1}}\\left( {x + 1} \\right){\/tex}<\/span>
                \n{tex}y = \\frac{3}{2}\\left( {x + 1} \\right){\/tex}<\/span>
                \nSimilarly,
                \nEquation of BC {tex}y = \\frac{{ – 1}}{2}\\left( {x – 7} \\right){\/tex}<\/span>
                \nEquation of AC {tex}= \\frac{1}{2}\\left( {x + 1} \\right){\/tex}<\/span>
                \nArea {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}<\/span> {tex} – \\int_{ – 1}^3 {\\frac{1}{2}\\left( {x + 1} \\right)dx} {\/tex}<\/span>
                \n{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}<\/span>{tex}- \\left[ {\\frac{x^2}{2}+x} \\right] _{-1}^3{\/tex}<\/span>
                \n{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}<\/span>
                \n{tex}-\\frac{1}{2} \\left[ {(\\frac{9}{2}+3)-(\\frac{1}{2}-1)} \\right] {\/tex}<\/span>
                \n{tex}=\\frac{3}{2}(2)+\\frac{1}{2}(10)-\\frac{1}{2}(8)=3+5-4{\/tex}<\/span>
                \n= 4 sq. units<\/li>\n
              11. y = x2<\/sup>
                \n\"CBSE
                \ny = x
                \n{tex} \\Rightarrow {\/tex}<\/span> x = 0, y = 0
                \nx = 1, y = 1
                \nArea {tex} = \\int_0^1 {xdx – \\int_0^1 {{x^2}dx} } {\/tex}<\/span>
                \n{tex}=\\int_0^1(x-x^2)dx{\/tex}<\/span>
                \n{tex}= \\left[ {\\frac{x^2}{2}-\\frac{x^3}{3}} \\right]_0^1{\/tex}<\/span>
                \n{tex}=\\frac{1}{2}-\\frac{1}{3}{\/tex}<\/span>
                \n{tex} = \\frac{1}{6}{\/tex}<\/span> sq. units<\/li>\n
              12. Given {tex}L=\\lim_{x\\to\\infty} \\left({\\frac{x^x}{x!}}\\right)^{1\/x}{\/tex}<\/span>
                \nTaking logarithm on both sides
                \n{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}<\/span>
                \n= {tex}\\lim_{x\\to\\infty} \\frac{1}{x}\\sum_{r=1}^xlog\\ \\frac{x}{r}{\/tex}<\/span>
                \n={tex}\\lim_{x\\to\\infty}\\frac{1}{x}\\sum_{r=1}^xlog\\ \\frac{1}{(r\/x)}{\/tex}<\/span>
                \n{tex}=\\int_0^1log\\frac{1}{x}\\ dx{\/tex}<\/span>
                \n{tex}=-\\int_0^1log\\ x\\ dx{\/tex}<\/span>
                \n{tex}=-[xlog\\ x+x]_0^1{\/tex}<\/span>
                \n{tex}=-[(1log\\ 1+1)-(0\\log0-0)]{\/tex}<\/span>\u00a0= 1
                \n{tex}\\therefore{\/tex}<\/span> {tex}Log\\ L=1\\ {\/tex}<\/span>
                \n{tex}\\Rightarrow L=e\\quad {\/tex}<\/span>
                \n{tex}\\Rightarrow \\lim_{x\\to\\infty} \\left({\\frac{x^x}{x!}}\\right)^{1\/x}=e{\/tex}<\/span><\/li>\n
              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}<\/span>
                \n{tex}=\\lim_{x\\to\\infty} \\sum_{r=0}^x\\frac{x^2}{(x+r)^3}{\/tex}<\/span>
                \n{tex}=\\lim_{x\\to\\infty} \\sum_{r=0}^x\\frac{1\/x}{(1+r\/x)^2}{\/tex}<\/span>
                \n{tex}=\\int_0^1\\frac{dy}{(1+y)^3},{\/tex}<\/span> replace {tex}\\ \\frac{r}{x}{\/tex}<\/span> by y and {tex}\\frac {1}{x}{\/tex}<\/span> by dy
                \n{tex}=\\left[{\\frac{-1}{2(1+y)^2}}\\right]_0^1{\/tex}<\/span>
                \n{tex}=\\left[\\frac{-1}{2(1+1^2)}-\\frac{-1}{2(1+0^2)}\\right]{\/tex}<\/span>
                \n{tex}=\\left[\\frac{-1}{2(2)}-\\frac{-1}{2(1)}\\right]{\/tex}<\/span>
                \n{tex}=\\left[\\frac{-1}{4}-\\frac{-1}{2}\\right]{\/tex}<\/span> {tex}=\\frac{1}{4}{\/tex}<\/span><\/li>\n
              14. We have, x2<\/sup> = y and y = x + 2
                \n{tex} \\Rightarrow {x^2} = x + 2{\/tex}<\/span>
                \n{tex}\\Rightarrow {x^2} – x – 2 = 0{\/tex}<\/span>
                \n{tex}\\Rightarrow {x^2} – 2x + x – 2 = 0{\/tex}<\/span>
                \n{tex}\\Rightarrow x\\left( {x – 2} \\right) + 1\\left( {x – 2} \\right) = 0{\/tex}<\/span>
                \n{tex}\\Rightarrow \\left( {x + 1} \\right)\\left( {x – 2} \\right) = 0{\/tex}<\/span>
                \n{tex}\\Rightarrow x = – 1,2{\/tex}<\/span>
                \n\"CBSE
                \n{tex}\\therefore{\/tex}<\/span> 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}<\/span>
                \n{tex}= \\left( {8 – 3 – \\frac{1}{2}} \\right) = \\frac{9}{2}{\/tex}<\/span><\/li>\n
              15. Given circles are {tex}x^2 + y^2 = 4{\/tex}<\/span>…(i)
                \n{tex}(x – 2)^2 + y^2 = 4{\/tex}<\/span>…(ii)
                \nEq. (i) is a circle with centre origin and
                \nRadius = 2.
                \nEq. (ii) is a circle with centre C (2, 0) and
                \nRadius = 2.
                \nOn solving Eqs. (i) and (ii), we get
                \n{tex}(x – 2)^2+ y^2\u00a0= x^2+ y^2{\/tex}<\/span>
                \n{tex} \\Rightarrow {x^2}{\\text{ – }}4x + 4 + {y^2} = {x^2} + {y^2}{\/tex}<\/span>
                \n{tex}\\Rightarrow x = 1{\/tex}<\/span>
                \nOn putting x = 1 in Eq. (i), we get
                \n{tex}y = \\pm \\sqrt { 3 }{\/tex}<\/span>
                \nThus, the points of intersection of the given circles are A (1, {tex}\\sqrt3{\/tex}<\/span>) and A'(1,-{tex}\\sqrt3{\/tex}<\/span>).
                \n\"CBSE
                \nClearly, required area= Area of the enclosed region OACA’O between circles
                \n= 2 [ Area of the region ODCAO]\n=2 [Area of the region ODAO + Area of the region DCAD]\n{tex}= 2 \\left[ \\int _ { 0 } ^ { 1 } y _ { 2 } d x + \\int _ { 1 } ^ { 2 } y _ { 1 } d x \\right]{\/tex}<\/span>
                \n{tex}= 2 \\left[ \\int _ { 0 } ^ { 1 } \\sqrt { 4 – ( x – 2 ) ^ { 2 } } d x + \\int _ { 1 } ^ { 2 } \\sqrt { 4 – x ^ { 2 } } d x \\right]{\/tex}<\/span>
                \n{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}<\/span>{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}<\/span>
                \n{tex}= \\left[ ( x – 2 ) \\sqrt { 4 – ( x – 2 ) ^ { 2 } } + 4 \\sin ^ { – 1 } \\left( \\frac { x – 2 } { 2 } \\right) \\right] _ { 0 } ^ { 1 }{\/tex}<\/span>{tex}+ \\left[ x \\sqrt { 4 – x ^ { 2 } } + 4 \\sin ^ { – 1 } \\frac { x } { 2 } \\right] _ { 1 } ^ { 2 } {\/tex}<\/span>
                \n{tex}= \\left[ \\left\\{ – \\sqrt { 3 } + 4 \\sin ^ { – 1 } \\left( \\frac { – 1 } { 2 } \\right) \\right\\} – 0 – 4 \\sin ^ { – 1 } ( – 1 ) \\right]{\/tex}<\/span>{tex}+ \\left[ 0 + 4 \\sin ^ { – 1 } 1 – \\sqrt { 3 } – 4 \\sin ^ { – 1 } \\frac { 1 } { 2 } \\right]{\/tex}<\/span>
                \n{tex}= \\left[ \\left( – \\sqrt { 3 } – 4 \\times \\frac { \\pi } { 6 } \\right) + 4 \\times \\frac { \\pi } { 2 } \\right] +{\/tex}<\/span>{tex}\\left[ 4 \\times \\frac { \\pi } { 2 } – \\sqrt { 3 } – 4 \\times \\frac { \\pi } { 6 } \\right]{\/tex}<\/span>
                \n{tex}= \\left( – \\sqrt { 3 } – \\frac { 2 \\pi } { 3 } + 2 \\pi \\right) + \\left( 2 \\pi – \\sqrt { 3 } – \\frac { 2 \\pi } { 3 } \\right){\/tex}<\/span>
                \n{tex}= \\frac { 8 \\pi } { 3 } – 2 \\sqrt { 3 }{\/tex}<\/span> sq units.<\/li>\n<\/ol>\n

                Chapter Wise Important Questions Class 12 Maths Part I and Part II<\/h2>\n
                  \n
                1. Relations and Functions<\/a><\/li>\n
                2. Inverse Trigonometric Functions<\/a><\/li>\n
                3. Matrices<\/a><\/li>\n
                4. Determinants<\/a><\/li>\n
                5. Continuity and Differentiability<\/a><\/li>\n
                6. Application of Derivatives<\/a><\/li>\n
                7. Integrals<\/a><\/li>\n
                8. Application of Integrals<\/a><\/li>\n
                9. Differential Equations<\/a><\/li>\n
                10. Vector Algebra<\/a><\/li>\n
                11. Three Dimensional Geometry<\/a><\/li>\n
                12. Linear Programming<\/a><\/li>\n
                13. Probability<\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                  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\u00a0myCBSEguide\u00a0website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1346,1432],"tags":[1867,1839,1838,1833,1832,1854],"class_list":["post-27921","post","type-post","status-publish","format-standard","hentry","category-cbse","category-mathematics-cbse-class-12","tag-cbse-class-12-mathematics","tag-extra-questions","tag-important-questions","tag-latest-exam-questions","tag-practice-questions","tag-practice-test"],"yoast_head":"\nCBSE Class 12 Maths Important Questions Application of Integrals<\/title>\n<meta name=\"description\" content=\"CBSE Class 12 Maths Important Questions Application of Integrals There are 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