{"id":27912,"date":"2019-10-18T17:57:32","date_gmt":"2019-10-18T12:27:32","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=27912"},"modified":"2019-10-25T11:50:34","modified_gmt":"2019-10-25T06:20:34","slug":"integrals-class-12-mathematics-chapter-7-important-question","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/integrals-class-12-mathematics-chapter-7-important-question\/","title":{"rendered":"Integrals Class 12 Mathematics Chapter 7 Important Questions"},"content":{"rendered":"

Integrals Class 12 Mathematics Chapter 7 Important Questions. <\/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 7 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 7 Maths Extra Questions<\/strong><\/p>\n

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

Integrals Class 12 Mathematics Extra Questions<\/h2>\n

Chapter 7 Integrals<\/strong><\/p>\n

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

    {tex}\\int {\\frac{1}{{{e^x} + 1}}dx} {\/tex}<\/span> is equal to<\/p>\n<\/div>\n<\/div>\n

    \n
    \n
    \n
      \n
    1. log {tex}(1 + {e^{ – 2x}})+C{\/tex}<\/span><\/li>\n
    2. {tex}\\log ({e^{ -2x}} -2 x)+C{\/tex}<\/span><\/li>\n
    3. \u2013 log(1 + e-x<\/sup>) + C<\/li>\n
    4. {tex}\\log ({e^{ 3x}} + x)+C{\/tex}<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
    5. \n
      \n
      \n

      The function{tex}f(x) = \\int\\limits_0^x {\\log (t + \\sqrt {1 + {t^2})} dt} {\/tex}<\/span> is<\/p>\n<\/div>\n<\/div>\n

      \n
      \n
      \n
        \n
      1. an odd function<\/li>\n
      2. an even function<\/li>\n
      3. Neither odd nor Even<\/li>\n
      4. a periodic function<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
      5. \n
        \n
        \n

        {tex}\\int\\limits_0^{\\pi \/2} {\\log \\left| {\\cos x} \\right|} {\/tex}<\/span>dx is equal to<\/p>\n<\/div>\n<\/div>\n

        \n
        \n
        \n
          \n
        1. {tex} – \\frac{\\pi }{2}\\log 2{\/tex}<\/span><\/li>\n
        2. {tex}\\pi \\;log{\\text{ }}2{\/tex}<\/span><\/li>\n
        3. {tex}\\frac{\\pi }{2}\\;\\log 3{\/tex}<\/span><\/li>\n
        4. {tex}-\\frac{\\pi }{3}\\;\\log 3{\/tex}<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
        5. \n
          \n
          \n

          {tex}\\int\\limits_a^b {\\frac{{\\log x}}{x}dx} {\/tex}<\/span> is equal to<\/p>\n<\/div>\n<\/div>\n

          \n
          \n
          \n
            \n
          1. {tex}\\frac{log\\;(b-a)}{b-a}{\/tex}<\/span><\/li>\n
          2. log (a+b).log (b\u2013a)<\/li>\n
          3. {tex}\\log(ab).log{(\\frac{b}{a}}){\/tex}<\/span><\/li>\n
          4. {tex}\\frac{1}{2}\\log(ab).log \\left( {\\frac{b}{a}} \\right){\/tex}<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
          5. \n
            \n
            \n

            If {tex}\\int {f(x)} {\/tex}<\/span> dx = f (x), then<\/p>\n<\/div>\n<\/div>\n

            \n
            \n
            \n
              \n
            1. f(x) = ax<\/sup><\/li>\n
            2. f (x) = x<\/li>\n
            3. f(x) = 0<\/li>\n
            4. {tex}f(x)=e^x{\/tex}<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
            5. The function A(x) denotes the ________ function and is given by A (x) =\u00a0{tex}\\int\\limits_a^x {f(x)} dx{\/tex}<\/span>.<\/li>\n
            6. The indefinite integral of\u00a0{tex}2x^{\\frac{1}{2}}{\/tex}<\/span>\u00a0is ________.<\/li>\n
            7. The indefinite integral of 2x2<\/sup> + 3 is ________.<\/li>\n
            8. \n
              \n
              \n

              Show that {tex}\\int {\\frac{{2x + 3}}{{{x^2} + 3x}}dx = \\log \\left| {{x^2} + 3x} \\right| + C} {\/tex}<\/span>.\u00a0<\/strong><\/p>\n<\/div>\n<\/div>\n<\/li>\n

            9. \n
              \n
              \n

              Evaluate {tex}\\int _ { 0 } ^ { 1 } \\frac { 1 } { \\sqrt { 1 – x ^ { 2 } } } d x.{\/tex}<\/span><\/p>\n<\/div>\n<\/div>\n<\/li>\n

            10. \n
              \n
              \n

              Evaluate {tex}\\int \\sin ^ { 3 } x d x.{\/tex}<\/span><\/p>\n<\/div>\n<\/div>\n<\/li>\n

            11. \n
              \n
              \n

              Evaluate the definite integral {tex}\\int\\limits_0^{\\frac{\\pi }{4}} {\\left( {2{{\\sec }^2}x + {x^3} + 2} \\right)dx} {\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            12. \n
              \n
              \n

              Integrate the following function {tex}\\frac{1}{{\\sqrt {7 – 6x – {x^2}} }}{\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            13. \n
              \n
              \n

              Integrate the function (x2<\/sup> + 1) log x<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            14. \n
              \n
              \n

              {tex}\\int_0^{\\pi \/4} {\\frac{{\\sin x\\cos x}}{{{{\\cos }^4}x + {{\\sin }^4}x}}dx} {\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            15. \n
              \n
              \n

              Evaluate {tex}\\int \\frac { \\sqrt { 1 – \\sin x } } { 1 + \\cos x } e ^ { \\frac { – x } { 2 } } d x.{\/tex}<\/span><\/p>\n<\/div>\n<\/div>\n<\/li>\n

            16. \n
              \n
              \n

              Evaluate {tex}\\int\\limits_0^1 {x\\log \\left( {1 + 2x} \\right)dx} {\/tex}<\/span><\/p>\n<\/div>\n<\/div>\n<\/li>\n

            17. \n
              \n
              \n

              Evaluate {tex}\\int _ { 1 } ^ { 3 } \\left( 2 x ^ { 2 } + 5 x \\right) d x{\/tex}<\/span> as a limit of a sum.<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n

              \n

              Chapter 7 Integrals<\/strong><\/p>\n

              Solution<\/b><\/p>\n

                \n
              1. \n
                  \n
                1. \u2013log(1 + e-x<\/sup>) + C,\u00a0Explanation:<\/strong> {tex}\\int {\\frac{{{e^{ – x}}}}{{1 + {e^{ – x}}}}} dx = – \\int {\\frac{{ – {e^{ – x}}}}{{1 + {e^{ – x}}}}} dx = – \\log (1 + {e^{ – x}})+C{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                2. \n
                    \n
                  1. an even function,\u00a0Explanation:<\/strong> t = -u
                    \nf(-x) = {tex}\\int\\limits_0^x {\\log ( – u + \\sqrt {1 + {u^2}} } )( – du){\/tex}<\/span>
                    \n= {tex} – \\int\\limits_0^\\pi {\\log \\left( {\\frac{{1 + u + {u^2}}}{{\\sqrt {1 + {u^2} + u} }}} \\right)} ( – du){\/tex}<\/span>
                    \n= {tex}\\int\\limits_0^x {\\log (u + \\sqrt {1 + {u^2}} } )du{\/tex}<\/span> = f(x)
                    \n{tex}\\Rightarrow{\/tex}<\/span> f(-x) = f(x) {tex}\\Rightarrow{\/tex}<\/span> f is an even function<\/li>\n<\/ol>\n<\/li>\n
                  2. \n
                      \n
                    1. {tex} – \\frac{\\pi }{2}\\log 2{\/tex}<\/span>,\u00a0Explanation:<\/strong> {tex}\\int\\limits_0^{\\pi \/2} {\\log \\left| {\\cos x} \\right|} dx{\/tex}<\/span>
                      \n{tex}= \\int\\limits_0^{\\pi \/2} {\\log \\left| {\\cos (\\frac{\\pi }{2} – x)} \\right|} dx {\/tex}<\/span>
                      \n{tex} = \\int\\limits_0^{\\pi \/2} {\\log \\left| {\\sin x} \\right|dx} {\/tex}<\/span>
                      \n{tex} = – \\frac{\\pi }{2}\\;\\log 2{\/tex}<\/span>(standard result)<\/li>\n<\/ol>\n<\/li>\n
                    2. \n
                        \n
                      1. {tex}\\frac{1}{2}\\log(ab).log \\left( {\\frac{b}{a}} \\right){\/tex}<\/span>,\u00a0Explanation:<\/strong> {tex}= {\\int\\limits_a^b {{{(log\\;x)}^{^1}}} ^{}}(\\frac{1}{x})dx{\/tex}<\/span>
                        \n{tex} \\\\(let\\; log\\;x=t\\;then\\;\\frac1xdx=dt) \\\\ {\/tex}<\/span>
                        \n{tex}= \\left[ {\\frac{{{{(log\\;x)}^2}}}{2}} \\right]_a^b \\\\ {\/tex}<\/span>
                        \n{tex}= \\frac{1}{2}\\left[ {{{(log\\;b)}^2} – {{(log\\;a)}^2}} \\right] {\/tex}<\/span>
                        \n{tex}\\\\ = \\frac{1}{2}(log\\;b + log\\;a)(log\\;b – log\\;a){\/tex}<\/span>
                        \n{tex}= \\frac{1}{2}\\log (ab)\\log \\frac{b}{a} {\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                      2. \n
                          \n
                        1. {tex}f(x)=e^x{\/tex}<\/span>, mExplanation:<\/strong> {tex}{\\frac{d}{{dx}}} (f(x)) = f(x){\/tex}<\/span>
                          \nIt implies that the function remains same after integrating or differentiating it. So the function must be ex<\/sup><\/li>\n<\/ol>\n<\/li>\n
                        2. area<\/li>\n
                        3. {tex}\\frac{4}{3} x^{\\frac{3}{2}}+c{\/tex}<\/span><\/li>\n
                        4. {tex}=\\frac{4}{5} x^{\\frac{5}{4}}+c{\/tex}<\/span><\/li>\n
                        5. Let {tex}I = \\int {\\frac{{2x + 3}}{{{x^2} + 3x}}dx} {\/tex}<\/span>
                          \nPut x2<\/sup> + 3x = t
                          \n{tex}\\Rightarrow{\/tex}<\/span> (2x + 3)dx = dt
                          \n{tex}\\therefore I = \\int {\\frac{1}{t}dt = \\log \\left| t \\right| + C} {\/tex}<\/span>
                          \n= log |(x2<\/sup> + 3x)| + C<\/li>\n
                        6. According to the question , {tex}I = \\int _ { 0 } ^ { 1 } \\frac { 1 } { \\sqrt { 1 – x ^ { 2 } } } d x{\/tex}<\/span>
                          \n= {tex}\\left[ \\sin ^ { – 1 } x \\right] _ { 0 } ^ { 1 } \\quad \\left[ \\because \\int_b^a \\frac { 1 } { \\sqrt { 1 – x ^ { 2 } } } d x = \\sin ^ { – 1 } (a) – \\sin ^ { – 1 } (b) \\right]{\/tex}<\/span>
                          \n{tex}= sin^{-1}(1)-sin^{-1}(0){\/tex}<\/span>
                          \n{tex}= \\sin ^ { – 1 } \\left( \\sin \\frac { \\pi } { 2 } \\right) – \\sin ^ { – 1 } ( \\sin 0 ) {\/tex}<\/span>
                          \n{tex}= \\frac { \\pi } { 2 } – 0 = \\frac { \\pi } { 2 }{\/tex}<\/span><\/li>\n
                        7. Let {tex} I = \\int \\sin ^ { 3 } x d x = \\int \\frac { 3 \\sin x – \\sin 3 x } { 4 } d x{\/tex}<\/span> {tex}\\left[ \\because \\sin 3 x = 3 \\sin x – 4 \\sin ^ { 3 } x \\right]{\/tex}<\/span>
                          \n{tex}= \\frac { 1 } { 4 } \\int 3 \\sin x d x – \\frac { 1 } { 4 } \\int \\sin 3 x d x{\/tex}<\/span>
                          \n{tex}= \\frac { 1 } { 4 } \\left( – 3 \\cos x + \\frac { \\cos 3 x } { 3 } \\right) + C{\/tex}<\/span><\/li>\n
                        8. {tex}\\int\\limits_0^{\\frac{\\pi }{4}} {\\left( {2{{\\sec }^2}x + {x^3} + 2} \\right)dx} {\/tex}<\/span>
                          \n{tex}= 2\\int\\limits_0^{\\frac{\\pi }{4}} {{{\\sec }^2}xdx + \\int\\limits_0^{\\frac{\\pi }{4}} {{x^3}dx + 2\\int\\limits_0^{\\frac{\\pi }{4}} {1dx} } } {\/tex}<\/span>
                          \n{tex}= 2\\left( {\\tan x} \\right)_0^{\\frac{\\pi }{4}} + \\left( {\\frac{{{x^4}}}{4}} \\right)_0^{\\frac{\\pi }{4}} + 2\\left( x \\right)_0^{\\frac{\\pi }{4}}{\/tex}<\/span>
                          \n{tex}= 2\\left( {\\tan \\frac{\\pi }{4} – \\tan {0^o}} \\right) + \\frac{{{{\\left( {\\frac{\\pi }{4}} \\right)}^4}}}{4} – 0 {\/tex}<\/span> {tex}+ 2\\left( {\\frac{\\pi }{4} – 0} \\right){\/tex}<\/span>
                          \n{tex}= 2\\left( {1 – 0} \\right) + \\frac{{\\left( {\\frac{{{\\pi ^4}}}{{256}}} \\right)}}{4} + \\frac{{2\\pi }}{4}{\/tex}<\/span>
                          \n{tex}= 2 + \\frac{{{\\pi ^4}}}{{1024}} + \\frac{\\pi }{2}{\/tex}<\/span>
                          \n{tex}= \\frac{{{\\pi ^4}}}{{1024}} + \\frac{\\pi }{2} + 2{\/tex}<\/span><\/li>\n
                        9. {tex}\\int {\\frac{1}{{\\sqrt {7 – 6x – {x^2}} }}dx} {\/tex}<\/span>
                          \n{tex} = \\int {\\frac{1}{{\\sqrt { – {x^2} – 6x + 7} }}dx} {\/tex}<\/span>
                          \n{tex}= \\int {\\frac{1}{{\\sqrt { – \\left( {{x^2} + 6x – 7} \\right)} }}dx} {\/tex}<\/span>
                          \n{tex}= \\int {\\frac{1}{{\\sqrt { – \\left( {{x^2} + 6x + 9 – 9 – 7} \\right)} }}dx} {\/tex}<\/span>
                          \n{tex}= \\int {\\frac{1}{{\\sqrt { – \\left\\{ {{{\\left( {x + 3} \\right)}^2} – 16} \\right\\}} }}dx} {\/tex}<\/span>
                          \n{tex}= \\int {\\frac{1}{{\\sqrt {{{\\left( 16 \\right)}} – {{\\left( {x + 3} \\right)}^2}} }}dx} {\/tex}<\/span>
                          \n{tex}= \\int {\\frac{1}{{\\sqrt {{{\\left( 4 \\right)}^2} – {{\\left( {x + 3} \\right)}^2}} }}dx} {\/tex}<\/span>
                          \n{tex}= {\\sin ^{ – 1}}\\left( {\\frac{{x + 3}}{4}} \\right) + c{\/tex}<\/span>
                          \n{tex}\\left[ {\\because \\int {\\frac{1}{{{a^2} – {x^2}}}dx = {{\\sin }^{ – 1}}\\frac{x}{a}} } \\right]{\/tex}<\/span><\/li>\n
                        10. {tex}\\int {\\left( {{x^2} + 1} \\right)\\log xdx} {\/tex}<\/span>
                          \n{tex} = \\int {\\left( {\\log x} \\right)\\left( {{x^2} + 1} \\right)dx} {\/tex}<\/span>
                          \n[Applying product rule]\n{tex}= \\log x\\left( {\\frac{{{x^3}}}{3} + x} \\right) – \\int {\\frac{1}{x}\\left( {\\frac{{{x^3}}}{3} + x} \\right)dx}{\/tex}<\/span>
                          \n{tex}= \\left( {\\frac{{{x^3}}}{3} + x} \\right)\\log x – \\int {\\left( {\\frac{{{x^2}}}{3} + 1} \\right)dx}{\/tex}<\/span>
                          \n{tex}= \\left( {\\frac{{{x^3}}}{3} + x} \\right)\\log x – \\frac{1}{3}\\int {{x^2}dx – \\int {1dx} } {\/tex}<\/span>
                          \n{tex}= \\left( {\\frac{{{x^3}}}{3} + x} \\right)\\log x – \\frac{1}{3}\\frac{{{x^3}}}{3} – x + c{\/tex}<\/span>
                          \n{tex}= \\left( {\\frac{{{x^3}}}{3} + x} \\right)\\log x – \\frac{{{x^3}}}{9} – x + c{\/tex}<\/span><\/li>\n
                        11. {tex}I = \\int_0^{\\pi \/4} {\\frac{{\\sin x\\cos x}}{{{{\\cos }^4}x + {{\\sin }^4}x}}dx} {\/tex}<\/span>
                          \nDividing Numerator and Denominator by {tex}{\\cos ^4}x{\/tex}<\/span>
                          \n{tex}= \\int_0^{\\pi \/4} {\\frac{{\\frac{{\\sin x.\\cos x}}{{{{\\cos }^4}x}}}}{{\\frac{{{{\\cos }^4}x}}{{{{\\cos }^4}x}} + \\frac{{{{\\sin }^4}x}}{{{{\\cos }^4}x}}}}} dx{\/tex}<\/span>
                          \n{tex}= \\int_0^{\\pi \/4} {\\frac{{\\tan x.{{\\sec }^2}x}}{{1 + {{\\tan }^4}x}}} dx{\/tex}<\/span>
                          \n{tex}= \\int_0^{\\pi \/4} {\\frac{{\\tan x.{{\\sec }^2}x}}{{1 + {{\\left( {{{\\tan }^2}x} \\right)}^2}}}} dx{\/tex}<\/span>
                          \nput {tex}{\\tan ^2}x = t{\/tex}<\/span>
                          \n{tex}2\\tan x.{\\sec ^2}xdx = dt{\/tex}<\/span>
                          \nwhen x=0, t=0 and when x={tex}\\frac{\u03c0}{4}{\/tex}<\/span> t=1
                          \n{tex} \\therefore I= \\frac{1}{2}\\int_0^1 {\\frac{{dt}}{{1 + {t^2}}}}{\/tex}<\/span>
                          \n{tex}= \\frac{1}{2}\\left[ {{{\\tan }^{ – 1}}t} \\right]_0^1{\/tex}<\/span>
                          \n{tex} = \\frac{1}{2}.\\frac{\\pi }{4} = \\frac{\\pi }{8}{\/tex}<\/span><\/li>\n
                        12. Given, {tex}I=\\int \\frac { \\sqrt { 1 – \\sin x } } { 1 + \\cos x } \\cdot e ^ { \\frac { – x } { 2 } } d x{\/tex}<\/span>
                          \nLet {tex}\\frac { – x } { 2 } = t \\Rightarrow d x = – 2 d t{\/tex}<\/span>
                          \n{tex} I = \\int \\frac { \\sqrt { 1 – \\sin ( – 2 t ) } } { 1 + \\cos ( – 2 t ) } e ^ { t } ( – 2 d t ){\/tex}<\/span> {tex}[\\because x = -2t]{\/tex}<\/span>
                          \n{tex}= – 2 \\int e ^ { t } \\frac { \\sqrt { 1 + \\sin 2 t } } { 1 + \\cos 2 t } d t{\/tex}<\/span>{tex}[ \\because \\sin ( – \\theta ) = – \\sin \\theta \\text { and } \\cos ( – \\theta ) = \\cos \\theta ]{\/tex}<\/span>
                          \n{tex}= – 2 \\int e ^ { t } \\frac { \\sqrt { cos^2x+sin^2x+2sinx cosx } } { 1 + \\cos 2 t } d t [\\because cos^2x+sin^2x=1 , sin2x=2sinx cosx ]{\/tex}<\/span>
                          \n{tex}= – 2 \\int e ^ { t } \\left( \\frac { \\sqrt { ( \\cos t + \\sin t ) ^ { 2 } } } { 2 \\cos ^ { 2 } t } \\right) d t [\\because (a+b)^2 =a^2+b^2+2ab]{\/tex}<\/span>
                          \n{tex}= – 2 \\int e ^ { t } \\left( \\frac { \\cos t + \\sin t } { 2 \\cos ^ { 2 } t } \\right) d t{\/tex}<\/span>
                          \n{tex}= – \\int e ^ { t } ( \\sec t + \\tan t \\sec t ) d t{\/tex}<\/span>{tex}[\\because{1\\over cosx} = secx , {sinx\\over cosx }= tanx]{\/tex}<\/span>
                          \nwe know that , {tex}\\int e ^ { t } \\left[ f ( t ) + f ^ { \\prime } ( t ) \\right] d t = e^tf(t)+C{\/tex}<\/span>
                          \nNow, consider {tex}f(t) = sec\\ t{\/tex}<\/span>
                          \nthen {tex}f'(t) = sec\\ t\\ tan\\ t{\/tex}<\/span>
                          \n{tex}\\therefore I= e ^ { t} \\sec t + C{\/tex}<\/span>
                          \n{tex}= – e ^ { – x \/ 2 } \\sec \\frac { x } { 2 } + C{\/tex}<\/span>{tex}\\left[ \\because t = \\frac { – x } { 2 } \\text { and } \\sec ( – \\theta ) = \\sec \\theta \\right]{\/tex}<\/span>
                          \n{tex}I = – e ^ { – x \/ 2 } \\sec \\frac { x } { 2 } + C{\/tex}<\/span><\/li>\n
                        13. {tex}I = \\int\\limits_0^1 {x\\log \\left( {1 + 2x} \\right)dx} {\/tex}<\/span>
                          \n{tex} = \\left[ {\\log \\left( {1 + 2x} \\right)\\frac{{{x^2}}}{2}} \\right]_0^1 – \\int {\\frac{1}{{1 + 2x}}.2.\\frac{{{x^2}}}{2}dx} {\/tex}<\/span>
                          \n{tex} = \\frac{1}{2}\\left[ {{x^2}\\log \\left( {1 + 2x} \\right)} \\right]_0^1 – \\int {\\frac{{{x^2}}}{{1 + 2x}}dx} {\/tex}<\/span>
                          \n{tex}= \\frac{1}{2}\\left[ {\\log 3 – 0} \\right]_0^1 – \\left[ {\\int_0^1 {\\left( {\\frac{x}{2} – \\frac{{\\frac{x}{2}}}{{1 + 2x}}} \\right)} dx} \\right]{\/tex}<\/span>
                          \n{tex} = \\frac{1}{2}\\log 3 – \\frac{1}{2}\\int_0^1 x dx + \\frac{1}{2}\\int_0^1 {\\frac{x}{{1 + 2x}}dx} {\/tex}<\/span>
                          \n{tex}= \\frac{1}{2}\\log 3 – \\frac{1}{2}\\left[ {\\frac{{{x^2}}}{2}} \\right]_0^1 + \\frac{1}{2}\\int_0^1 {\\frac{{\\frac{1}{2}\\left( {2x + 1 – 1} \\right)}}{{\\left( {2x + 1} \\right)}}} dx{\/tex}<\/span>
                          \n{tex}= \\frac{1}{2}\\log 3 – \\frac{1}{2}\\left[ {\\frac{1}{2} – 0} \\right] {\/tex}<\/span> {tex}+ \\frac{1}{4}\\int_0^1 {dx – \\frac{1}{4}\\int_0^1 {\\frac{1}{{1 + 2x}}dx} } {\/tex}<\/span>
                          \n{tex}= \\frac{1}{2}\\log 3 – \\frac{1}{4} + \\frac{1}{4}\\left[ x \\right]_0^1 – \\frac{1}{8}\\left[ {\\log \\left| {\\left( {1 + 2x} \\right)} \\right|} \\right]_0^1{\/tex}<\/span>
                          \n{tex} = \\frac{1}{2}\\log 3 – \\frac{1}{4} + \\frac{1}{4} – \\frac{1}{8}\\left[ {\\log 3 – \\log 1} \\right]{\/tex}<\/span>
                          \n{tex} = \\frac{1}{2}\\log 3 – \\frac{1}{8}\\log 3{\/tex}<\/span>
                          \n{tex}= \\frac{3}{8}\\log 3{\/tex}<\/span><\/li>\n
                        14. According to the question , {tex}I = \\int _ { 1 } ^ { 3 } \\left( 2 x ^ { 2 } + 5 x \\right) d x{\/tex}<\/span>
                          \nOn comparing the given integral with {tex} \\int _ { a } ^ { b } f ( x ) d x,{\/tex}<\/span> we get
                          \na = 1, b = 3 and f(x) =2x2<\/sup> + 5x
                          \nWe know that , {tex} \\int_a^b f (x)dx = \\mathop {\\lim }\\limits_{h \\to 0} h[f(a) + f(a + h)+ f(a + 2h) + … + f(a + (n – 1)h) \\ ]{\/tex}<\/span>…(i)
                          \nwhere, {tex}h = \\frac {b – a}{n} \\Rightarrow nh = b – a \\\\ \\ \\ \\quad \\quad \\quad \\quad \\quad \\quad = 3-1=2{\/tex}<\/span>
                          \nf(a) = f(1) = 2(1)2<\/sup> + 5(1) = 2 + 5 = 7
                          \nf(a+h) = f(1 + h) = 2(1 + h)2<\/sup> + 5(1 + h) = 2 + 2h2<\/sup> + 4h + 5 + 5h = 2h2<\/sup>+ 9h + 7
                          \nf(a + 2h) = 2(1 + 2h)2<\/sup> + 5(1 + 2h) = 2 + 8h2<\/sup> + 8h + 5 +10h = 8h2<\/sup> + 18h + 7
                          \nso on
                          \nf(a+(n-1)h) = f{1 + (n – 1)h} = 2{1 + (n – 1)h}2<\/sup> + 5{1 + (n – 1)h}2 <\/sup>= 2 + 2(n – 1)2h2<\/sup> + 4(n – 1)h + 5 + 5(n – 1)h = 2(n – 1)2h2<\/sup> + 9(n – 1)h + 7
                          \nOn putting all above values in (i), we get
                          \n{tex} {\\int_1^3 {\\left( {2{x^2} + 5x} \\right)} dx = \\mathop {\\lim }\\limits_{h \\to 0} }{\/tex}<\/span>h[7 + (2h2<\/sup> + 9h + 7) +(8h2<\/sup> + 18h + 7)…….+2(n – 1)2h2<\/sup> + 9(n – 1)h + 7]\nOn rearranging terms , we get
                          \n{tex}= \\mathop {\\lim }\\limits_{h \\to 0} {\/tex}<\/span>h[7 + 7+ 7……..+ 7] + {tex} \\mathop {\\lim }\\limits_{h \\to 0} {\/tex}<\/span>h[2h2<\/sup> + 8h2<\/sup>+………+2(n – 1)2<\/sup>h2<\/sup>] + {tex} \\mathop {\\lim }\\limits_{h \\to 0} {\/tex}<\/span>h[9h + 18h +……+9(n -1 )h]\n{tex}= \\mathop {\\lim }\\limits_{h \\to 0} {\/tex}<\/span>7nh + {tex} \\mathop {\\lim }\\limits_{h \\to 0} {\/tex}<\/span>2h3<\/sup>[ 12<\/sup> + 22<\/sup> + 32<\/sup> +………+(n -1)2<\/sup>] +{tex} \\mathop {\\lim }\\limits_{h \\to 0} {\/tex}<\/span>9h2<\/sup>[ 1+ 2 +……..+(n – 1)]\n{tex} = \\mathop {\\lim }\\limits_{h \\to 0} 7(2) + \\mathop {\\lim }\\limits_{h \\to 0} \\frac{{2{h^3} \\cdot n(n – 1)(2n – 1)}}{6}{\/tex}<\/span>{tex} + \\mathop {\\lim }\\limits_{h \\to 0} \\frac{{9{h^2} \\cdot n(n – 1)}}{2}{\/tex}<\/span>{tex}[\\because \\sum n = \\frac {n(n+1)}{2} , \\sum n^2 = \\frac {n(n+1)(2n+1)}{6}]{\/tex}<\/span>
                          \n{tex} = 14 + \\mathop {\\lim }\\limits_{h \\to 0} \\frac{{nh(nh – h)(2nh – h)}}{3}{\/tex}<\/span>+ {tex}\\mathop {\\lim }\\limits_{h \\to 0} \\frac{9}{2} \\cdot nh(nh – h){\/tex}<\/span>
                          \n{tex}= 14 + \\frac { 2 ( 2 – 0 ) ( 4 – 0 ) } { 3 } + \\frac { 9 } { 2 } \\cdot 2 ( 2 – 0 ){\/tex}<\/span>
                          \n{tex}= 14 + \\frac { 16 } { 3 } + 18 {\/tex}<\/span>
                          \n{tex}= \\frac { 42 + 16 + 54 } { 3 } {\/tex}<\/span>
                          \n{tex}= \\frac { 112 } { 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":"

                            Integrals Class 12 Mathematics Chapter 7 Important Questions. 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 schools … 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-27912","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":"\nIntegrals Class 12 Mathematics Chapter 7 Important Questions<\/title>\n<meta name=\"description\" content=\"Integrals Class 12 Mathematics Chapter 7 Important Questions These 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