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NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.9

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NCERT Solutions for Class 12 Maths Chapter 7 (Ex 7.9)

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.9 (Ex 7.9) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 7 Integrals Exercise 7.9 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 7 - Integrals

Exercise:

Exercise - 7.9

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

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Competitive Exams after 12th Science

Access NCERT Solutions for Class 12 Maths Chapter 7- Integrals

Exercise 7.9

1.Integrate the given integral $\mathbf{\int\limits_{ - 1}^1 {(x + 1)dx} }$.

Ans: To simplify the question let us suppose, $I = \int\limits_{ - 1}^1 {(x + 1)dx} $

$\int {(x + 1)dx = \dfrac{{{x^2}}}{2}}  + x$

Again, let us suppose the function is $F(x) = \dfrac{{{x^2}}}{2} + x$

By second fundamental theorem of calculus, we obtain

$  I = F(1) - F( - 1) $

   $= \left( {\dfrac{1}{2} + 1} \right) - \left( {\dfrac{1}{2} - 1} \right) $

$   = \dfrac{1}{2} + 1 - \dfrac{1}{2} + 1 $

$   = 2 $ 

 

2.Integrate the given integral $\mathbf{\int\limits_2^3 {\dfrac{1}{x}dx} }$

Ans: To simplify the question let us suppose, $I = \int\limits_2^3 {\dfrac{1}{x}dx} $

Thus, $\int {\dfrac{1}{x}dx = \log \left| x \right|} $

Again, let us suppose the function is$F(x) = \log \left| x \right|$

By second fundamental theorem of calculus, we obtain

$  I = F(3) - F(2) $

  $ = \log \left| 3 \right| - \log \left| 2 \right| $

$   = \log \dfrac{3}{2} $ 


3.Integrate the given integral $\mathbf{\int\limits_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx} }$

Ans: To simplify the question let us suppose,

$I = \int\limits_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx}  $

  $\int\limits_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx}  = 4\left( {\dfrac{{{x^4}}}{4}} \right) - 5\left( {\dfrac{{{x^3}}}{3}} \right) + 6\left( {\dfrac{{{x^2}}}{2}} \right) + 9(x) $ 

Again, let us suppose the function is 

$ \Rightarrow {x^4} - \dfrac{{5{x^3}}}{3} + 3{x^2} + 9x = F(x)$

By second fundamental theorem of calculus, we obtain

$I = F(2) - F(1)$

$ I = \left\{ {{2^4} - \dfrac{{5{{(2)}^3}}}{3} + 3{{(2)}^2} + 9(2)} \right\} - \left\{ {{{(1)}^4} - \dfrac{{5{{(1)}^3}}}{3} + 3{{(1)}^2} + 9(1)} \right\} $

  $ = \left( {16 - \dfrac{{40}}{3} + 12 + 18} \right) - \left( {1 - \dfrac{5}{3} + 3 + 9} \right) $

  $ = 16 - \dfrac{{40}}{3} + 12 + 18 - 1 + \dfrac{5}{3} - 3 - 9 $

 $  = 33 - \dfrac{{35}}{3} $

  $ = \dfrac{{99 - 35}}{3} $

  $ = \dfrac{{64}}{3} $

 

4.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{4}} {\left( {\sin 2xdx} \right)}} $

Ans: To simplify the question let us suppose,

$I = \int\limits_0^{\dfrac{\pi }{4}} {\left( {\sin 2xdx} \right)}  $

  $\int {\sin 2xdx = \left( {\dfrac{{ - \cos 2x}}{2}} \right)}  $ 

Again, let us suppose the function is$\left( {\dfrac{{ - \cos 2x}}{2}} \right) = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{4}} \right) - F(0) $

 $  =  - \dfrac{1}{2}\left[ {\cos 2\left( {\dfrac{\pi }{4}} \right) - \cos 0} \right] $

 $  =  - \dfrac{1}{2}\left[ {\cos 2\left( {\dfrac{\pi }{4}} \right) - \cos 0} \right] $

$   =  - \dfrac{1}{2}\left[ {\cos 2\left( {\dfrac{\pi }{4}} \right) - \cos 0} \right] $

 $  =  - \dfrac{1}{2}\left[ {0 - 1} \right] $

$   = \dfrac{1}{2} $ 


5.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{2}} {\left( {\cos 2xdx} \right)} }$

Ans: To simplify the question let us suppose,

$  I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\cos 2xdx} \right)}  $

$  \int {\cos 2xdx = \left( {\dfrac{{\sin 2x}}{2}} \right)}  $ 

Again, let us suppose the function is$\left( {\dfrac{{\sin 2x}}{2}} \right) = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{2}} \right) - F(0) $

$   = \dfrac{1}{2}\left[ {\sin 2\left( {\dfrac{\pi }{2}} \right) - \sin 0} \right] $

$   = \dfrac{1}{2}\left[ {\sin \pi  - \sin 0} \right] $

$   = \dfrac{1}{2}\left[ {0 - 0} \right] $

$   = 0 $

 

6.Integrate the given integral $\mathbf{\int\limits_4^5 {{e^x}dx} }$

Ans: To simplify the question let us suppose,

$  I = \int\limits_4^5 {{e^x}dx}  $

$  \int\limits_4^5 {{e^x}dx}  = {e^x} = F(x) $ 

By second fundamental theorem of calculus, we obtain

$I = F(5) - F(4) $

  $ = {e^5} - {e^4} $

 $  = {e^4}\left( {e - 1} \right) $ 


7.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{4}} {\tan xdx} }$

Ans: To simplify the question let us suppose,

$I = \int\limits_0^{\dfrac{\pi }{4}} {\tan xdx}  $

 $ \int {\tan xdx =  - \log \left| {\cos x} \right|}  $ 

Again, let us suppose the function is$ - \log \left| {\cos x} \right| = F(x)$

By second fundamental theorem of calculus, we obtain

$ I = F\left( {\dfrac{\pi }{4}} \right) - F(0) $

 $  =  - \log \left| {\cos \dfrac{\pi }{4}} \right| + \log \left| {\cos 0} \right| $

$   =  - \log \left| {\dfrac{1}{{\sqrt 2 }}} \right| + \log \left| 1 \right| $

$   =  - \log {\left( 2 \right)^{ - \dfrac{1}{2}}} $

$   = \dfrac{1}{2}\log 2 $ 


8.Integrate the given integral $\mathbf{\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\cos ecxdx} }$

Ans: To simplify the question let us suppose,

$ I = \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}} {\cos ecxdx}  $

$  \int {\cos ecxdx = \log \left| {\cos ecx\dfrac{\pi }{6} - \cot \dfrac{\pi }{6}} \right|}  $ 

Again, let us suppose the function is$F(x) = \log \left| {\cos x} \right|$

By second fundamental theorem of calculus, we obtain

$I = F\left( {\dfrac{\pi }{4}} \right) - F\left( {\dfrac{\pi }{6}} \right) $

$   = \log \left| {\cos ec\dfrac{\pi }{4} - \cot \dfrac{\pi }{4}} \right| - \log \left| {\cos ec\dfrac{\pi }{6} - \cot \dfrac{\pi }{6}} \right| $

$   = \log \left| {\sqrt 2  - 1} \right| - \log \left| {2 - \sqrt 3 } \right| $

$   = \log \left( {\dfrac{{\sqrt 2  - 1}}{{2 - \sqrt 3 }}} \right) $ 


9.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }}} }$

Ans: To simplify the question let us suppose,

$I = \int\limits_0^1 {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }}} $

Again, let us suppose the function is

$ I = \int {\dfrac{{dx}}{{\sqrt {1 - {x^2}} }} = F(x)}  $

 $ {\sin ^{ - 1}}x = F(x) $ 

By second fundamental theorem of calculus, we obtain

$ I = F(1) - F(0) $

 $  = {\sin ^{ - 1}}(1) - {\sin ^{ - 1}}(0) $

  $ = \dfrac{\pi }{2} - 0 $

$   = \dfrac{\pi }{2} $ 


10.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{dx}}{{\sqrt {1 + {x^2}} }}}} $

Ans: To simplify the question let us suppose,

$  I = \int\limits_0^1 {\dfrac{{dx}}{{1 + {x^2}}}}  $

$  \int {\dfrac{{dx}}{{1 + {x^2}}} = {{\tan }^{ - 1}}}  $ 

Again, let us suppose the function is

${\tan ^{ - 1}}x = F(x)$

By second fundamental theorem of calculus, we obtain

$I = F(1) - F(0) $

   $= {\tan ^{ - 1}}(1) - {\tan ^{ - 1}}(0) $

  $ = \dfrac{\pi }{4} $ 


11.Integrate the given integral $\mathbf{\int\limits_2^3 {\dfrac{{dx}}{{{x^2} - 1}}} }$

Ans: To simplify the question let us suppose,

$  I = \int\limits_2^3 {\dfrac{{dx}}{{{x^2} - 1}}}  $

 $ \int {\dfrac{{dx}}{{{x^2} - 1}} = \dfrac{1}{2}\log \left| {\dfrac{{x - 1}}{{x + 1}}} \right|}  $ 

Again, let us suppose the function is

$\dfrac{1}{2}\log \left| {\dfrac{{x - 1}}{{x + 1}}} \right| = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F(3) - F(2) $

$   = \dfrac{1}{2}\left[ {\log \left| {\dfrac{{3 - 1}}{{3 + 1}}} \right| - \log \left| {\dfrac{{2 - 1}}{{2 + 1}}} \right|} \right] $

$   = \dfrac{1}{2}\left[ {\log \left| {\dfrac{2}{4}} \right| - \log \left| {\dfrac{1}{3}} \right|} \right] $

$   = \dfrac{1}{2}\left[ {\log \dfrac{1}{2} - \log \dfrac{1}{3}} \right] $

 $  = \dfrac{1}{2}\left[ {\log \dfrac{3}{2}} \right] $ 


12.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} }$

Ans: To simplify the question let us suppose,

$\int {{{\cos }^2}xdx = \int {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)dx} }  $

 $ \dfrac{x}{2} + \dfrac{{\sin 2x}}{4} = \dfrac{1}{2}\left( {x + \dfrac{{\sin 2x}}{2}} \right) $ 

Again, let us suppose the function is

$\dfrac{1}{2}\left( {x + \dfrac{{\sin 2x}}{2}} \right) = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{2}} \right) - F(0) $

  $ = \dfrac{1}{2}\left[ {\left( {\dfrac{\pi }{2} + \dfrac{{\sin \pi }}{2}} \right) - \left( {0 + \dfrac{{\sin 0}}{2}} \right)} \right] $

$   = \dfrac{1}{2}\left[ {\dfrac{\pi }{2} + 0 - 0 + 0} \right] $

$   = \dfrac{\pi }{4} $ 


13.Integrate the given integral $\mathbf{\int\limits_2^3 {\dfrac{{xdx}}{{{x^2} + 1}}}} $

Ans: To simplify the question let us suppose,

$ I = \int\limits_2^3 {\dfrac{{xdx}}{{{x^2} + 1}}}  $

$  \int {\dfrac{{xdx}}{{{x^2} + 1}}}  $

$   = \dfrac{1}{2}\int {\dfrac{{2x}}{{{x^2} + 1}}dx}  $

$   = \dfrac{1}{2}\log \left( {1 + {x^2}} \right) $ 

Again, let us suppose the function is

$\dfrac{1}{2}\log \left( {1 + {x^2}} \right) = F(x)$

By second fundamental theorem of calculus, we obtain

$  I = F(3) - F(2) $

  $ = \dfrac{1}{2}\left[ {\log \left( {1 + {{\left( 3 \right)}^2}} \right) - \log \left( {1 + {{(2)}^2}} \right)} \right] $

$   = \dfrac{1}{2}\left[ {\log 10 - \log 5} \right] $

  $ = \dfrac{1}{2}\left[ {\log \dfrac{{10}}{5}} \right] $

   $= \dfrac{1}{2}\left[ {\log 2} \right] $ 


14.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{2x + 3}}{{5{x^2} + 1}}dx} }$

Ans: To simplify the question let us suppose,

$  I = \int\limits_0^1 {\dfrac{{2x + 3}}{{5{x^2} + 1}}dx}  $

$  \int {\dfrac{{2x + 3}}{{5{x^2} + 1}}dx}  $

  $ = \dfrac{1}{5}\int {\dfrac{{5\left( {2x + 3} \right)}}{{5{x^2} + 1}}dx}  $

$   = \dfrac{1}{5}\int {\dfrac{{\left( {10x + 15} \right)}}{{5{x^2} + 1}}dx}  $

 $  = \dfrac{1}{5}\int {\dfrac{{10x}}{{5{x^2} + 1}}dx}  + 3\int {\dfrac{1}{{5{x^2} + 1}}dx}  $

   $= \dfrac{1}{5}\int {\dfrac{{10x}}{{5{x^2} + 1}}dx}  + 3\int {\dfrac{1}{{5\left( {{x^2} + \dfrac{1}{5}} \right)}}dx}  $

$   = \dfrac{1}{5}\log \left( {5{x^2} + 1} \right) + \dfrac{3}{5}\dfrac{1}{{\dfrac{1}{{\sqrt 5 }}}}{\tan ^{ - 1}}\dfrac{x}{{\dfrac{1}{{\sqrt 5 }}}} $

$   = \dfrac{1}{5}\log \left( {5{x^2} + 1} \right) + \dfrac{3}{{\sqrt 5 }}{\tan ^{ - 1}}\left( {\sqrt 5 } \right)x $

  $ = F(x) $ 

By second fundamental theorem of calculus, we obtain

$  I = F(3) - F(2) $

$   = \dfrac{1}{5}\left[ {\log \left( {1 + 5} \right) - \dfrac{3}{{\sqrt 5 }}\log \left( {\sqrt 5 } \right)} \right] - \left[ {\dfrac{1}{5}\log (1) + \dfrac{3}{{\sqrt 5 }}{{\tan }^{ - 1}}(0)} \right] $

$   = \dfrac{1}{5}\left[ {\log 6} \right] + \dfrac{3}{{\sqrt 5 }}{\tan ^{ - 1}}\sqrt 5 $ 


15.Integrate the given integral $\mathbf{\int\limits_0^1 {x{e^{{x^2}}}dx} }$

Ans: To simplify the question let us suppose,

$= \int\limits_0^1 {x{e^{{x^2}}}dx}$

$  {x^2} = t $

  $ \Rightarrow 2xdx = dt $ 

As $x \to 0,t \to 0$ ,

Again $x \to 1,t \to 1,$, 

  $\therefore I = \dfrac{1}{2}\int\limits_0^1 {{e^t}dt}  $

$  \dfrac{1}{2}\int\limits_0^1 {{e^t}dt}  = \dfrac{1}{2}{e^t}dt $

$  \dfrac{1}{2}{e^t}dt = F(t) $ 

By second fundamental theorem of calculus, we obtain

$  I = F(1) - F(0) $

$   = \dfrac{1}{2}e - \dfrac{1}{2}{e^0} $

$   = \dfrac{1}{2}\left( {e - 1} \right) $ 


16.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{5{x^2}}}{{{x^2} + 4x + 3}}dx}} $

Ans: To simplify the question let us suppose,

$I = \int\limits_0^1 {\dfrac{{5{x^2}}}{{{x^2} + 4x + 3}}dx} $

Dividing $5{x^2}$by ${x^2} + 4x + 3$, we obtain

 $ I = \int\limits_1^2 {\left\{ {5 - \dfrac{{20x + 15}}{{{x^2} + 4x + 3}}} \right\}} dx $

  $ = \int\limits_1^2 {5dx - \int\limits_1^2 {\dfrac{{20x + 15}}{{{x^2} + 4x + 3}}} } dx $

  $ = \left[ {5x} \right]_1^2 - \int\limits_1^2 {\dfrac{{20x + 15}}{{{x^2} + 4x + 3}}dx}  $

$  I = 5 - {I_1} $ 

 Consider,

Let,

$20x + 15 = A\dfrac{d}{{dx}}\left( {{x^2} + 4x + 3} \right) + B $

   $= 2Ax + (4A + B) $ 

Equating the coefficients of $x$ and constant term, we obtain

$A = 10$ and $B =  - 25$

  Let,${x^2} + 4x + 3 = t $

 $  \Rightarrow \left( {2x + 4} \right)dx = dt $

$   \Rightarrow {I_1} = 10\int {\dfrac{{dt}}{t} - 25\int {\dfrac{{dx}}{{{{\left( {x + 2} \right)}^2} - {1^2}}}} }  $

 $  = 10\log t - 25\left[ {\dfrac{1}{2}\log \left( {\dfrac{{x + 1}}{{x + 3}}} \right)} \right]_1^2 $

 $  = \left[ {10\log 15 - 10\log 18} \right] - 25\left[ {\dfrac{1}{2}\log \dfrac{3}{5} - \dfrac{1}{2}\log \dfrac{2}{4}} \right] $

 $  = \left[ {10\log 5 + 10\log 3 - 10\log 4 - 10\log 2} \right] - \dfrac{{25}}{2}\left[ {\log 3 - \log 5 - \log 2 + \log 4} \right] $

 $  = \left[ {10 + \dfrac{{25}}{2}} \right]\log 5 + \left[ { - 10 - \dfrac{{25}}{2}} \right]\log 4 + \left[ {10 + \dfrac{{25}}{2}} \right]\log 3 + \left[ { - 10 + \dfrac{{25}}{2}} \right]\log 2 $

$   = \dfrac{{45}}{2}\log 5 - \dfrac{{45}}{2}\log 4 - \dfrac{5}{2}\log 3 + \dfrac{5}{2}\log 2 $

$   = \dfrac{{45}}{2}\log \dfrac{5}{4} - \dfrac{5}{2}\log \dfrac{3}{2} $ 

Substituting the value of ${I_1}$in (1), we obtain

$ I = 5 - \left[ {\dfrac{{45}}{2}\log \dfrac{5}{2} - \dfrac{5}{2}\log \dfrac{3}{2}} \right] $

  $ = 5 - \dfrac{5}{2}\left[ {9\log \dfrac{5}{2} - \log \dfrac{3}{2}} \right] $ 


17.Integrate the given integral $\mathbf{\int\limits_0^{\dfrac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx} }$

Ans: To simplify the question let us suppose,

$ I = \int\limits_0^{\dfrac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx}  $

 $ \int {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx = 2\tan x + \dfrac{{{x^4}}}{4} + 2x}  $

$  2\tan x + \dfrac{{{x^4}}}{4} + 2x = F(x) $ 

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{4}} \right) + F\left( 0 \right) $

$   = \left\{ {\left( {2\tan \dfrac{\pi }{4} + \dfrac{1}{4}{{\left( {\dfrac{\pi }{4}} \right)}^2} + 2\left( {\dfrac{\pi }{4}} \right)} \right) - \left( {2\tan 0 + 0 + 0} \right)} \right\} $

  $ = 2\tan \dfrac{\pi }{4} + {\dfrac{\pi }{{{4^5}}}^4} + \dfrac{\pi }{2} $

   $= 2 + \dfrac{\pi }{2} + \dfrac{{{\pi ^4}}}{{1024}} $ 


18.Integrate the given integral $\mathbf{\int\limits_0^\pi  {\left( {{{\sin }^2}\dfrac{x}{2} - {{\cos }^2}\dfrac{x}{2}} \right)dx} }$

Ans: To simplify the question let us suppose,

$ I = \int\limits_0^\pi  {\left( {{{\sin }^2}\dfrac{x}{2} - {{\cos }^2}\dfrac{x}{2}} \right)dx}  $

 $  =  - \int\limits_0^\pi  {\left( {{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}} \right)dx}  $

   $=  - \int\limits_0^\pi  {\cos xdx}  $

   $=  - \int\limits_0^\pi  {\cos xdx}  =  - \sin x $ 

By second fundamental theorem of calculus, we obtain

$  I = F\left( {\dfrac{\pi }{4}} \right) + F\left( 0 \right) $

  $ =  - \sin \pi  + \sin 0 $

$   = 0 $ 


19.Integrate the given integral $\mathbf{\int\limits_0^1 {\dfrac{{6x + 3}}{{{x^2} + 4}}dx} }$

Ans: To simplify the question let us suppose,

$ I = \int\limits_0^2 {\dfrac{{6x + 3}}{{{x^2} + 4}}dx}  $

$  \int {\dfrac{{6x + 3}}{{{x^2} + 4}}dx}  $

$   = 3\int {\dfrac{{2x + 1}}{{{x^2} + 4}}dx}  $

$   = 3\int {\dfrac{{2x + 1}}{{{x^2} + 4}}dx}  $

$   = 3\int {\dfrac{{2x}}{{{x^2} + 4}}dx}  + 3\int {\dfrac{1}{{{x^2} + 4}}dx}  $

$   = 3\log ({x^2} + 4) + \dfrac{3}{2}{\tan ^{ - 1}}\dfrac{x}{2} $

  $ = F(x) $ 

By second fundamental theorem of calculus, we obtain

 $ I = F(3) - F(2) $

   $= 3\left[ {\log \left( {{2^2} + 4} \right) - \dfrac{3}{2}{{\tan }^{ - 1}}\left( {\dfrac{2}{2}} \right)} \right] - \left[ {3\log (0 + 4) + \dfrac{3}{2}{{\tan }^{ - 1}}(\dfrac{0}{2})} \right] $

 $  = 3\left[ {\log 8} \right] + \dfrac{3}{2}{\tan ^{ - 1}}1 - 3\log 4 - \dfrac{3}{2}{\tan ^{ - 1}}0 $

 $  = 3\log 8 + \dfrac{3}{2}{\tan ^{ - 1}}1 - 3\log 4 - \dfrac{3}{2}{\tan ^{ - 1}}0 $

   $= 3\log 8 + \dfrac{3}{2}\left( {\dfrac{\pi }{4}} \right) - 3\log 4 - 0 $

   $= 3\log \left( {\dfrac{8}{4}} \right) + \dfrac{{3\pi }}{8} $

   $= 3\log 2 + \dfrac{{3\pi }}{8} $ 


20.Integrate the given integral $\int\limits_0^1 {\left( {x{e^x} - \sin \dfrac{{\pi x}}{4}} \right)dx} $

Ans: To simplify the question let us suppose,

$I = \int\limits_0^1 {\left( {x{e^x} - \sin \dfrac{{\pi x}}{4}} \right)dx}  $

$  \int\limits_0^1 {\left( {x{e^x} - \sin \dfrac{{\pi x}}{4}} \right)dx}  = x\int {{e^x}dx - \int {\left\{ {\left( {\dfrac{d}{{dx}}x} \right)\int {{e^x}dx} } \right\}dx + \left\{ {\dfrac{{ - \cos \dfrac{{\pi x}}{4}}}{{\dfrac{\pi }{4}}}} \right\}} }  $

  $ = x{e^x} - x\int {{e^x}dx - \dfrac{4}{\pi }\cos \pi \dfrac{x}{4}}  $

$   = F(x) $ 

By second fundamental theorem of calculus, we obtain

 $ I = F(1) - F(0) $

   $= \left( {{e^1} - {e^1} - \dfrac{4}{\pi }\cos \pi \dfrac{1}{4}} \right) - \left( {0.{e^0} - {e^0} - \dfrac{4}{\pi }\cos 0} \right) $

 $  = e - e - \dfrac{4}{\pi }\left( {\dfrac{1}{{\sqrt 2 }}} \right) + 1 + \dfrac{4}{\pi } $

   $= 1 + \dfrac{4}{\pi } - \dfrac{{2\sqrt 2 }}{\pi } $ 


21.Integrate the given integral $\int\limits_1^{\sqrt 3 } {\dfrac{{dx}}{{1 + {x^2}}}} $

  1. \[\dfrac{\pi }{3}\]

  2. $\dfrac{{2\pi }}{3}$

  3. $\dfrac{\pi }{6}$

  4. $\dfrac{\pi }{{12}}$

Ans: To simplify the question let us suppose,

$  \int {\dfrac{{dx}}{{1 + {x^2}}}}  = {\tan ^{ - 1}}x $

By second fundamental theorem of calculus, we obtain

$ \int\limits_1^{\sqrt 3 } {\dfrac{{dx}}{{1 + {x^2}}}}  = F(\sqrt 3 ) - F(1) $

 $  = {\tan ^{ - 1}}\sqrt 3  - {\tan ^{^{ - 1}}}1 $

   $= \dfrac{\pi }{3} - \dfrac{\pi }{4} $

$   = \dfrac{\pi }{{12}} $ 

Hence, the correct Answer is $\dfrac{\pi }{{12}}$


22.Integrate the given integral$\int {\dfrac{{dx}}{{4 + 9{x^2}}}}  = \int {\dfrac{{dx}}{{{{(2)}^2} + {{(3x)}^2}}}}$ 

  1. $\dfrac{\pi }{6} $

  2. $\dfrac{\pi }{{12}} $

  3. $\dfrac{\pi }{{24}} $

  4. $\dfrac{\pi }{4} $ 

equals

Ans: To simplify the question let us suppose,

 3x = t

  $\Rightarrow 3dx = dt$ 

$\int {\dfrac{{dx}}{{{{(2)}^2} + {{(3x)}^2}}}}  = \dfrac{1}{3}\int {\dfrac{{dt}}{{{{(2)}^2} + {{(t)}^2}}}}  $

  $ = \dfrac{1}{3}\left[ {\dfrac{1}{2}{{\tan }^{ - 1}}\dfrac{t}{2}} \right] $

  $ = \dfrac{1}{6}{\tan ^{ - 1}}\left( {\dfrac{{3x}}{2}} \right) $

   $= F(x) $ 

By second fundamental theorem of calculus,

$\int\limits_0^{\dfrac{2}{3}} {\dfrac{{dx}}{{4 + 9{x^2}}}}  = F\left( {\dfrac{2}{3}} \right) - F(0) $

   $= \dfrac{1}{6}{\tan ^{ - 1}}\left( {\dfrac{3}{2} \times \dfrac{2}{3}} \right) - \dfrac{1}{6}{\tan ^{^{ - 1}}}0 $

 $  = \dfrac{1}{6}{\tan ^{ - 1}}1 - 0 $

 $  = \dfrac{1}{6} \times \dfrac{\pi }{4} $

 $  = \dfrac{\pi }{{24}} $ 

Hence, the correct Answer is C.$\dfrac{\pi }{{24}}$

NCERT Solutions for Class 12 Maths PDF Download

Important Questions for Class 12 Maths Chapter 7

Given below are few important questions from Chapter 7 Integrals that came in Class 12 Board Exam:

  1. Let A = {1, 2, 3… 12} and R be a relation in A × A defined by (a, b) R (c, d) if ad = bc ∀ (a, b), (c, d) ∈ A × A. Prove that R is an equivalence relation. Also obtain the equivalence class [(3, 4)].

  2. A trust invested some money in two types of bonds. The first bond pays 10% interest and the second bond pays 12% interest. The trust received ₹ 2800 as interest. However, if the trust had interchanged money in bonds, they would have got ₹ 100 less as interest using matrix method, finding the amount invested in each bond by the trust.

  3. If ‘*’ is a binary operation on R defined by a * b = a + b + ab. Prove that * is commutative and associative. Find the identity element. Also show that every element of R is invertible except –1.

NCERT Solution Class 12 Maths of Chapter 7 All Exercises


Chapter 7 - Integrals Exercises in PDF Format

Exercise 7.1

22 Questions & Solutions (21 Short Answers, 1 MCQs)

Exercise 7.2

39 Questions & Solutions (37 Short Answers, 2 MCQs)

Exercise 7.3

24 Questions & Solutions (22 Short Answers, 2 MCQs)

Exercise 7.4

25 Questions & Solutions (23 Short Answers, 2 MCQs)

Exercise 7.5

23 Questions & Solutions (21 Short Answers, 2 MCQs)

Exercise 7.6

24 Questions & Solutions (22 Short Answers, 2 MCQs)

Exercise 7.7

11 Questions & Solutions (9 Short Answers, 2 MCQs)

Exercise 7.8

6 Questions & Solutions (6 Short Answers)

Exercise 7.9

22 Questions & Solutions (20 Short Answers, 2 MCQs)

Exercise 7.10

10 Questions & Solutions (8 Short Answers, 2 MCQs)

Exercise 7.11

21 Questions & Solutions (19 Short Answers, 2 MCQs)

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.9

Opting for the NCERT solutions for Ex 7.9 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.9 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

 

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 7 Exercise 7.9 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

 

Besides these NCERT solutions for Class 12 Maths Chapter 7 Exercise 7.9, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

 

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 7 Exercise 7.9 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well. 

FAQs on NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.9

1. What is the meaning of NCERT solutions for class 12 maths chapter 7 integral Exercise 7.9

NCERT solution for class 12 maths chapter 7 integral Exercise 7.9  contains an explanation of each concept and solution and is taken from the topics named “The second fundamental theorem of integral calculus”, “The first fundamental theorem of integral calculus”, Area function. This chapter concerns the fundamental theorem of calculus, having all the different aspects of integral calculus. These solutions help the students to understand the method of applying the concepts of integral in each problem they are unable to solve.

2. What do you know by a definite integral?

A definite integral has a unique value. It is the area under a curve between two fixed points. It is denoted by int ab f(x)dx Where a is called the lower limit and b is called the upper limit. The definite integral is the antiderivative of the function to obtain the function f(x) and the upper and lower limit is applied to find the value f(b) - f(a). The central operation we use the summing up of an infinite number of infinitesimally thin things is one way to visualize it as integral calculus.

3. Where can I download the NCERT solution for class 12 maths chapter 7 integral Exercise 7.9?

I can download the NCERT solution for class 12 maths chapter 7 integral Exercise 7.9 from vedantu.com. Solutions are designed by expert teachers of Vedantu. It helps in the preparation for board exams and competitive exams. It contains the concept of integral. Students can learn about definite integral calculus and its properties. These NCERT Solutions are straightforward and can be understood easily.  This solution can be downloaded For free to practice them offline as well.

4. What is the importance of definite integral?

We can use definite integrals to find the area under, over, or between curves in calculus. When the area between the curve of the function and the x-axis is equal to the definite integration of the function in the given interval the function is said to be strictly positive. It is used in architecture and electrical engineering. In economics, it helps to find out the total cost function and total revenue function from the marginal cost. A definite integral can be used to determine the mass of the object if its density function is known. Work can also be calculated by integrating a force function.

5. Why should I practice NCERT solutions class 12 maths chapter 7 integral Exercise 7.9?

We should practice NCERT solutions class 12 chapter 7 integral exercise 7.9 because it helps to learn all the rules for the definite integral and to find the area, volume, etc for a defined range as a limit of a sum. We also learn the properties, formulas, and how to find the definite integral. The practice of their proofs gives a better understanding. Students practice to perform well in higher grades. practice help to solve other related problems. It also helps to understand the concept of topics and develop problem-solving skills.