Chapter 7: Integrals

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 $ 

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} $ 

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} $

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} $ 

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 $

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) $ 

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 $ 

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) $ 

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} $ 

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} $