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Evaluation of Definite Integrals by substitution

When the variable in a definite integral is changed, the substitution in terms of new variable should be effected at three place

(i) In the integral,

(ii) In the differential, say , dx

(iii) In the limits

The limits of the new variable, say, t are simply the values of t corresponding to the original variable, say , x , and so they can be easily obtained by putting the values of x in the substitutional relation between x and t.  The method is illustrated in the following example

Example

\displaystyle \int _{ 0 }^{ 4 }{ \frac { 1 }{ x+\sqrt { x } } } dx

Solution:

\displaystyle Let\quad I=\int _{ 0 }^{ 4 }{ \frac { 1 }{ x+\sqrt { x } } } dx\\ Put\quad x={ t }^{ 2 }\quad so\quad that\quad dx=2t\quad dt.\quad when\quad x=0,\quad x={ t }^{ 2 }\Rightarrow t=0.\\ When\quad x=4,{ t }^{ 2 }=x\Rightarrow { t }^{ 2 }=4\Rightarrow t=2

\displaystyle \therefore I=\int _{ 0 }^{ 4 }{ \frac { 1 }{ x+\sqrt { x } } } dx\\ =\int _{ 0 }^{ 2 }{ \frac { 2t }{ { t }^{ 2 }+t } } dt\\ =2\int _{ 0 }^{ 2 }{ \frac { t }{ { t }+1 } } dt\\ =2{ \left[ \log { \left( t+1 \right) } \right] }_{ 0 }^{ 2 }\\ =2\left[ \log { 3 } -\log { 1 } \right] \\ =2\log { 3 }  

April 18, 2019
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