No products in the cart.
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
Solution:
![\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 } Â](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Ctherefore+I%3D%5Cint+_%7B+0+%7D%5E%7B+4+%7D%7B+%5Cfrac+%7B+1+%7D%7B+x%2B%5Csqrt+%7B+x+%7D+%7D+%7D+dx%5C%5C+%3D%5Cint+_%7B+0+%7D%5E%7B+2+%7D%7B+%5Cfrac+%7B+2t+%7D%7B+%7B+t+%7D%5E%7B+2+%7D%2Bt+%7D+%7D+dt%5C%5C+%3D2%5Cint+_%7B+0+%7D%5E%7B+2+%7D%7B+%5Cfrac+%7B+t+%7D%7B+%7B+t+%7D%2B1+%7D+%7D+dt%5C%5C+%3D2%7B+%5Cleft%5B+%5Clog+%7B+%5Cleft%28+t%2B1+%5Cright%29+%7D+%5Cright%5D+%7D_%7B+0+%7D%5E%7B+2+%7D%5C%5C+%3D2%5Cleft%5B+%5Clog+%7B+3+%7D+-%5Clog+%7B+1+%7D+%5Cright%5D+%5C%5C+%3D2%5Clog+%7B+3+%7D+%C2%A0+&bg=ffffff&fg=000&s=1&c=20201002)
