"Let "int e^(x){f(x)-f'(x)}dx=phi(x). " ...

`"Let "int e^(x){f(x)-f'(x)}dx=phi(x). " Then "int e^(x) f(x)dx " is "`

`phi(x)+e^(x)f(x)`

`phi(x)-e^(x)f(x)`

`(1)/(2){phi(x)-e^(x)f'(x)}`

`(1)/(2){phi(x)+e^(x)f'(x)}`

Text Solution

Verified by Experts

The correct Answer is:

We know that `inte^(x){f(x)+f'(x)}dx=e^(x)f(x)` It is given that
`int e^(x){f(x)-f'(x)}dx=phi(x)`
Adding these two , we get
`2inte^(x)f(x)dx=phi(x)+e^(x)f(x)`
`rArrinte^(x)f(x)dx=(1)/(2){phi(x)+e^(x)f(x)}`

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