Sum

If u and v are two functions of x then prove that

`intuvdx=uintvdx-int[du/dxintvdx]dx`

Hence evaluate, `int xe^xdx`

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#### Solution

Let ` int vdx=w.....(1)`

`then " " (dw)/dx=v.....(2)`

`Now d/dx(u,w)=u.d/dx(w)+wd/dx(u)`

`=u.v+w(du)/dx......."from"(2)`

By definition of integration.

`u.w=int[u.v+w(du)/dx]dx`

`=intu.vdx+intw.(du)/dx dx`

`int u.v dx=u.w-int w (du)/dx dx`

`=u int v dx-int [(du)/dxintv.dx]dx`

[next section only required for question 2]

Hence, `int xe^xdx = x.inte^xdx-int[d/dx x.inte^xdx]dx`

`=xe^x-int1xxe^xdx`

`=xe^x-e^x+c`

Concept: Methods of Integration: Integration by Parts

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