Why does integration by parts work?

Question

3250 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T13:19:36

Because of the product rule for differentiation.

Recall that the integral of a function is the function (family) that has that derivative.

#int f(x) dx = F(x) +C# if and only if #F'(x) = f(x)#.

That is to say

#int F'(x) dx = F(x) +C#.

We know from our study of derivatives that

#d/dx(f(x)g(x)) = f(x) g'(x) + g(x) f'(x)#.

Written in terms of differentials, we have:

#d(uv) = u dv + v du#.

So, with #uv# in the role of #F(x)# above, we have

#int d(uv) = int (udv+vdu)#.

So,

#uv = intudv + intvdu#.

And

#intudv + intvdu = uv#.

Consequently,

#intudv = uv-intvdu#.

Written using prime notation, we have

#d/dx(f(x)g(x)) = f(x) g'(x) + g(x) f'(x)#.

#f(x)g(x) = int f(x) g'(x)dx + intg(x) f'(x)dx#

#int f(x) g'(x)dx + intg(x) f'(x)dx = f(x)g(x) #

#int f(x) g'(x)dx = f(x)g(x)- intg(x) f'(x)dx #