integral

Kibu · New Member Offline Joined: 02 Dec 2008 Posts: 16

Is that correct?
$\int_{f(0)}^{f(k)} g(f(x)) df = \int_0^k g(f(x)) dx$

**Posted:** Wed Nov 04, 2009 9:42 am

bookworm_vn

**Posted:** Wed Nov 04, 2009 10:03 am

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Kibu · New Member Offline Joined: 02 Dec 2008 Posts: 16

So e.x. $\int_{f(0)}^{f(2)} (1 + f(x)) df = \int_0^2 (1 + x) dx = 2 + 0.5\cdot 2^2 = 4$ and it doesn't matter how does f(x) look?

**Posted:** Wed Nov 04, 2009 10:16 am

fedja

It should really be $\int_0^k g(f(x))\,df=\int_{f(0)}^{f(k)}g(x)\,dx$ . Also, be aware that there may be cases where the right hand side makes sense but the left hand side does not. As written, it is true if $f,g$ are defined on the entire line and are reasonably nice (say, $g$ continuous and $f$ continuously differentiable).

**Posted:** Wed Nov 04, 2009 5:30 pm

Kibu · New Member Offline Joined: 02 Dec 2008 Posts: 16

Thanks!!!

**Posted:** Thu Nov 05, 2009 1:16 am