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Let $h(x)=cx+n$ with $c>0$. Defining $g(x)=cf(h(x))$. Q: Prove that $$\int^b_a f (x)dx=\int^{h^{-1}(b)}_{h^{-1}(a)} g(x)dx $$

How can I prove it?

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edited: c instead of m. – 17SI.34SA Apr 15 '13 at 18:19
Maybe I'm overlooking something, but doesn't it follow immediately from the Riemann sum definition of the integral? – Hurkyl Apr 15 '13 at 18:33
yes -- same Riemann sums.. – Halil Duru Apr 15 '13 at 19:26
up vote 1 down vote accepted

Hint: Use Fundamental Theorem of Calculus (evaluation version) for both sides or simply write

the Riemann sum for both integrals and establish a correspondence..

The setting is :

[Let $ cA+n=a$ and $cB+n =b$ and P a partition of $[a,b]$ with $ a=x_0<x_1<x_2 ...<x_k=b$ and

the corresponding partition D of [A,B] is $A=X_0<X_1<X_2...<X_K=B$ where $h(X_i)=x_i$.]

$I=\int_a^b f(x) dx $ means that $\forall $$\varepsilon >0 $ $\exists \delta$ such that $\Vert P \Vert<\delta $ implies |$\sum _{i=1}^k f(x^*_i)\Delta x_i -I|<\varepsilon$ .

So we can use this information to state the following :

$\Vert D \Vert < \delta/c$ implies |$\sum _{i=1}^k c f(h(X^*_i)) \Delta X_i -I|=|\sum _{i=1}^k f(x^*_i)\Delta x_i -I|<\varepsilon $ (using $c . \Delta X_i=\Delta x_i$)

But this assures us that $\int_A^B cf(h(x)) dx=I $ , too.

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Could you elaborate more on the way with Riemann sums? – 17SI.34SA Apr 15 '13 at 20:35
i tried to explain above .. – Halil Duru Apr 16 '13 at 18:40

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