Q:

Calculus by Apostol Exercise 1.26 number 26

What I did first is to use theorem 1.18 to subtract a from the limits of integration of the integral of f(x), leaving the new limits of integration to be from 0 to (b-a). In that case I can use theorem 1.19 to assume that the k is the (b-a), by algebraic manipulation, I ended up having integral of f[(b-a)a+(b-a)x], which left me stumped, I just can't think of a way to remove that (b-a) beside a. Any suggestions?  

A:

First you need to apply theorem 1.19 by considering k to be 1/(b-a) and then apply theorem 1.18 for c = - a/(b-a). You'll get what you are looking for !!
I guess you need full workout. Here it is:
For k = 1/(b-a) Theorem 1.19 yields
$$\int_a^b f(x)dx = (b-a)\int_{\frac{a}{b-a}}^{\frac{b}{b-a}} f((b-a)x) dx$$ 
Now theorem 1.18 comes into play. Consider c = - a/(b-a) and apply it to the RHS:
$$\int_a^b f(x)dx = (b-a)\int_{\frac{a}{b-a}-\frac{a}{b-a}}^{\frac{b}{b-a}-\frac{a}{b-a}} f((b-a)(x+\frac{a}{b-a})) dx \\= (b-a)\int_{0}^{1} f(a+(b-a)x)dx$$ 
Hope it helps.

