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: Whenever applying Theorem 2 you must only take into account the variable of integration ($a$ is fixed):
$$
\int_a^b f(x)\, dx = \int_0^{b-a} f(x+a)\, dx = (b-a)\int_0^1 f((b-a)x +a ) \, dx.
$$
To see it more clearly, just take an $x \in [0,1]$. Then, $(b-a)x$ lies in $[0,b-a]$, and $(b-a)x+a$ lies in $[a,b]$, which was the original interval. Or you can also write $f_a(x)=f(x+a)$, so that you have
\begin{align*}
\int_0^{b-a} f(x+a)\, dx &= \int_0^{b-a} f_a(x)\, dx = (b-a)\int_0^1 f_a((b-a)x)\, dx \\
& = (b-a)\int_0^1 f((b-a)x +a ) \, dx.
\end{align*}
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.
