show that $\int_{0}^{1}[a-(a-b)x]dx = \frac{1}{b-a}\int_{a}^{b}(x)dx$

Question

I am currently studying for my cal final and I found this problem in an exam of a previous year.

If is continuous on [a,b] show that $\int_{0}^{1}[a-(a-b)x]dx = \frac{1}{b-a}\int_{a}^{b}(x)dx$

How can it be solved?

Answer 1

Let $u=a-(a-b)x$

$du = -(a-b)dx$

Then you have $$\int_{0}^{1}[a-(a-b)x]dx = \frac{1}{b-a}\int_{a}^{b}(u)du=$$

$$\frac{1}{b-a}\int_{a}^{b}(x)dx$$

Answer 2

Hint: Substitute $a-(a-b)x$ with a suitable variable.