Change of Limits of Integration

Theorem
Let $f : \R \to \R$ be a real function.

Let $f$ be integrable.

Let $a$, $b$, and $c$ be real numbers.

Then:


 * $\displaystyle \int_a^b f \left({x}\right) \ \mathrm d x = \int_{a + c}^{b + c} f \left({x - c}\right) \ \mathrm d x$

Proof
Let $F$ be the primitive of $f \left({x}\right)$, i.e.:


 * $\displaystyle \int f \left({x}\right) = F \left({x}\right)$

By Fundamental Theorem of Calculus:


 * $\displaystyle \int_a^b f \left({x}\right) \ \mathrm d x = F \left({b}\right) - F \left({a}\right)$

Let $G$ be the primitive of $f \left({x - c}\right)$, i.e.:


 * $\displaystyle \int f \left({x - c}\right) = G \left({x}\right)$

By Fundamental Theorem of Calculus:


 * $\displaystyle \int_{a + c}^{b + c} f \left({x - c}\right) \ \mathrm d x = G \left({b + c}\right) - G \left({a + c}\right)$

By Primitive of Function of $a x + b$:


 * $\displaystyle F \left({x}\right) = G \left({x + c}\right)$

Therefore, we have:


 * $F \left({b}\right) = G \left({b + c}\right)$
 * $F \left({a}\right) = G \left({a + c}\right)$

The result follows.