Prove a formula for the integral from a to b of f(Ax+B)

Claim:

$\int_a^b f(Ax+B) \, dx = \begin{dcases} \frac{1}{A} \cdot \int_{Aa + B}^{Ab + B} f(x) \, dx & \text{if } A \neq 0 \\ (b-a) f(B) \phantom{\frac{1}{2}} & \text{if } A = 0. \end{dcases}$

Proof. First, for the case $A = 0$ , we have,

$\int_a^b f(Ax+B) \, dx = \int_a^b f(B) \, dx = f(B) \int_a^b \, dx = f(B) \cdot (b-a).$

(Since $f(B)$ is just a constant.)
Next, for the case $A \neq 0$ we use the expansion/contraction of the interval of integration (Theorem 1.19 of Apostol) to get,

$\int_a^b f(Ax+B) \, dx = \frac{1}{A} \int_{Aa}^{Ab} f(x+B) \, dx.$

Then we use invariance of the integral under translation (Theorem 1.18 of Apostol) to finish proving the claimed identity,

$\frac{1}{A} \int_{Aa}^{Ab} f(x+B) \, dx = \frac{1}{A} \int_{Aa + B}^{Ab + B} f(x) \, dx. \qquad \blacksquare$

Stumbling Robot

Prove a formula for the integral from a to b of f(Ax+B)

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