Use translation and expansion/contraction to prove an integral identity

by RoRi

July 26, 2015

Use the expansion/contraction theorem and the translation property to prove

$\int_a^b f(x-c) \, dx = \int_{c-b}^{c-a} f(x) \, dx.$

Proof. By the expansion/contraction property (with $k = -1$ ) we have,

$\int_a^b f(c-x) \, dx = - \int_{-a}^{-b} f(x+c) \, dx = \int_{-b}^{-a} f(x+c) \, dx.$

Then, using the translation property we have,

$\int_{-b}^{-a} f(x+c) \, dx = \int_{c-b}^{c-a} f(x) \, dx. \qquad \blacksquare$

Stumbling Robot