Derive an identity of integral formulas

by RoRi

July 26, 2015

For $f$ an integrable function on $[a,b]$ prove the following identity:

$\int_a^b f(x) \, dx = (b-a) \int_0^1 f(a+(b-a)x) \, dx.$

Proof. First, we use the theorem on invariance of the integral under translation (Theorem 1.18 of Apostol), to write,

$\int_a^b f(x) \, dx = \int_0^{b-a} f(x+a) \, dx.$

Then, using the expansion/contraction of the interval of integration (Theorem 1.19 of Apostol) we obtain

$\int_0^{b-a} f(x+a) \, dx = (b-a) \int_0^1 f(x(b-a) + a) \, dx = (b-a) \int_0^1 f(a + (b-a)x). \qquad \blacksquare$

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