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Prove an equivalent form of the expansion/contraction of interval of integration

by RoRi

By Theorem 1.8 (Apostol, p. 68) we may expand or contract the interval of integration for a definite integral by the following formula:

    \[ \int_{ka}^{kb} s \left( \frac{x}{k} \right) \, dx = k \int_a^b s(x) \, dx \qquad \text{for all } k \in \mathbb{R}_{>0}. \]

Prove that the following formula is equivalent for all k \in \mathbb{R}_{>0}:

    \[ \int_{ka}^{kb} f(x) \, dx = k \int_a^b f(kx) \, dx. \]

Proof. From the theorem we have, for any j > 0,

    \[ \int_{a/j}^{b/j} f\left( j\frac{x}{j}\right) \, dx = \frac{1}{j} \cdot \int_a^b f \left( \frac{x}{j} \right) \, dx. \]

Which gives us,

    \[ j \int_{a/j}^{b/j} f(x) \, dx = \int_a^b f \left( \frac{x}{j} \right) \, dx. \]

Then, if j \in \mathbb{R}_{>0}, then we also have \frac{1}{j} \in \mathbb{R}_{>0}, so we may apply the theorem with k = 1/j,

    \begin{align*} \implies && \frac{1}{k} \cdot \int_{ka}^{kb} f(x) \, dx &= \int_a^b f(kx) \, dx \\ \implies && k \cdot \int_a^b f(kx) \, dx &= \int_{ka}^{kb} f(x) \, dx. \qquad \blacksquare \end{align*}

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