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Prove expansion/contraction and translation invariance of interval of integration using method of substitution

Use the method of substitution to prove invariance under translation (Theorem 1.18 on page 81 of Apostol) and to prove expansion or contraction of the interval of integration (Theorem 1.19 on page 81 of Apostol).


Theorem: (Invariance Under Translation) For a function f integrable on an interval [a,b] and for every c \in \mathbb{R} we have

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

Proof. If P is a primitive of f, then

    \[ \int_a^b f(x) \, dx = P(b) - P(a). \]

Let

    \[ u = g(x) = x-c \quad \implies \quad du = g'(x) \, dx = dx. \]

So,

    \begin{align*}  \int_{a+c}^{b+c} f(x-c) \, dx &= \int_{a+c}^{b+c} f(g(x)) g'(x) \, dx \\  &= P(g(b+c)) - P(g(a+c)) \\  &= P(b) - P(a). \end{align*}

Hence, we indeed have

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

Theorem: (Expansion or Contraction of the Interval of Integration) For a function f integrable on an interval [a,b] and for every k \in \mathbb{R} with k \neq 0,

    \[ \int_a^b f(x) \, dx = \frac{1}{k} \int_{ka}^{kb} f \left(\frac{x}{k} \right) \, dx. \]

Proof. Let

    \[ u = g(x) = \frac{x}{k} \quad \implies \quad du = g'(x) \, dx = \frac{1}{k} \, dx. \]

Then we have

    \begin{align*}  \frac{1}{k} \int_{ka}^{kb} f \left( \frac{x}{k} \right) \, dx &= \int_{ka}^{kb} f(g(x)) g'(x) \, dx \\  &= \int_{g(ka)}^{g(kb)} f(u) \, du \\  &= P(g(b)) - P(g(a)) \\  &= P(b) - P(a). \end{align*}

Thus, we indeed have

    \[ \int_a^b f(x) \, dx = \frac{1}{k} \int_{ka}^{kb} f \left( \frac{x}{k} \right) \, dx. \qquad \blacksquare \]

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