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(C) Let $I = \frac{1}{c}\int_{ac}^{bc} {f\left( \frac{x}{c} \right)} \,dx$. Substitute $\frac{x}{c} = t$,which implies $x = ct$. Then,the differential

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$\int_a^b {f(x)\,dx}$

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(C) Let $I = \frac{1}{c}\int_{ac}^{bc} {f\left( \frac{x}{c} \right)} \,dx$. Substitute $\frac{x}{c} = t$,which implies $x = ct$. Then,the differential $dx = c \, dt$. Now,change the limits of integration: When $x = ac$,$t = \frac{ac}{c} = a$. When $x = bc$,$t = \frac{bc}{c} = b$. Substituting these into the integral,we get: $I = \frac{1}{c} \int_{a}^{b} f(t) \cdot (c \, dt)$ $I = \int_{a}^{b} f(t) \, dt$. Since the variable of integration is a dummy variable,we can replace $t$ with $x$: $I = \int_{a}^{b} f(x) \, dx$. Thus,the correct option is $C$.

Assume that $f$ is continuous everywhere,then $\frac{1}{c}\int_{ac}^{bc} {f\left( {\frac{x}{c}} \right)} \,dx = $

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