When $f$ is integrable, for what $g$ does $\int_{g(a)}^{g(b)} f(u) du = \int_a^b f(g(x))\cdot g'(x) dx$ hold?

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I am reading "Calculus 4th Edition" by Michael Spivak.

There are the following problems and theorem in this book.

On p. 274 Problem 14.
If $f$ is integrable, then $$\int_a^b f(x) dx = \int_{a+c}^{b+c} f(x-c) dx.$$

On p. 275 Problem 16.
If $f$ is integrable, then $$\int_{ca}^{cb} f(t) dt = c \int_a^b f(ct) dt.$$

On p. 369 Theorem 2.
If $f$ and $g'$ are continuous, then $$\int_{g(a)}^{g(b)} f(u) du = \int_a^b f(g(x))\cdot g'(x) dx.$$

By Problems 14 and 16, if $f$ is integrable, then
$$\int_{ca+d}^{cb+d} f(u) du = \int_{ca}^{cb} f(u+d) du = c \int_a^b f(cx+d) dx.$$ Of course, if $f$ is continuous, the above equation holds by Theorem 2.

When $f$ is integrable, for what $g$ does $$\int_{g(a)}^{g(b)} f(u) du = \int_a^b f(g(x))\cdot g'(x) dx$$ hold?