Indefinite integral of f(ax + b)
By Sarah Scott •
I was studying Calculus and I ran into this following expression:
$$\eqalign{\int f(ax + b)dx =\dfrac{1}{a}F(ax+b)}$$
I was wondering what it means, I couldn't figure out though. Also, what is the name of this theorem? So I can look up on the internet its proof and its usage
Thanks in advance
$\endgroup$ 21 Answer
$\begingroup$Substitute $y = ax + b$. Then $dy = adx$, so $$\int f(ax + b) \ dx = \int f(y) \ \frac{dy}{a} = \frac{1}{a}\int f(y) dy.$$
$F(x)$ commonly denotes the antiderivative of $f$. That is, $F'(x) = f(x)$ and $\int f(x) \ dx = F(x) + c$.
Put these two statements together and you're there. It's too straightforward a result to deserve a name, I would just call it a simple case of integration by substitution.
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