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My teacher used this property in class and said that it's the "scalar multiple rule" or something like that, but I looked it up and couldn't find anything. I'm not sure why $\frac{1}{k}$ is necessary.

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    Try taking the derivative of the right hand side using the chain rule and see that it gives the right result (this shows why you need $1/k$). To derive it try to evaluate the integral using the substitution $u = kx + b$ – 2017-01-06

4 Answers 4

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This looks like the second part (the formula per se) of the complete statement of this rule. The complete rule is as follows:

If $\displaystyle \int f(x)\,dx=F(x)+C$, then $\displaystyle \int f(kx+b)\,dx=\frac{1}{k}F(kx+b)+C$.

And then, since you asked two questions, let's answer them both.

(1) Why does it work? We can verify any integration formula by taking the derivative. Here it will involve the Chain Rule: $$\left[\frac{1}{k}F(kx+b)+C\right]'=\frac{1}{k}f(kx+b)\cdot[kx+b]'+0=\frac{1}{k}f(kx+b)\cdot k=f(kx+b),$$ where we used the given fact that $F$ is an antiderivative of $f$.

(2) Where does it come from? From integration with a substitution. To integrate $\displaystyle \int f(kx+b)\,dx$, we substitute $u=kx+b$; then $du=k\,dx$ and therefore $\displaystyle dx=\frac{1}{k}\,du$. Substituting into the given integral, we get: $$\displaystyle \int f(kx+b)\,dx=\int f(u)\cdot\frac{1}{k}\,du=\frac{1}{k}\int f(u)\,du=\frac{1}{k}F(u)+C=\frac{1}{k}F(kx+b)+C.$$

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Differentiate both sides and apply chain rule to see that

$$f(kx+b)=\frac d{dx}\left(\frac1kF(kx+b)+c\right)$$

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Let $u = kx + b$. Then, $x = \dfrac{u-b}{k}$, so $\text{d}x = \dfrac{1}{k}\text{ d}u$.

Then we have $$\int f(kx+b)\text{ d}x = \int f(u) \cdot \dfrac{1}{k}\text{ d}u = \dfrac{1}{k}\int f(u)\text{ d}u = \dfrac{1}{k}[F(u)+c] = \dfrac{1}{k}F(u)+\dfrac{1}{k}\cdot c\text{.} $$ Substitute back in $u =kx + b$, and note that $\dfrac{1}{k} \cdot c$ is a constant, so we can write $$\int f(kx+b)\text{ d}x = \dfrac{1}{k}\cdot F(kx+b)+C$$

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Won't add to the formal derivation of what's been written before as all in order, but worth thinking with this style of question what it is getting at i.e. that integration is "reverse differentiation" loosely and going from there.