To illustrate my question, consider the function given by the following equation:


My AP Calculus AB teacher says the anti-derivative is $F(x) = 3(x-8)^{12}+C$. I understand easily using the reverse power rule that if we simply had $g(x)=36x^{11}$, then we could easily use the reverse power rule as well as the constant multiple rule to get an anti-derivative of $G(x) = 3x^{12}+C$. But what to do about the translation -- and why does what the teacher did work?

I know from class that $[f(x+a)]' = f'(x+a)$, which was justified to us graphically and seemed to make sense. But that is a differentiation rule, not an integration rule. Obviously, the two operations are related, but I've been itching to see a more formal proof of why the "horizontal translation" rule apparently holds in both cases... IF it universally holds, of course -- maybe it's in fact specific to certain types of functions, and just happens to hold in the particular problem my teacher gave us!

Is anyone willing to attempt to justify why what my teacher did to find the anti-derivative works, in a satisfying way? (The chain rule is, I suppose, okay if you must use it, although a proof without would be better because I haven't really learned the chain rule much yet.)

  • $\begingroup$ The antiderivative equivalent to the chain rule is called change of variable. For $\int36(x-8)^{11}\,dx$ set $y=x-8$. Then $\frac{dy}{dx} = 1$ (derivative of $x-8$), so $dy = dx$ and so, by substituting $x-8$ by $y$ and $dx$ by $dy$, we get $\int36(x-8)^{11}\,dx = \int 36y^{11}\,dy = 36\cdot \int y^{11}\,dy = 36\cdot \frac 1{12}y^{12} = 3y^{12} = 3(x-8)^{12}$. $\endgroup$
    – amsmath
    Oct 14, 2019 at 4:22
  • 1
    $\begingroup$ Once you know that $[f(x+a)]' = f'(x+a)$, you can get the integral result using Jake Mirra's answer. To know that $[f(x+a)]' = f'(x+a)$ you can use the chain rule, or justify it graphically. The reason translations work like this is because the derivative of $x-a$ with respect to $x$ is 1. $\endgroup$ Oct 14, 2019 at 4:50
  • $\begingroup$ I do think understanding the chain rule is essential before understanding why the horizontal translation works. $\endgroup$ Oct 14, 2019 at 4:54
  • $\begingroup$ @Ameet Sharma, incidentally, how would you apply the chain rule here to prove the differentiation property? I seem to recall that $dy/dx = dy/du * du/dx$, and so it makes sense to me why in this case if we let $u=x+a$ the derivative of $u$ with respect to $x$ is simply 1. But how do we find $dy/du$ = $f'(x+a)$? $\endgroup$
    – Will
    Oct 14, 2019 at 4:58
  • $\begingroup$ Suppose $y = 3(x-8)^{12} + C$. Let $u= x-8$. Then $y = 3u^{12}+C$. $\frac{dy}{du} = 36u^{11}$. $\frac{du}{dx} = 1$. $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = 36u^{11} = 36(x-8)^{11}$ $\endgroup$ Oct 14, 2019 at 5:04

1 Answer 1


It really is as simple as, if $ g'(x) = h(x) $, then $ \int{h(x)dx} = g(x) $.

Now, in the above sentence, replace $ g(x) $ with $ f(x-c) $ and $ h(x) $ with $ f'(x-c) $. Theres nothing more to say.

  • $\begingroup$ Would you be able to expand on your answer a bit? As things stand right now, I feel like it kind of leaves me in the place I started. Also, we haven't used integrals very much in class - we've stuck to the notion of anti-derivatives, which we denote with a capital letter for the function instead of a lower-case one. $\endgroup$
    – Will
    Oct 14, 2019 at 4:49
  • 2
    $\begingroup$ Are you comfortable with the notion that $ \cfrac{d}{dx} \left[ 3(x-8)^{12} \right] = 36(x-8)^{11} $? If so that means the thing on the left is the anti-derivative of the thing on the right. If you're not comfortable with that fact, then that's where your real confusion lies. $\endgroup$
    – Jake Mirra
    Oct 14, 2019 at 5:06
  • $\begingroup$ So, to confirm, it's true, as a general rule, that the anti-derivative of $f(x+a)$ equals $F(x+a)$, where $F(x)$ is the anti-derivative of $f(x)$? $\endgroup$
    – Will
    Oct 14, 2019 at 6:08
  • 1
    $\begingroup$ Other than some pedantic gesticulations about how the equality of anti derivatives is only up to an additive constant, the answer is, "Yes." $\endgroup$
    – Jake Mirra
    Oct 14, 2019 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.