# Why does an anti-derivative "preserve" a translation from the function being integrated?

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

$$f(x)=36(x-8)^{11}$$

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.)

• 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 $\int36(x-8)^{11}\,dx = \int 36y^{11}\,dy = 36\cdot \int y^{11}\,dy = 36\cdot \frac 1{12}y^{12} = 3y^{12} = 3(x-8)^{12}$. Commented Oct 14, 2019 at 4:22
• 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. Commented Oct 14, 2019 at 4:50
• I do think understanding the chain rule is essential before understanding why the horizontal translation works. Commented Oct 14, 2019 at 4:54
• @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 $dy/du$ = $f'(x+a)$?
– Will
Commented Oct 14, 2019 at 4:58
• Suppose $y = 3(x-8)^{12} + C$. Let $u= x-8$. Then $y = 3u^{12}+C$. $\frac{dy}{du} = 36u^{11}$. $\frac{du}{dx} = 1$. $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = 36u^{11} = 36(x-8)^{11}$ Commented Oct 14, 2019 at 5:04

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.
• 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. Commented Oct 14, 2019 at 5:06
• 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 $f(x)$?