I understand how to do this when the x is on top but stuck on how to do it with it on the bottom

Question

anonymous · Answer

If $$f(x)=\int\limits_{x}^{17}t^2dt$$
 then
f'(x) =

anonymous · Answer

t^2

anonymous · Answer

oh wait maybe not.

anonymous · Answer

can't you just put in a negative and switch the lims of int.?

anonymous · Answer

$$
f(x)=\int\limits_{x}^{17}t^2dt = -\int\limits_{17}^{x}t^2dt
$$

anonymous · Answer

duh, ok so then how do I use the 17? the last one was 0 on the bottom and didn't matter

anonymous · Answer

You could also consider that: $$
\int\limits_{x}^{17}t^2dt = \int\limits_{0}^{17}t^2dt - \int\limits_{0}^{x}t^2dt
$$Since: $$
\frac{d}{dx}\int\limits_{0}^{17}t^2dt  = 0
$$It would still work out.

anonymous · Answer

do it like you would a normal integration problem 
(-t^3/3) from (17 to x)

anonymous · Answer

then differentiate

anonymous · Answer

@heradog Since $17$ is constant with respect to $x$, it doesn't matter.

anonymous · Answer

The bottom can be any constant with respect to $x$ and it still works.

anonymous · Answer

I didn't think it did, but where does -t^3/3 come from?

anonymous · Answer

It's an anti-derivative of $t^2$, don't worry about it because you don't need to use it here.

anonymous · Answer

Just use $$ g(x) = \int_a^x f(t)dt \implies g'(x) = f(x) $$

anonymous · Answer

I got the right answer of -x^2 which is where I was hung up.  I thought x^2 was the answer but because it was switched it had to be a negative.  I've got this one now, so can you help me with one that's more complicated?