Question

How do you use part I of the Fundamental Theorem of Calculus to find the derivative of `f(x) = int {1} / {1+t^{2}} dt` from x to 5?

Answer 1

See the explanation section.

Explanation:

FTC 1 says (among other things) that if `g` is continuous on interval `[a,b]` and we define

`f(x) = int_a^x g(t) dt` for `x` in `[a,b]` then `f'(x) = g(x)`

If you meant `f(x) = int_5^x {1} / {1+t^{2}} dt`
(which is read "the integral from `5` to `x`")

the `f'(x) = 1/(1+x^2)`.

If you really meant the integral from `x` to `5`, we will first rewrite so that the lower limit is a constant.

`f(x) = int_x^5 1 / {1+t^{2}} dt = -int_5^x {1} / {1+t^{2}} dt`, so

`f'(x) = -1/(1+x^2)`.

Note 1
In some ways, the hardest part of the FTC1 question on an exam is convincing yourself that is really is that easy!

Note 2

Some courses apparently give a slightly different version of Part 1 of the FTC.

For `f(x) = int_(g(x))^(h(x)) F(t) dt`,

we get

`f'(x) = F(h(x)) h'(x) - F(g(x)) g'(x)`

So you may see that sometimes here on Socratic.