# 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?

Sep 11, 2015

See the explanation section.

#### Explanation:

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

$f \left(x\right) = {\int}_{a}^{x} g \left(t\right) \mathrm{dt}$ for $x$ in $\left[a , b\right]$ then $f ' \left(x\right) = g \left(x\right)$

If you meant $f \left(x\right) = {\int}_{5}^{x} \frac{1}{1 + {t}^{2}} \mathrm{dt}$
(which is read "the integral from $5$ to $x$")

the $f ' \left(x\right) = \frac{1}{1 + {x}^{2}}$.

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

$f \left(x\right) = {\int}_{x}^{5} \frac{1}{1 + {t}^{2}} \mathrm{dt} = - {\int}_{5}^{x} \frac{1}{1 + {t}^{2}} \mathrm{dt}$, so

$f ' \left(x\right) = - \frac{1}{1 + {x}^{2}}$.

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 $f \left(x\right) = {\int}_{g \left(x\right)}^{h \left(x\right)} F \left(t\right) \mathrm{dt}$,

we get

$f ' \left(x\right) = F \left(h \left(x\right)\right) h ' \left(x\right) - F \left(g \left(x\right)\right) g ' \left(x\right)$

So you may see that sometimes here on Socratic.