# How do you use the second fundamental theorem of Calculus to find the derivative of given int t^(1/4) dt from [1,x]?

Mar 10, 2016

$\frac{d}{\mathrm{dx}} {\int}_{1}^{x} {t}^{\frac{1}{4}} \mathrm{dt} = {x}^{\frac{1}{4}}$

#### Explanation:

The second fundamental theorem of calculus states that if $f$ is a continuous function on an open interval $I$ and $a \in I$, then if
$F \left(x\right) = {\int}_{a}^{x} f \left(t\right) \mathrm{dt}$
then
$F ' \left(x\right) = f \left(x\right)$ for all $x \in I$

As $f \left(x\right) = {x}^{\frac{1}{4}}$ is continuous on $\left(0 , \infty\right)$ and $1 \in \left(0 , \infty\right)$, then if we set
$F \left(x\right) = {\int}_{1}^{x} f \left(t\right) \mathrm{dt}$
then, by the second fundamental theorem of calculus,
$F ' \left(x\right) = f \left(x\right) = {x}^{1} / 4$ for all $x \in \left(0 , \infty\right)$.

That is to say

$\frac{d}{\mathrm{dx}} {\int}_{1}^{x} {t}^{\frac{1}{4}} \mathrm{dt} = {x}^{\frac{1}{4}}$