# How do you use the second fundamental theorem of Calculus to find the derivative of given int 1/(t^5) dt from [1,x]?

Apr 7, 2016

You could either evaluate the integral and then differentiate or reason as follows:

#### Explanation:

The Second Fundamental Theorem of Calculus says that

If $f$ is continuous on $\left[a , b\right]$, then

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = F \left(b\right) - F \left(a\right)$
where $F$ is a function for which $F ' \left(x\right) = f \left(x\right)$ for all $x$ in $\left[a , b\right]$.

In this case we are using the variable $t$ in the integrand and the variable $x$ as the upper limit of integration.

We want the derivative of ${\int}_{1}^{x} \frac{1}{t} ^ 5 \mathrm{dt}$.

Note that since we are asked about the interval $\left[1 , x\right]$, we must have $x > 1$ (otherwise the interval is either empty or undefined).

So, $\frac{1}{t} ^ 5$ is continuous on $\left[1 , x\right]$, and

${\int}_{1}^{x} \frac{1}{t} ^ 5 \mathrm{dt} = F \left(x\right) - F \left(1\right)$ where $F$ is a function such that $F ' \left(x\right) = \frac{1}{x} ^ 5$.

And there is our answer. The derivative of ${\int}_{1}^{x} \frac{1}{t} ^ 5 \mathrm{dt}$ is $\frac{1}{x} ^ 5$.