# How do you use the second fundamental theorem of Calculus to find the derivative of given int [(ln(t)^(2))/t]dt from [3,x]?

Apr 7, 2016

${\int}_{3}^{x} {\ln}^{2} \frac{t}{t} \mathrm{dt} = \frac{{\ln}^{3} \left(x\right) - {\ln}^{3} \left(3\right)}{3}$

#### Explanation:

First, note that using the substitution $u = \ln \left(t\right) \implies \mathrm{du} = \frac{1}{t} \mathrm{dt}$ we have

$\int {\ln}^{2} \frac{t}{t} \mathrm{dt} = \int {u}^{2} \mathrm{du} = {u}^{3} / 3 + C = {\ln}^{3} \frac{t}{3} + C$

The second fundamental theorem of calculus states that
$F ' \left(x\right) = f \left(x\right) \implies {\int}_{a}^{b} f \left(x\right) \mathrm{dx} = F \left(b\right) - F \left(a\right)$

By the above, we have $\frac{d}{\mathrm{dt}} {\ln}^{3} \frac{t}{3} = {\ln}^{2} \frac{t}{t}$, and so

${\int}_{3}^{x} {\ln}^{2} \frac{t}{t} \mathrm{dt} = {\ln}^{3} \frac{x}{3} - {\ln}^{3} \frac{3}{3}$

$= \frac{{\ln}^{3} \left(x\right) - {\ln}^{3} \left(3\right)}{3}$

Apr 7, 2016

Instead of actually finding the function and then differentiating, one could 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 ${\int}_{3}^{x} \left[\ln {\left(t\right)}^{2} / t\right] \mathrm{dt}$.

Note that since we are asked about the interval $\left[3 , x\right]$, we must have $x > 3$ (otherwise the interval is either empty or undefined).

So, $\ln {\left(t\right)}^{2} / t$ is continuous on $\left[3 , x\right]$, and

${\int}_{3}^{x} \left[\ln {\left(t\right)}^{2} / t\right] \mathrm{dt} = F \left(x\right) - F \left(3\right)$ where $F$ is a function such that $F ' \left(x\right) = \ln {\left(x\right)}^{2} / x$.

And there is our answer. The derivative of ${\int}_{3}^{x} \left[\ln {\left(t\right)}^{2} / t\right] \mathrm{dt}$ is $\ln {\left(x\right)}^{2} / x$.

NOTE

The First Fundamental Theorem of Calculus says this directly. It says:

If $f$ is continuous on $\left[a , b\right]$ and $g$ is defined for all $x$ in $\left[a , b\right]$ by,

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