# What is the integral from 0 to 1: x 5^x dx?

May 15, 2015

The answer is : ${\int}_{0}^{1} x {5}^{x} \mathrm{dx} \approx 1.56$.

We have $h \left(x\right) = {5}^{x} x = f ' \left(x\right) g \left(x\right)$.

Here, $f ' \left(x\right) = {5}^{x}$ and $g \left(x\right) = x$.

The antiderivative of $h \left(x\right)$ is given by :
$f \left(x\right) g \left(x\right) - \int f \left(x\right) g ' \left(x\right) \mathrm{dx}$.

Firstly, let's calculate the derivative of ${5}^{x}$ by using the limit definition, it will help us to get its antiderivative :

$j ' \left(x\right) = {\lim}_{h \to 0} \frac{j \left(x + h\right) - j \left(x\right)}{h}$

$\left({5}^{x}\right) ' = {\lim}_{h \to 0} \frac{{5}^{x + h} - {5}^{x}}{h} = {\lim}_{h \to 0} \frac{{5}^{x} \cdot \left({5}^{h} - 1\right)}{h}$

$= {5}^{x} {\lim}_{h \to 0} \frac{{5}^{h} - 1}{h} = {5}^{x} {\lim}_{h \to 0} \frac{{e}^{h \ln \left(5\right)} - 1}{h}$

$= {5}^{x} {\lim}_{h \to 0} \ln \left(5\right) \left(\frac{{e}^{h \ln \left(5\right)} - 1}{h \ln \left(5\right)}\right)$

$= {5}^{x} \ln \left(5\right) {\lim}_{h \to 0} \frac{{e}^{h \ln \left(5\right)} - 1}{h \ln \left(5\right)}$

$= {5}^{x} \ln \left(5\right) {\lim}_{k \to 0} \frac{{e}^{k} - 1}{k}$ , where $k = h \ln \left(5\right)$

By the definition of $e$, ${\lim}_{k \to 0} \frac{{e}^{k} - 1}{k} = 1$.

So $\left({5}^{x}\right) ' = {5}^{x} \ln \left(5\right)$.

Therefore, the antiderivative of $f ' \left(x\right) = {5}^{x}$ is $f \left(x\right) = {5}^{x} / \ln \left(5\right)$ because :

$\left({5}^{x} / \ln \left(5\right)\right) ' = \frac{\left({5}^{x}\right) '}{\ln} \left(5\right) = \frac{{5}^{x} \ln \left(5\right)}{\ln} \left(5\right) = {5}^{x}$

And the derivative of $g \left(x\right) = x$ is $g ' \left(x\right) = 1$.

Thus, the antiderivative of $h \left(x\right)$ is :

$H \left(x\right) = f \left(x\right) g \left(x\right) - \int f \left(x\right) g ' \left(x\right) \mathrm{dx} = \frac{{5}^{x} x}{\ln} \left(5\right) - \int {5}^{x} / \ln \left(5\right)$

$= \frac{{5}^{x} x}{\ln} \left(5\right) - \frac{1}{\ln} \left(5\right) \int {5}^{x} = \frac{{5}^{x} x}{\ln} \left(5\right) - {5}^{x} / {\ln}^{2} \left(5\right)$

Now, we can calculate the integral :

${\int}_{0}^{1} h \left(x\right) \mathrm{dx} = {\left[H \left(x\right)\right]}_{0}^{1} = H \left(1\right) - H \left(0\right)$

$= \frac{5}{\ln} \left(5\right) - \frac{5}{\ln} ^ 2 \left(5\right) - 0 + \frac{1}{\ln} ^ 2 \left(5\right) = \frac{5}{\ln} \left(5\right) - \frac{4}{\ln} ^ 2 \left(5\right)$

$= \frac{5 \ln \left(5\right) - 4}{\ln} ^ 2 \left(5\right) \approx 1.56$.

That's it!