# What is the net area between f(x) = xe^x-3x  and the x-axis over x in [1, 5 ]?

Mar 4, 2018

The net area between $x {e}^{x} - 3 x$ and the x-axis over $x \in \left[1 , 5\right]$ is equal to color(purple)(5(ln5 - 1) - ln(3)(ln(ln3) - 1) + (3ln^2 3 - 75)/2.

#### Explanation:

Let $A$ be the given net area. Then,

$A = {\int}_{{x}_{1}}^{{x}_{2}} x {e}^{x} - 3 x \mathrm{dx}$.

Now, since the area wanted is the area above the x-axis, that means we have find the minimum and maximum values of $x \in \left[1 , 5\right]$, ${x}_{1}$ and ${x}_{2}$ respectively, such that $f \left({x}_{1}\right)$ and $f \left({x}_{2}\right)$ are both bigger than $0$.

Let's plug in ${x}_{1} = 1$ and ${x}_{2} = 5$ firstly;

$f \left(1\right) = e - 3 < 0$
$f \left(5\right) = 5 {e}^{5} - 15 > 0$

Clearly, ${x}_{1}$ has to be bigger than $1$. Because $x {e}^{x}$ increases exponentially at a much faster rate then $3 x$, it means that $f \left(x\right)$ is monotonic on the interval $\left[1 , + \infty\right]$. We don't need the exact interval of monotony to be able to find $A$.

In order to find the exact value of ${x}_{1}$, we must solve the equation $\alpha {e}^{\alpha} - 3 \alpha = 0$ :

$\alpha \left({e}^{\alpha} - 3\right) = 0$

The two possible solutions are $0$ and $\ln 3$. However, ${x}_{1} \in \left[1 , 5\right] \implies {x}_{1} = \ln 3$. $A$ has now become

$A = {\int}_{\ln} {3}^{5} x {e}^{x} - 3 x \mathrm{dx}$.

Given the basic properties of integrals, we have

$A = \textcolor{red}{{\int}_{\ln} {3}^{5} x {e}^{x} \mathrm{dx}} - 3 \left(\textcolor{b l u e}{{\int}_{0}^{5} x \mathrm{dx} - {\int}_{0}^{\ln} 3 x \mathrm{dx}}\right)$

• $\textcolor{red}{\text{Solving the integral highlighted in red}}$ :

Let $u = {e}^{x} \implies x = \ln u$.

$\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{x} \implies \mathrm{dx} = \frac{\mathrm{du}}{e} ^ x = \frac{\mathrm{du}}{u}$

${\int}_{\ln} {3}^{5} x {e}^{x} \mathrm{dx} = {\int}_{\ln} {3}^{5} \ln \left(u\right) \cdot u \frac{\mathrm{du}}{u}$

$= {\int}_{\ln} {3}^{5} \ln u \mathrm{du}$

We will solve this with integration by parts :

${\int}_{a}^{b} f \left(u\right) g ' \left(u\right) = f \left(u\right) g \left(u\right) {|}_{a}^{b} - {\int}_{a}^{b} g \left(u\right) f ' \left(u\right)$

In our case, $f \left(u\right) = \ln \left(u\right) \implies f ' \left(u\right) = \frac{1}{u}$ and $g ' \left(u\right) = 1 \implies g \left(u\right) = u$, $b = 5$ and $a = \ln 3$.

${\int}_{\ln} {3}^{5} \ln u \mathrm{du} = 5 \ln \left(5\right) - \ln \left(3\right) \ln \left(\ln 3\right) - {\int}_{\ln} {3}^{5} u \cdot \frac{1}{u} \mathrm{du}$

$= 5 \ln \left(5\right) - \ln \left(3\right) \ln \left(\ln 3\right) - {\int}_{0}^{5} 1 \mathrm{du} + {\int}_{0}^{\ln} 3 1 \mathrm{du}$

$= 5 \ln \left(5\right) - \ln \left(3\right) \ln \left(\ln 3\right) - 5 + \ln 3$

$\textcolor{red}{= 5 \left(\ln 5 - 1\right) - \ln \left(3\right) \left(\ln \left(\ln 3\right) - 1\right)}$

• $\textcolor{b l u e}{\text{Solving the difference of integrals :}}$

We know ${\int}_{0}^{k} x \mathrm{dx} = {k}^{2} / 2$, so

$\textcolor{b l u e}{{\int}_{0}^{5} x \mathrm{dx} - {\int}_{0}^{\ln} 3 x \mathrm{dx} = \frac{25 - {\ln}^{2} 3}{2}}$

And finally,

color(purple) (A) = color(red)(5(ln5 - 1) - ln(3)(ln(ln3) - 1)) + color(blue)((3ln^2 3 - 75)/2)