What is the net area between f(x) = e^(3x)-4x and the x-axis over x in [1, 2 ]?

Dec 4, 2017

$\frac{1}{3} {e}^{6} - \frac{1}{3} {e}^{3} - 6 \cong 121.78$

Explanation:

We can represent this area as the integral of $f \left(x\right)$ on the interval $\left[1 , 2\right]$:
${\int}_{1}^{2} {e}^{3 x} - 4 x \setminus \mathrm{dx}$

We start by computing the antiderivative:
$\int \setminus {e}^{3 x} - 4 x \setminus \mathrm{dx} = \int \setminus {e}^{3 x} \setminus \mathrm{dx} - \int \setminus 4 x \setminus \mathrm{dx}$

The left one can be solved using a u-substitution with $u = 3 x$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 3$:
$\int \setminus {e}^{3 x} \setminus \mathrm{dx} = \int \setminus {e}^{u} / 3 \setminus \mathrm{du} = {e}^{u} / 3 = {e}^{3 x} / 3$

So now we know:
$\int \setminus {e}^{3 x} - 4 x \setminus \mathrm{dx} = {e}^{3 x} / 3 - 2 {x}^{2}$

We can then evaluate the definite integral:
${\int}_{1}^{2} {e}^{3 x} - 4 x \setminus \mathrm{dx} = {\left[{e}^{3 x} / 3 - 2 {x}^{2}\right]}_{1}^{2} = \left(\frac{1}{3} {e}^{6} - 8\right) - \left(\frac{1}{3} {e}^{3} - 2\right)$

$= \frac{1}{3} {e}^{6} - \frac{1}{3} {e}^{3} - 8 + 2 = \frac{1}{3} {e}^{6} - \frac{1}{3} {e}^{3} - 6 \cong 121.78$