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

Mar 23, 2016

The area is $\frac{{e}^{6} - {e}^{3} - 6}{3} \approx 125.7811$.

#### Explanation:

The area between a curve and the $x$-axis can be found through integrating the function.

Here, we have

$\text{area} = {\int}_{1}^{2} \left({e}^{3 x} - 2 x + 1\right) \mathrm{dx}$

We can integrate term by term and then evaluate from $1$ to $2$.

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

The first integral is the most difficult to find. We will want to get it into the form

$\int {e}^{u} \mathrm{du} = {e}^{u} + C$

Thus, we set $u = 3 x$, so $\mathrm{du} = 3 \mathrm{dx}$.

Manipulating just the first integral, we see:

${\int}_{1}^{2} {e}^{3 x} \mathrm{dx} = \frac{1}{3} {\int}_{1}^{2} {e}^{3 x} \cdot 3 \mathrm{dx} = \frac{1}{3} {\int}_{3}^{6} {e}^{u} \mathrm{du}$

It is very important to note that switching from $x$ to $u$ requires us to switch the bounds of the definite integral. The $1$ and $2$ have both been plugged into the $u = 3 x$ expression, becoming $3$ and $6$, respectively.

Now that we've found this integral, we can simply integrate the other two as well. All together, we have the expression:

$= \frac{1}{3} {\left[{e}^{u}\right]}_{3}^{6} - 2 {\left[{x}^{2} / 2\right]}_{1}^{2} + {\left[x\right]}_{1}^{2}$

Which gives:

$= \frac{1}{3} \left({e}^{6} - {e}^{3}\right) - 2 \left({2}^{2} / 2 - {1}^{2} / 2\right) + \left(2 - 1\right)$

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

Approximately, this is $\approx 125.7811$.