# What is the net area between f(x) = x^2-x+8  and the x-axis over x in [2, 4 ]?

Apr 20, 2016

$\frac{86}{3}$

#### Explanation:

The integral of the function $f \left(x\right)$ on the interval $x \in \left[a , b\right]$, that is, ${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$ is equal to the area between the curve and the $x$-axis. Thus, the answer to your question can be written as

${\int}_{2}^{4} \left({x}^{2} - x + 8\right) \mathrm{dx}$

We can integrate the first two terms using the following rule of integration:

$\int {x}^{n} = {x}^{n + 1} / \left(n + 1\right) + C$

The last, $8$, can be integrated using the following rule where $k$ is a constant:

$\int k \mathrm{dx} = k x + C$

So, when we integrate this function (momentarily ignoring the integral's bounds), we get the antiderivative

$F \left(x\right) = {x}^{3} / 3 - {x}^{2} / 2 + 8 x + C$

So, to find the integral of $f \left(x\right)$ evaluated from $2$ to $4$, we must take $F \left(4\right) - F \left(2\right)$ (this is the first fundamental theorem of calculus).

This can be notated as

${\int}_{2}^{4} \left({x}^{2} - x + 8\right) \mathrm{dx} = {\left[{x}^{3} / 3 - {x}^{2} / 2 + 8 x\right]}_{2}^{4}$

$= \left({4}^{3} / 3 - {4}^{2} / 2 + 8 \left(4\right)\right) - \left({2}^{3} / 3 - {2}^{2} / 2 + 8 \left(2\right)\right)$

$= \left(\frac{64}{3} - 8 + 32\right) - \left(\frac{8}{3} - 2 + 16\right)$

=(136/3)-(50/3)=color(blue)(86/3