# What is the net area between f(x) = sinx - cosx and the x-axis over x in [0, 3pi ]?

Feb 26, 2018

${\int}_{0}^{3 \pi} \sin \left(x\right) - \cos \left(x\right) \mathrm{dx} = 2$

#### Explanation:

We know that the lower and upper bounds are $0 , 3 \pi$ and that the function is $f \left(x\right) = \sin \left(x\right) - \cos \left(x\right)$, so we can rewrite the question as the definite integral of $\sin \left(x\right) - \cos \left(x\right)$ from $0$ to $3 \pi$.
${\int}_{0}^{3 \pi} \sin \left(x\right) - \cos \left(x\right) \mathrm{dx}$

We know that the anti-derivative of $\sin \left(x\right)$ is $- \cos \left(x\right)$ and that the anti-derivative of $\cos \left(x\right)$ is equal to $\sin \left(x\right)$. So we can find the anti-derivative of $\sin \left(x\right) - \cos \left(x\right)$ by substituting their anti-derivatives to get that $F \left(x\right) = - \cos \left(x\right) - \sin \left(x\right)$.

We can then evaluate the anti-derivative at its bounds $0 , 3 \pi$.

So the area $A$ is equal to $F \left(3 \pi\right) - F \left(0\right)$

$F \left(3 \pi\right) = - \cos \left(3 \pi\right) - \sin \left(3 \pi\right) = 1$
$F \left(0\right) = - \cos \left(0\right) - \sin \left(0\right) = - 1$

So the area $A$ is equal to $1 - \left(- 1\right) = 2$

So ${\int}_{0}^{3 \pi} \sin \left(x\right) - \cos \left(x\right) \mathrm{dx} = 2$.