# How do you find the total area between the curve f(x)=sin 2x and g(x)=cos x, pi/6 ≤ x ≤ 5pi/6?

Feb 1, 2018

0.5

#### Explanation:

Area between two curves that don't intersect at a domain is
${\int}_{a}^{b} f \left(x\right) - g \left(x\right)$, where f(x) is the function with the larger y values. Either using a graphing calculator or some double angle identities, you find that the graphs intersect at $\frac{\pi}{6} , \frac{\pi}{2} , \mathmr{and} \frac{5 \pi}{6}$. Because of the intersection at $\frac{\pi}{2}$ you'll have to split the area into two parts to use the formula above into: $\left[\frac{\pi}{6} , \frac{\pi}{2}\right]$ and $\left[\frac{\pi}{2} , \frac{5 \pi}{6}\right]$. But notice that in $\left[\frac{\pi}{6} , \frac{\pi}{2}\right]$, sin(2x)>cos(x), but its the opposite for the other domain, so make sure to put the larger function in the correct place into the formula:

$\left[{\int}_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin \left(2 x\right) - \cos \left(x\right)\right]$+$\left[{\int}_{\frac{\pi}{2}}^{\frac{5 \pi}{6}} \cos \left(x\right) - \sin \left(2 x\right)\right]$

I don't have the time right now to integrate this step by step (maybe someone else can?), although I'll tell you that ${\int}_{^} \sin \left(2 x\right) = - \frac{1}{2} \cos \left(2 x\right)$, and ${\int}_{^} \cos \left(x\right) = \sin \left(x\right)$. I hope you can carry on from there, it's just a lot of algebra (or plugging into calculator) You'll get -0.5, but area is positive, so it's just 0.5. Let me know if you have any questions.