# Question #e31c1

Nov 24, 2016

#### Explanation:

Here is a graph of the 3 boundaries:

Decide whether to integrate $\mathrm{dy}$ or $\mathrm{dx}$:

You want to find the area between the y axis and green curve until it intersects with the blue curve. At first, you might want to do it with respect to y but you will be integrating inverse sine and inverse cosine function. I say no. I think that it is easier to find the area of the same region as the difference between the functions with respect to x.

Find the limits of integration:

We know that we want to start integrating from $x = 0$ but we need to find where we stop by finding the x coordinate of the point of intersection. Set the right side of the two equations equal to each other:

$6 \cos \left(x\right) = 7 \sin \left(2 x\right)$

Use $\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$

$6 \cos \left(x\right) = 14 \sin \left(x\right) \cos \left(x\right)$

$6 = 14 \sin \left(x\right)$

$\sin \left(x\right) = \frac{3}{7}$

$x = {\sin}^{-} 1 \left(\frac{3}{7}\right)$

$x \approx 0.4429$

$A r e a = {\int}_{0}^{0.4429} \left(6 \cos \left(x\right) - 7 \sin \left(2 x\right)\right) \mathrm{dx}$
$A r e a = 6 \sin \left(x\right) + \frac{7}{2} \cos \left(2 x\right) {|}_{0}^{0.4429}$
$A r e a \approx 1.28571$