Answer
$(\frac{\sqrt{3}}{2} +1) + \frac{7\pi}{24} $
Work Step by Step
In order to find this area, we first need to solve for the line between the points $(0,0)$ and $(\frac{7\pi}{6}, -\frac{1}{2})$. We can do this by using slope-intercept form.
In this case, the slope is: $m = \frac{(-\frac{1}{2} - 0)}{(\frac{7\pi}{6} - 0)} = \frac{(-\frac{1}{2})}{(\frac{7\pi}{6})} = -\frac{3}{7\pi}$
Since the line intersects the y- axis at $(0,0)$, the slope intercept form $(y = mx +b )$ for the line is then:
$ y = -\frac{3}{7\pi}x$
Now, since we are trying to find the highlighted area, we want to integrate the upper function (in this case $\sin(x)$) minus the lower function (in this case $-\frac{3}{7\pi}x$) between the two intersection points.
We do thus the following integral;
$\int_{0}^{\frac{7\pi}{6}} (\sin(x) - (-\frac{3}{7\pi}x)) dx$ = $\int_{0}^{\frac{7\pi}{6}} (\sin(x) + \frac{3}{7\pi}x)) dx$
= $(-\cos(x) + \frac{3}{7\pi}x^{2}*\frac{1}{2})) \Bigr\rvert_{x = 0}^{x =\frac{7\pi}{6}}$
(The anti-derivative of sine is negative cosine, and the anti-derivative of the second term can be found using the power rule for integration)
=$(\frac{\sqrt{3}}{2} +1) + (\frac{3}{7\pi})(\frac{7\pi}{6})^{2}\frac{1}{2} $
(since $\cos(\frac{7\pi}{6}) = - \frac{\sqrt{3}}{2} )$
= $(\frac{\sqrt{3}}{2} +1) + \frac{7\pi}{24} $
