Find the area of the finite region enclosed between the following curves:
y= 0.3x + 5 sin x and y = 2x
Thank you for helping!!
It helps if you have a visual of what is going on. First we find the intersection points to determine the limits of integration:
$\displaystyle 0.3x + 5\sin x = 2x \: \: \iff \: \: \sin x = 0.34 x$
This is impossible to solve algebraically so you're going have to obtain the intersection points via your graphing calculator. Let's call these intersection points x = a, 0, and b.
If you graph it out, you'll see y = 2x is greater than 0.3x + 5sin x over the interval $\displaystyle [a,0]$ and the other way around for the interval $\displaystyle [0,b]$. (Remember, area enclosed by two curves is given by the integral of the upper minus the lower function) So,
$\displaystyle A = \int_{a}^{0} \left[2x - \left(0.3x + 5\sin x\right)\right] dx + \int_{0}^{b} \left[(0.3x + 5\sin x) - 2x\right]dx$