1.
Find the area enclosed between and , on the interval from to .
Area =????
2. Find the area of the region enclosed between and from to . Hint: Notice that this region consists of two parts.
If $\displaystyle f(x) \geq g(x)$ on the intervall I, then the area is $\displaystyle \int_I (f(x)-g(x))dx$
The procedure is the same for the second problem, only that you you might need to split the integral into smaller intervals and switch the functions around (try to graph the functions to get a better overview).
Notice how $\displaystyle f(x)>g(x)$ on $\displaystyle [-3,7]$. So the area is evaluated as:
$\displaystyle \int_{-3}^{7}\frac{x^{2}}{4}+10-x\,\,dx$
Simply evaluate the integral from there.
For the next problem, find where the two graphs intersect and knowing how sine and cosine curves are on the given interval (which is $\displaystyle [0,\frac{7\pi}{10}]$), you'll know that to the left of the intersection $\displaystyle 3cos(x)>2sin(x)$ and to the right of it $\displaystyle 2sin(x)>3cos(x)$
$\displaystyle 3cos(x)=2sin(x)$
$\displaystyle tan(x)=\frac{3}{2}$
$\displaystyle x=tan^{-1}(\frac{3}{2})$
So the two intergrals needed are:
$\displaystyle \int_{0}^{tan^{-1}(\frac{3}{2})}3cos(x)-2sin(x)\,\,dx + \int_{tan^{-1}(\frac{3}{2})}^{\frac{7\pi}{10}}2sin(x)-3cos(x)\,\,dx$