The area between two curves Hi there's one problem on my study guide that my teacher didn't go over and I don't know how to approach/solve it. Here's the problem:
Find the area between the curves on the given interval. Draw a graph of the functions and the region.
$$y=x^4 , y=x-1, -2 \le x \le 0.$$
 A: Hint: determine if there are any intersection points of these two curves on the specified intervals. If there are none, great! Determine which one of the two is greater on the interval (WLOG call the greater of the two $y_{1}$, and the smaller $y_{2}$; graphing should help you determine this!), and then find $$\int_{-2}^{0}y_{1}-y_{2} dx$$ If there are intersection points, break up the integral into several subintervals of $[0, 2]$ and again determine which curve is greater on each portion of the integral. Then integrate each piece exactly as you did for the case that there are no intersection points.
Edit: at intersection points, the function which is greater might change, and so you would have to break up the integral accordingly to get the right subtraction of functions. In this case, $x^{4}$ and $x-1$ don't intersect each other on the interval $[-2, 0]$, and so you don't have to worry about it; $x^{4} > x - 1$ for all $-2 \leq x \leq 0$. 
A: Let $E$ the region given by two curves $f$, $g$, which are functions with domain $[a,b]$ in the $x$ axis and $f>g$. Then:
$$A(E)=\iint_E 1\, dA=\int_a^b\int_{g(x)}^{f(x)} 1\, dydx=\int_a^b (f(x)-g(x))\,dx=\int_a^b f-\int_a^b g $$
Separe in cases when $f>g$ and $f\leq g$.
