Recall how we were able to find the area under a non-negative function over an interval by slicing the area into a bunch of vertical strips; approximating each of these with various rectangles; and finding the limit of the sum of the rectangular areas as their widths all approached zero. This limit, of course, was the limit of a Riemann sum, and thus expressible as a definite integral.

We can use the same idea to find the area between *two* curves, as the below example suggests

Find the area bounded by the curves $y=2x^2+10$ and $y=4x+16$

First, we will want to graph both curves. The first is an upwards-opening parabola with $y$-intercept of $(0,10)$. The second is a line of positive slope whose $y$-intercept is higher, at $(0,16)$. Drawing these will give us a good idea of the general shape of the region bound between them -- but to get an even representation, let us also find out where these curves intersect. We do this in the normal way, setting the right-hand sides equal and solving for $x$:

$$\begin{array}{rcl} 2x^2+10 &=& 4x+16\\ 2x^2-4x-6 &=& 0\\ x^2-2x-3 &=& 0\\ (x+1)(x-3) &=& 0\\ x &=& -1 \textrm{ or } 3 \end{array}$$In addition to graphing both curves, notice we have sliced the area bounded between these two functions into vertical strips -- highlighting one of these strips with its approximating rectangle, colored magenta. Think about the area the magenta rectangle contributes to the Riemann sum that approximates the area between the curves.

This rectangle's width (a horizontal distance) is some small difference of $x$-coordinates -- a $\Delta x$.

It's height (a vertical distance) is the difference of a $y$-coordinate on the upper (red) function and the corresponding $y$-coordinate on the lower (blue) function. This difference is thus given by $(4x+16) - (2x^2+10)$ for some $x$ in the interval $[-1,3]$.

Importantly, remember that vertical distances (like all distances) must always be positive. Consequently, we *always* subtract the bottom $y$-coordinate from the top $y$-coordinate so that the result is positive.

Multiplying the height and width together for each such rectangle, the sum of the areas of the approximating rectangles for the strips shown can be represented by a Riemann sum of the form

$$\sum_{i=1}^n \left[(4(x_i^*)+16) - (2(x_i^*)^2+10)\right] \Delta x$$In the limit, of course, this becomes the definite integral shown below.

$$\int_{-1}^3 [(4x+16) - (2x^2+10)]\,dx$$From this point, finding the area is routine -- we simply find an antiderivative and appeal to the Fundamental Theorem of Calculus to evaluate the definite integral:

$$\begin{array}{rcl} \textrm{Area } &=& \displaystyle{\int_{-1}^3 [(4x+16) - (2x^2+10)]\,dx}\\ &=& \displaystyle{(-2x^2 + 4x + 6)\,dx}\\ &=& \displaystyle{\left( -\frac{2}{3}x^3 + 2x^2 + 6x \right) \bigg\rvert_{-1}^3}\\ &=& \displaystyle{\frac{64}{3}} \end{array}$$