Example

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

Solution

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 better 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 $x$:

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

In the limit, of course, this becomes the definite integral shown below.

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:

Example

Find the area bound by the curves $y=x^2$ and $y=-x^2 + 4x$ in two different ways.

Solution

As before, we will want to first graph both functions (which both graph as parabolas down) and find any points of intersection.

To graph the parabolas, we observe the first has a vertex at $(0,0)$ and opens up, while the secondopens down and can be rewritten (by completing the square) as $$\begin{array}{rcl} y &=& -(x^2-4x)\\ &=&-(x^2-4x+4)+4\\ &=&-(x-2)^2+4 \end{array}$$ and thus has a vertex at $(2,4)$.

Finding points of intersection in the normal way, we discover $$\begin{array}{rcl} x^2 &=& -x^2 + 4x\\ 2x^2 - 4x &=& 0\\ 2x(x-2) &=& 0\\ x &=& 0 \textrm{ or } 2 \end{array}$$

Nicely, the intersections then just happen to be at the vertices earlier found. Thus, graphing them together we have

As was done in the first example, if we slice the bounded area vertically into roughly rectangular strips, we make the width (a horizontal distance) of each of the approximating rectangles a small $\Delta x$, which will eventually turn into a $dx$ in a definite integral that gives this area.

Similarly, we find the height (a vertical distance) of each rectangle by subtracting the $y$-coordinate on the lower (red) function from the $y$-coordinate of the upper (blue) function. When expressed in terms of $x$ (so as to match our $dx$) this height will be given by $(-x^2 + 4x) - (x^2)$, provided $x$ is in the interval $[0,2]$ (otherwise we get a negative distance as the height).

Consequently (and assuming a regular partition), this tells us the area can be approximated by the Riemann sum shown below. $$\sum_{i=1}^n \left[(-(x_i^*)^2+4(x_i^*)) - (x_i^*)^2\right] \Delta x$$

This in turn, as we take the limit of the above sum as $n \rightarrow \infty$, becomes the definite integral $$\int_0^2 \left[(-x^2 + 4x) - x^2 \right] dx$$ Computing the value of this integral is straight-forward: $$\left. \left[- \frac{1}{3}x^3 + 2x^2 - \frac{1}{3} x^3 \right] \right|_0^2 = \frac{8}{3}$$

However, what if we slice the area bound by these curves horizontally instead? This is shown below.

Notice the small dimension of each of the approximating rectangles is now a vertical distance/height and thus a difference of $y$-coordinates -- a $\Delta y$.

One might be curious about the addition above of solutions for $x$ in terms of $y$ for both equations in question. The reason for their inclusion is this: for a given $y$-coordinate, if we find the approximating rectangle's width (a horizontal distance, and thus a difference in $x$ coordinates) in terms of $y$, we will be able to setup a Riemann sum using a (regular) partition of size $n$ of an appropriate $y$-interval that approximates the bound area. Then, taking a limit as $n \rightarrow \infty$, we can find the area bound by the curves as a definite integral in terms of $y$.

Remembering to subtract the left (blue) $x$-coordinate from the right (red) $x$ coordinate so that the resulting horizontal width is positive, we can express the aforementioned limit of a Riemann sum as $$\lim_{n \rightarrow \infty} \sum_{i=1}^n \left[\left(\sqrt{y_i^*}\right) - \left(2 - \sqrt{4-y_i^*}\right)\right] \Delta y$$ which then becomes the definite integral given by $$\int_0^4 \left[\left(\sqrt{y}\right) - \left(2 - \sqrt{4-y}\right)\right] dy$$ Again, the computation of this definite integral is straight-forward and unsurprisingly results in $\frac{8}{3}$ once more.