Area between curves

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We compute the area of a region between two curves using the definite integral.

We have seen how integration can be used to find signed area between a curve and the $x$ -axis. With very little change we can find areas bounded between curves. Generally we should interpret ‘‘area’’ in the usual sense, as a necessarily positive quantity. Suppose $f(x)>g(x)$ for all $x$ on the interval $[a,b]$ and we wish to know the geometric area bounded between $f$ and $g$ on this interval:

From the graph above we see that the area we want is the area under $f$ minus the area under $g$ , which is to say $\begin{align*} \mathrm{Area} &= \int_a^b f(x)\d x-\int_a^b g(x)\d x \\ &= \int_a^b f(x)-g(x) \d x. \end{align*}$

It will also be useful to adopt the following intuitive way of thinking about this problem: In a somewhat nonrigorous way, we can think of an integral as ‘‘summing up” an infinite number of infinitesimal quantities. In this case we are summing rectangles of width $\d x$ and height $f(x)-g(x)$ :

In this case, the integral

could be interpreted as:

The blackbluesum of the area of rectangles whose blueblackheights are the difference between the top curve and bottom curve, and whose greenwidths are infinitesimal.

This way of thinking about integrals will be very useful to us in later, more complex, applications.

Integrating with respect to x

We’ll start with basic examples and gradually build to more difficult ones.

Let $f(x) = -x^2+4x+3$ and $g(x)=-x^3+7x^2-10x+5$ . Compute the area between them on the interval $1 \le x \le 2$ .

Write with me: $\int _1^2 f(x)-g(x)\d x = \answer [given]{\frac {49}{12}}.$

$\int _1^2 -x^2+4x+3-(-x^3+7x^2-10x+5) \d x$ $\begin{align*} &=\int_1^2 x^3-8x^2+14x-2 \d x \\ &=\eval{\frac{x^4}{4}-\frac{8x^3}{3}+7x^2-2x}_1^2 \\ &= \frac{16}{4} - \frac{64}{3}+28-4-(\frac{1}{4}- \frac{8}{3}+7-2) \\ &=23-\frac{56}{3}-\frac{1}{4}\\ &=\frac{49}{12} \end{align*}$

In our first example, one curve was higher than the other over the entire interval. This does not always happen.

Find the area between $f(x)= -x^2+4x$ and $g(x)=x^2-6x+5$ over the interval $0 \le x \le 1$ .

Since the two curves cross, we need to compute two areas and add them. First we find the intersection point of the curves. Letting $a$ be the $x$ -coordinate of the point of intersection, we have $a = \answer [given]{\frac {5-\sqrt {15}}{2}}$

$\begin{align*} -a^2+4a &= a^2-6a+5 \\ 0 &= 2a^2-10a+5 \\ a &= \frac{10 \pm \sqrt{100-40}}{4}\\ a &= \frac{5 \pm \sqrt{15}}{2}. \end{align*}$

Of the two solutions, only $\frac {5-\sqrt {15}}{2}$ is within the region of interest.

Then the total area is $\textrm {Area} = \int _0^a g(x) - f(x) \d x + \int _a^1 f(x) - g(x) \d x$ since $g$ is above $f$ on the interval $[0,a]$ , and $f$ is above $g$ on the interval $[a,1]$ . Computing this area we find $\textrm {Area} = \answer [given]{5\sqrt {15} - \frac {52}{3}}.$

$\int _0^a g(x) - f(x) \d x + \int _a^1 f(x) - g(x) \d x$ $\begin{align*} = \int_0^a & \answer{x^2-6x+5}-(-x^2+4x)\d x \\ &+\int_a^1 \answer{-x^2+4x}-(x^2-6x+5)\d x \end{align*}$

Simplifying we find

$\begin{align*} &=\int_0^a 2x^2-10x+5\d x+\int_a^1 -2x^2+10x-5\d x \\ &= \eval{\frac{2x^3}{3}-5x^2+5x}_0^a + \eval{ \frac{-2x^3}{3}+5x^2-5x}_a^1 \\ &= 5\sqrt{15}-\frac{52}{3}. \end{align*}$

In both of our examples above, we gave you the limits of integration by bounding the $x$ -values between $0$ and $1$ . However, in some problems you will have to do more work to determine these bounds.

Find the area bounded between $f(x)= -x^2+4x$ and $g(x)=x^2-6x+5$ .

Here we are not given a specific interval, so it must be the case that there is a ‘‘natural’’ region involved. Since the curves are both parabolas, the only reasonable interpretation is the region between the two intersection points, which we found in the previous example to be $a=\answer [given]{(5-\sqrt {15})/2}$ and $b=\answer [given]{(5+\sqrt {15})/2}$ . Since $f \ge g$ on $[a,b]$ , the geometric area is $\int _a^b f(x) - g(x) \d x = \answer [given]{5\sqrt {15}}.$

$\begin{align*} \int_a^b -x^2+4x-(x^2-6x+5)\d x &=\int_a^b -2x^2+10x-5\d x \\ &=\eval{\answer{\frac{-2x^3}{3}+5x^2-5x}}_a^b \\ &=\answer{5\sqrt{15}}. \end{align*}$

Integrating with respect to y

Consider the region bounded by the function $f(x) = x-3$ , $g(x) = \sqrt {x-1}$ and the horizontal axis:

While we could find the area of this region by breaking the computation up into the two regions $[1,3]$ and $[3,5]$ (check that $5$ is the intersection point!), there is another approach. We can think of splitting the region up into horizontal rectangles.

We can rewrite $y = \sqrt {x-1}$ as $x = y^2+1$ , and we can rewrite $y = x-3$ as $x = y+3$ , so we have the following picture:

In this case, the integral

could be interpreted as:

The blackbluesum of the area of rectangles whose blueblackwidths are the difference between the right curve and left curve, and whose greenheights are infinitesimal.

Let’s see an example.

Compute the area of the region bounded by the function $f(x) = x-3$ , $g(x) = \sqrt {x-1}$ and the horizontal axis:

This area will be easier to compute if we look at $f^{-1}$ and $g^{-1}$ . We have that $\begin{align*} f^{-1}(y) &= \answer[given]{y+3}\\ g^{-1}(y) &= \answer[given]{y^2 +1} \end{align*}$

We also must find where $f^{-1}$ and $g^{-1}$ intersect. Write with me

$\begin{align*} y+3 &= y^2 +1,\\ \answer[given]{y^2-y-2} &= 0\\ y &= -1 \text{ or }\answer[given]{2}. \end{align*}$

Note that $-1$ is not relevant for this problem. Thus the area is given by $\int _{0}^{2}(y+3) - (y^2+1)\d y = \answer [given]{\frac {10}{3}}.$

$\begin{align*} \int_0^2 (y+3) - (y^2+1) \d y &= \int_0^2 -y^2+y+2 \d y\\ &=\eval{\answer{\frac{-y^3}{3} + \frac{y^2}{2}+2y}}_0^2\\ &=\answer{\frac{10}{3}}. \end{align*}$