The area between two curves

So far we can determine the area between a curve and the $x$ -axis. Now we generalise this a bit, and determine the area between two curves.

Example 1

Consider the two functions

f(x)=3.5-x^2

and

g(x)=0.5x+1.5

What is the area $A$ of the region that is enclosed by these two functions (shaded area in the figure below)?

The idea is straight forward, we simple determine the area under the upper curve (figure below, left) and subtract from it the area under the lower curve (figure below). Try yourself, then go to the solution for help.

Solution

To find the integration limits $a$ and $b$ , we have to determine the $x$ -coordinates of the intersection points between the graphs $f$ and $g$ . So, find $x$ with

\begin{array}{rll} f(x) & = &g(x) \\ 3.5-x^2 & =& 0.5x+1.5\\ -x^2-0.5x+2 & =& 0 \end{array}

Applying the midnight formula, we obtain $a=-1.686$ and $b=1.186$ . We also need the antiderivative of

h(x)=f(x)-g(x)=3.5-x^2-0.5x-1.5

and this is

H(x)=3.5x-\frac{1}{3}x^3-0.25 x^2-1.5x

The area of the shaded region

\begin{array}{lll} A &=& A_{upper}-A_{lower}\\[1em] &=&\int_{-1.686}^{1.186} (f(x)-g(x))\, dx\\[1em] &=& H(1.2)- H(-1.7))\\[1em] &=&\underline{3.94} \end{array}

But what if the region between the curves is given as in the example below, to the left? We cannot subtract anymore the area beneath the lower graph from the area beneath the upper graph. In principle we could just go ahead and divide the region into smaller subregions in such a way that we could determine their areas using integrals. While this works, it is normally time consuming. Luckily, there is a much simpler way (see the figure below):

Recipe 1

Shift the two graphs (and therefore the region) upwards by the same constant $c$ , where $c$ has to be large enough so that the whole region will be above the $x$ -axis. See the figure below (right). The area stays the same, but now we can work out the area as already discussed above.

The shifted functions are

\begin{array}{lll} f(x) &=& h(x)+c\\ g(x) &=& k(x)+c \end{array}

The shifted area is then:

\begin{array}{lll} A^{shifted} &=& \int_{a}^{b} f(x)- \int_a^b g(x))\, dx\\[1em] &=& \int_{a}^{b} (f(x)-g(x))\, dx\\[1em] &=& \int_{a}^{b} (h(x)+c-k(x)-c)\, dx\\[1em] &=& \int_{a}^{b} (h(x)-k(x))\, dx \end{array}

Note that we have used the theorem about weighted sums to get from the first line to the second line in the equation.

The constant $c$ has no influence on the result at all, as it cancels. So, we actually do not have to shift at all!

Let's summarise all these findings in a theorem.

Theorem 1

Consider two functions, where the graph of $f_{upper}$ is above the graph of $f_{lower}$ from $a$ to $b$ on the $x$ -axis. The (normal) area $A$ between the two graphs is

Equation 1

\begin{array}{ll} A&=&\int_a^b f_{upper}(x)\, dx - \int_a^b f_{lower}(x)\, dx\\[1em] &=&\int_a^b \left(f_{upper}(x)-f_{lower}(x)\right)\, dx\\[1em] \end{array}

Area between upper and lower curve.

Exercise 1

Determine the area enclosed by the functions $f(x)=x^2-0.5$ and $g(x)=1.5-x^2$ .
Determine the area enclosed by the functions $\sin(x)$ and $\cos(x)$ (see figure).
Consider two polynomials of degree $2$ (shown below). Determine the area enclosed by these two functions.
Find the area enclosed by the two polynomials $f$ and $g$ of degree $2$ , where
- $f$ has the $x$ -intercepts $-1$ and $2$ , and the $y$ -intercept $1$
- $g$ has the $x$ -intercepts $0$ and $2$ , and the graph passes through the point $(1|-2)$ .

Solution

It is $f_{upper}(x)=1.5-x^2$ and $f_{lower}(x)=x^2-0.5$ , and they intersect at $a=-1$ and $b=1$ . We have to determine the antiderivative of
$h(x)=f_{upper}(x)-f_{lower}(x)=1.5-x^2-x^2+0.5=2-2x^2$
which is
$H(x)=2x-\frac{2}{3}x^3$
Thus, we have
$\begin{array}{lll} A &=& \int_{-1}^1 (f_{upper}(x)-f_{lower}(x))\, dx \\ &=& \int_{-1}^1 (2-2x^2)\, dx\\ &=& H(1)-H(-1)\\ &=&(2-\frac{2}{3})-(-2+\frac{2}{3})\\ &=&\underline{\frac{8}{3}} \end{array}$
Find the point of intersection between the graphs. First the left one: Find $x$ with
$\begin{array}{lll} \sin(x) &=&\cos(x) \quad \vert :\cos(x)\\ \frac{\sin(x)}{\cos(x)} &=& 1\\ \tan(x) &=& 1\\ x &=& \tan^{-1}(1)=\frac{\pi}{4} \end{array}$
We guess the right one: $x=\frac{5\pi}{4}$ (symmetry argument). Verify that this is correct: Indeed: $\sin(\frac{5\pi}{4})=\cos(\frac{5\pi}{4})$ .

We have to find the antiderivative of
$h(x)=f_{upper}(x)-f_{lower}(x)=\sin(x)-\cos(x)$
which is
$H(x)=-\cos(x)-\sin(x)$
Thus we have
$\begin{array}{lll} A &=&\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin(x)-\cos(x))\, dx \\ &=& H(\frac{5\pi}{4})-H(\frac{\pi}{4})\\ &=&=-\cos(\frac{5\pi}{4})-\sin(\frac{5\pi}{4})+\cos(\frac{\pi}{4})+\sin(\frac{\pi}{4})\\ &=&\underline{2\sqrt{2}} \end{array}$
1. We first need to find the function equations of $f$ and $g$ . As $f$ is a polynomial of degree $2$ and has the $x$ -intercepts $-1$ and $3$ , we write
  $f(x)=a(x-(-1))(x-3)=a(x+1)(x-3)$
  Because of $f(1)=2$ it follows $a(2)(-2)=-4a=2$ and thus $a=-\frac{1}{2}$ . Thus, we have
  $\begin{array}{lll} f(x) &=& -\frac{1}{2}(x+1)(x-3)\\ &=&\underline{-\frac{1}{2}x^2+x+1.5} \end{array}$
  As $g$ is also a polynomial of degree $2$ with the $x$ -intercepts $-2$ and $3$ , we write
  $g(x)=a(x-(-2))(x-3)=a(x+2)(x-3)$
  Because of $g(1)=-1$ it follows $a(3)(-2)=6a=1$ and thus $a=\frac{1}{6}$ . Thus, we have
  $\begin{array}{lll} g(x) &=& \frac{1}{6}(x+2)(x-3)\\ &=&\underline{\frac{1}{6}x^2-\frac{1}{6}x-1} \end{array}$
2. We need to find the point of intersections between $f$ and $g$ . So find $x$ with
  $\begin{array}{lll} f(x) &=& g(x)\\ -\frac{1}{2}x^2+x+1.5 &=& \frac{1}{6}x^2-\frac{1}{6}x-1\\ -\frac{4}{6}x^2+\frac{7}{6}x+2.5 &=& 0 \quad \vert \cdot(-6)\\ 4x^2-7x-15 &=& 0 \end{array}$
  Using the midnight formula, we get $x_1=-1.25$ and $x_2=3$ .
3. With
  $f_{upper}(x)=f(x)=-\frac{1}{2}x^2+x+1.5$
  and
  $f_{lower}(x)=g(x)=\frac{1}{6}x^2-\frac{1}{6}x-1$
  we get
  $\begin{array}{lll} h(x)&=&f_{upper}(x)-f_{lower}(x)\\ &= & -\frac{1}{2}x^2+x+1.5-\frac{1}{6}x^2+\frac{1}{6}x+1\\ &=& -\frac{2}{3}x^2+\frac{7}{6}x+2.5 \end{array}$
  The antiderivative is
  $H(x)=-\frac{2}{9}x^3+\frac{7}{12}x^2+2.5x$
  The area is therefore
  $\begin{array}{lll} A &=& \int_{-1.25}^{3} (f_{upper}(x)-f_{lower}(x))\, dx\\ &=& \int_{-1.25}^{3} (-\frac{2}{3}x^2+\frac{7}{6}x+2.5)\, dx\\ &=& H(3)-H(-1.25)\\ &=&6.75-(-1.779)\\ &=&\underline{8.5295}\\ \end{array}$
Draw the figure. For a change, we do not use the formula, but solve it directly by moving the region up by e.g. $3$ , so that the region will be above the $x$ -axis. But first, find the equations of $f$ and $g$ :
- Find $f$ : $f(x)=a(x+1)(x-2)$ and because of $f(0)=1$ it follows $a=-0.5$ .
- Find $g$ : $g(x)=b(x-0)(x-2)$ and because of $g(1)=-2$ , we get $b=2$
Moving the region up by $3$ , we thus get
- $f_{upper}(x)=-0.5(x+1)(x-2)+3$
- $f_{lower}(x)=2x(x-2)+3$
We need to find the point of intersection between $f_{upper}$ and $g_{upper}$ (or between $f$ and $g$ ). So find $x$ with $f_{upper}(x)=f_{lower}(x)$ :
$-0.5(x+1)(x-2)+3 = 2x(x-2)+3$
Expanding and using the midnight formula, we get two intersection points with $x$ -coordinates $x_1=-0.2$ and $x_2=2$ . Thus, the area $A$ of the region is
$\begin{array}{lll} A &=& \int_{-0.2}^2 f_{upper}(x) dx - \int_{-0.2}^2 f_{lower}(x) dx\\ &=& 8.45 - 4.01\\ &=& \underline{4.44} \end{array}$