Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 7

Applications of the Definite Integral

Areas Between Curves

In the first remark in Unit 6, we indicated that a definite integral gives the area under a curve only if the integrand function is positive on the interval given by the limits of integration. If the integrand function is not positive, then the definite integral is the areas under the curve on the interval where the function is positive minus the areas under the curve where the function is negative. In this section, we find areas between curves. We start with a single curve only.

Example 7.5. Consider the function ${\displaystyle f\left(x\right)=x^2-4x}$ on the interval ${\displaystyle \left[-2,5\right]\text{.}}$ We want to find the area under this curve. Since the graph of this function is a parabola that is concave upwards, we find the ${\displaystyle x}$-intercepts to see where the function is positive and where it is negative. The ${\displaystyle x}$-intercepts are ${\displaystyle \left(0,0\right)}$ and ${\displaystyle \left(4,0\right)\text{,}}$ it is positive on the intervals ${\displaystyle \left[-2,0\right)}$ and ${\displaystyle \left(4,5\right]\text{,}}$ and it is negative on ${\displaystyle \left(0,4\right)\text{.}}$ So, the area is given by the following definite integrals

${{\displaystyle {\int }_{-2}^0{x}^2-4xdx={\displaystyle \frac{x^3}{3}}-2x^2}|}_{-2}^0=-\left({\displaystyle \frac{-8}{3}}-8\right)={\displaystyle \frac{32}{3}}$.

${{\displaystyle {\int }_4^5{x}^2-4xdx={\displaystyle \frac{x^3}{3}}-2x^2}|}_4^5=\left({\displaystyle \frac{125}{3}}-2(25)\right)-\left({\displaystyle \frac{64}{3}}-2(16)\right)={\displaystyle \frac{7}{3}}$.

${{\displaystyle {\int }_0^4{x}^2-4xdx=-\left({\displaystyle \frac{x^3}{3}}-2x^2\right)}|}_0^4=-\left({\displaystyle \frac{64}{3}}-2(16)\right)={\displaystyle \frac{32}{3}}$.

The area is

${\displaystyle {\displaystyle \frac{32}{3}}+{\displaystyle \frac{7}{3}}+{\displaystyle \frac{32}{3}}={\displaystyle \frac{71}{3}}.}$

It is always useful to use symmetry when solving a problem. If a function ${\displaystyle f\left(x\right)}$ is even, then on an interval ${\displaystyle \left[-a,a\right]\text{,}}$ the areas under the curve on the left and right of the ${\displaystyle y}$-axis are equal. Therefore,

${\displaystyle {\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx.}$

The integral on the right may be easier than the integral on the left.

If a function ${\displaystyle f\left(x\right)}$ is odd, then on an interval ${\displaystyle \left[-a,a\right]\text{,}}$ the areas under the curve on the left and right of the ${\displaystyle y}$-axis are equal, but one is under the ${\displaystyle x}$-axis and the other is above it. Therefore,

${\displaystyle {\int }_{-a}^{a}f\left(x\right)dx=0.}$

and the area under the curve on the interval ${\displaystyle \left[-a,a\right]}$ is again equal to

${\displaystyle 2{\int }_0^{a}f\left(x\right)dx.}$

Example 7.6. The function

${\displaystyle g\left(x\right)=x{\left(x^2+1\right)}^2}$

is odd, and the area under this curve on the interval ${\displaystyle \left[-1,1\right]}$ is

${2{\displaystyle {\int }_0^{1}x{\left(x^2+1\right)}^{2}dx={\displaystyle {\int }_0^{1}2x}{\left(x^2+1\right)}^{2}dx={\displaystyle \frac{{\left(x^2+1\right)}^3}{3}}}|}_0^1={\displaystyle \frac{7}{3}}$.

Example 7.7. Analyse the following definite integral

${\displaystyle {\int }_{-3}^{2}x\cos \left(x^2\right)dx.}$

The function ${\displaystyle f\left(x\right)=x\cos \left(x^2\right)}$ is odd, and the definite integral on the interval ${\displaystyle \left[-2,2\right]}$ is zero, so the integral is equal to

$\begin{array}{rl}{\displaystyle {\int }_{-3}^{-2}x\cos (x^2)dx}& ={\displaystyle \frac{1}{2}}{\displaystyle {\int }_{-3}^{-2}2x\cos (x^2)dx}\\ & ={{\displaystyle \frac{\sin (x^2)}{2}}|}_{-3}^{-2}\\ & ={\displaystyle \frac{\sin (4)-\sin (9)}{2}}\text{.}\end{array}$

We can also consider functions in terms of ${\displaystyle y\text{,}}$ and apply a definite integral to find the area between the ${\displaystyle y}$-axis and the curve, as we did with functions in terms of ${\displaystyle x\text{.}}$

Example 7.8. The function ${\displaystyle g\left(y\right)=y^2-3y}$ is a parabola open to the right, and its ${\displaystyle y}$-intercepts are ${\displaystyle \left(0,0\right)}$ and ${\displaystyle \left(0,3\right)\text{.}}$ Therefore, the parabola is to the left of the ${\displaystyle y}$-axis on the interval ${\displaystyle \left[0,3\right]\text{,}}$ that is ${\displaystyle g\left(y\right)<0}$ for ${\displaystyle y}$ in the interval ${\displaystyle \left(0,3\right)\text{,}}$ and it is on the right of the ${\displaystyle y}$-axis on ${\displaystyle \left[3,5\right]\text{,}}$ that is ${\displaystyle g\left(y\right)>0}$ for ${\displaystyle y}$ in the interval ${\displaystyle \left(3,5\right)\text{,}}$ thus the area under this curve on the interval ${\displaystyle \left[0,5\right]}$ is

$-{\displaystyle {\int }_0^3{y}^2-3ydy}={-\left({\displaystyle \frac{y^3}{3}}-{\displaystyle \frac{3y^2}{2}}\right)|}_0^3=-9+{\displaystyle \frac{27}{2}}$.

${\displaystyle {\int }_3^5{y}^2-3ydy}={\left({\displaystyle \frac{y^3}{3}}-{\displaystyle \frac{3y^2}{2}}\right)|}_3^5={\displaystyle \frac{125}{3}}-{\displaystyle \frac{75}{2}}-9+{\displaystyle \frac{27}{2}}$.

The area is

${\displaystyle -9+{\displaystyle \frac{27}{2}}+{\displaystyle \frac{125}{3}}-{\displaystyle \frac{75}{2}}-9+{\displaystyle \frac{27}{2}}=-18-{\displaystyle \frac{21}{2}}+{\displaystyle \frac{125}{3}}={\displaystyle \frac{79}{6}}.}$

If two functions ${\displaystyle f}$ and ${\displaystyle g}$ are positive and ${\displaystyle g\left(x\right)\le f\left(x\right)}$ for all ${\displaystyle x}$ in the interval ${\displaystyle \left[a,b\right]\text{,}}$ then the graph between them is equal to the area under ${\displaystyle f}$ minus the area under ${\displaystyle g\text{,}}$ as is shown in Figure 7.1, below.

Figure 7.1: Area between functions ${\displaystyle f}$ and ${\displaystyle g\text{,}}$ where ${\displaystyle f}$ and ${\displaystyle g}$ are positive and ${\displaystyle g\left(x\right)\le f\left(x\right)}$ for all ${\displaystyle x}$ in the interval ${\displaystyle \left[a,b\right]}$

Then by Property 2, the area is

${\displaystyle {\int }_a^{b}f\left(x\right)-{\int }_a^{b}g\left(x\right)dx={\int }_a^{b}f\left(x\right)-g\left(x\right)dx.}$

Example 7.9. The area between ${\displaystyle \sin x}$ and ${\displaystyle \cos x}$ on the interval ${\displaystyle \left[0,\pi ∕4\right]\text{,}}$ is shown in Figure 7.2, below.

Figure 7.2: Area between ${\displaystyle \sin x}$ and ${\displaystyle \cos x}$ on the interval ${\displaystyle \left[0,{\displaystyle \frac{\pi }{4}}\right]}$

Since ${\displaystyle \cos x\ge \sin x}$ on the interval ${\displaystyle \left[0,\pi ∕4\right]\text{,}}$ the area is given by the definite integral

${\displaystyle {\int }_0^{\pi ∕4}\cos x-\sin xdx=\sin x+\cos x{|}_0^{\pi ∕4}=\sqrt{2}-1.}$

If two functions ${\displaystyle f}$ and ${\displaystyle g}$ are not positive and ${\displaystyle g\left(x\right)\le f\left(x\right)}$ for all ${\displaystyle x}$ in the interval ${\displaystyle \left[a,b\right]\text{,}}$ as in Figure 7.3, below, then to obtain the area between their graphs, we consider the area between the graphs of the functions ${\displaystyle f\left(x\right)+C}$ and ${\displaystyle g\left(x\right)+C\text{,}}$ where ${\displaystyle C}$ is a positive constant such that the functions ${\displaystyle f\left(x\right)+C}$ and ${\displaystyle g\left(x\right)+C}$ are positive (see Figure 7.3).

Figure 7.3: Technique for determining the area between functions ${\displaystyle f}$ and ${\displaystyle g\text{,}}$ where ${\displaystyle f}$ and ${\displaystyle g}$ are not positive and ${\displaystyle g\left(x\right)\le f\left(x\right)}$ for all ${\displaystyle x}$ in the interval ${\displaystyle \left[a,b\right]}$

Since the functions ${\displaystyle f\left(x\right)+C}$ and ${\displaystyle g\left(x\right)+C}$ are a shift of the functions ${\displaystyle f\left(x\right)}$ and ${\displaystyle g\left(x\right)\text{,}}$ the areas between them are equal. But now ${\displaystyle f\left(x\right)+C}$ and ${\displaystyle g\left(x\right)+C}$ are positive, and the area between them is

${\displaystyle {\int }_a^{b}f\left(x\right)+C-\left(g\left(x\right)+C\right)dx={\int }_a^{b}f\left(x\right)-g\left(x\right)dx.}$

This example shows that the area between any two function ${\displaystyle f\left(x\right)}$ and ${\displaystyle g\left(x\right)\text{,}}$ such that ${\displaystyle g\left(x\right)\le f\left(x\right)}$ on the interval ${\displaystyle \left[a,b\right]\text{,}}$ is equal to

${\displaystyle {\int }_a^{b}f\left(x\right)-g\left(x\right)dx.}$

Example 7.10. The area between the functions ${\displaystyle f\left(x\right)=x-x^2}$ and ${\displaystyle g\left(x\right)=3x-1}$ is given by a definite integral. To find it, we must find their points of intersection; that is, ${\displaystyle x-x^2=3x-1\text{.}}$

Solving for ${\displaystyle x\text{,}}$ ${\displaystyle x=-1-\sqrt{2}\text{,}}$ and ${\displaystyle x=-1+\sqrt{2}\text{.}}$

From their graphs, we see that ${\displaystyle g\left(x\right)\le f\left(x\right)}$ on the interval ${\displaystyle \left[-1-\sqrt{2},-1+\sqrt{2}\right]}$ (see Figure 7.4, below).

Figure 7.4: Area between the functions ${\displaystyle f\left(x\right)=x-x^2}$ and ${\displaystyle g\left(x\right)=3x-1}$ (not to scale)

Hence the area is given by

$\begin{array}{rl}{\displaystyle {\int }_{-1-\sqrt{2}}^{-1+\sqrt{2}}x-x^2-(3x-1)dx}& ={\displaystyle {\int }_{-1-\sqrt{2}}^{-1+\sqrt{2}}x-x^2-3x+1dx}\\ & ={\displaystyle {\int }_{-1-\sqrt{2}}^{-1+\sqrt{2}}-2x-x^2+1dx}\\ & ={-x^2-{\displaystyle \frac{x^3}{3}}+x|}_{-1-\sqrt{2}}^{-1+\sqrt{2}}\\ & ={x\left(-x-{\displaystyle \frac{x^2}{3}}+1\right)|}_{-1-\sqrt{2}}^{-1+\sqrt{2}}\\ & =\left(-1+\sqrt{2}\right)\left[2-\sqrt{2}-{\displaystyle \frac{{\left(-1+\sqrt{2}\right)}^2}{3}}\right]\\ & +\left(1+\sqrt{2}\right)\left[2+\sqrt{2}-{\displaystyle \frac{{\left(1+\sqrt{2}\right)}^2}{3}}\right]\\ & ={\displaystyle \frac{8\sqrt{2}}{3}}\text{.}\end{array}$

Example 7.11. To find the area between the parabola ${\displaystyle f\left(x\right)=x^2-6x+9}$ and the line ${\displaystyle g\left(x\right)=4x}$ on the interval ${\displaystyle \left[0,3\right]\text{,}}$ we must make a sketch of these functions (see Figure 7.5, below).

Figure 7.5: Area between the the parabola ${\displaystyle f\left(x\right)=x^2-6x+9}$ and the line ${\displaystyle g\left(x\right)=4x}$ on the interval ${\displaystyle \left[0,3\right]}$ (not to scale)

We find the point of intersection of these two functions on the interval ${\displaystyle \left[0,3\right]}$; thus, ${\displaystyle x^2-6x+9=4x\text{.}}$

Solving for ${\displaystyle x\text{,}}$ we find that${\displaystyle x=1}$ and ${\displaystyle x=9.}$

We see that we have two different areas to consider, one on the interval ${\displaystyle \left[0,1\right]}$ where ${\displaystyle f\left(x\right)\ge g\left(x\right)\text{,}}$ and another on ${\displaystyle \left[1,3\right]}$ where ${\displaystyle f\left(x\right)\le g\left(x\right)\text{.}}$ The areas are given by the following definite integrals:

${\displaystyle \begin{array}{ll}{\displaystyle {\int }_0^1{x}^2}\hfill & -6x+9-4xdx+{\displaystyle {\int }_1^{3}4x-\left(x^2-6x+9\right)dx}\hfill \\ \hfill & ={\displaystyle {\int }_0^1{x}^2-10x+9dx+{\displaystyle {\int }_1^3-x^2+10x-9dx\text{.}}}\hfill \end{array}}$

Evaluating each of these integrals, we obtain

${\displaystyle {\int }_0^1{x}^2-10x+9dx={\displaystyle \frac{x^3}{3}}-5x^2+9x{|}_0^1={\displaystyle \frac{13}{3}},}$

and

${\displaystyle {\int }_1^3-x^2+10x-9dx=-{\displaystyle \frac{x^3}{3}}+5x^2-9x{|}_1^3={\displaystyle \frac{40}{3}}.}$

The area is ${\displaystyle {\displaystyle \frac{13}{3}}+{\displaystyle \frac{40}{3}}={\displaystyle \frac{53}{3}}.}$

The area between these functions is given by the integral

$\begin{array}{rl}{\displaystyle {\int }_1^{9}4x-(x^2-6x+9)dx}& ={\displaystyle {\int }_1^{9}10x-x^2-9dx}\\ & ={-{\displaystyle \frac{x^3}{3}}+5x^2-9x|}_1^9\\ & ={\displaystyle \frac{256}{3}}\end{array}$

For functions in terms of the variable ${\displaystyle y\text{,}}$ we also know that if ${\displaystyle g\left(y\right)\le f\left(y\right)}$ on the interval ${\displaystyle \left[c,d\right]\text{,}}$ then the graph of the function ${\displaystyle g}$ is on the left of the graph of ${\displaystyle f\text{,}}$ and the area between these functions is given by the definite integral

${\displaystyle {\int }_c^{d}f\left(y\right)-g\left(y\right)dy.}$

Example 7.12. See Exercise 18 on page 325 of the textbook.

The sketch of the curves ${\displaystyle 4x+y^2=12}$ and ${\displaystyle y=x}$ is shown in Figure 7.6, below.

Figure 7.6: Curves ${\displaystyle 4x+y^2=12}$ and ${\displaystyle y=x}$

The points of intersection of these two curves are the solutions of the equation ${\displaystyle 4y+y^2=12\text{,}}$ since ${\displaystyle y=x\text{.}}$ Solving for ${\displaystyle y\text{,}}$ we find that ${\displaystyle y=-6}$ and ${\displaystyle y=2\text{.}}$ [Check this.]

Since the parabola is on the right of the line, the area is given by the integral

Observe that solving for ${\displaystyle x}$ in the equation ${\displaystyle 4x+y^2=12}$ gives the function

${\displaystyle g\left(y\right)=3-{\displaystyle \frac{y^2}{4}}\text{.}}$

Note that you will find the theorem below helpful as you complete the assigned exercises.

Theorem 7.2. Integrals of Symmetric Functions

Suppose ${\displaystyle f}$ is continuous on ${\displaystyle \left[-a,a\right]\text{,}}$ then

1. if ${\displaystyle f}$ is even ${\displaystyle \left[f\left(-x\right)=f\left(x\right)\right]\text{,}}$ then ${\displaystyle {\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx\text{.}}$
2. if ${\displaystyle f}$ is odd ${\displaystyle \left[f\left(-x\right)=-f\left(x\right)\right]\text{,}}$ then ${\displaystyle {\int }_{-a}^{a}f\left(x\right)dx=0.}$