Areas of Regions Involving Curves

Learning Outcomes

Determine the area of a region between two curves by integrating with respect to the independent variable
Find the area of a compound region

Area of a Region between Two Curves

Let $f(x)$ and $g(x)$ be continuous functions over an interval $\left[a,b\right]$ such that $f(x)\ge g(x)$ on $\left[a,b\right].$ We want to find the area between the graphs of the functions, as shown in the following figure.

This figure is a graph in the first quadrant. There are two curves on the graph. The higher curve is labeled “f(x)” and the lower curve is labeled “g(x)”. There are two boundaries on the x-axis labeled a and b. There is shaded area between the two curves bounded by lines at x=a and x=b.

Figure 1. The area between the graphs of two functions, $f(x)$ and $g(x),$ on the interval $\left[a,b\right].$

As we did before, we are going to partition the interval on the $x\text{-axis}$ and approximate the area between the graphs of the functions with rectangles. So, for $i=0,1,2\text{,…},n,$ let $P=\left\{{x}_{i}\right\}$ be a regular partition of $\left[a,b\right].$ Then, for $i=1,2\text{,…},n,$ choose a point ${x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right],$ and on each interval $\left[{x}_{i-1},{x}_{i}\right]$ construct a rectangle that extends vertically from $g({x}_{i}^{*})$ to $f({x}_{i}^{*}).$ Figure 2(a) shows the rectangles when ${x}_{i}^{*}$ is selected to be the left endpoint of the interval and $n=10.$ Figure 2(b) shows a representative rectangle in detail.

Interactive

Use this calculator to learn more about the areas between two curves.

This figure has three graphs. The first graph has two curves, one over the other. In between the curves is a rectangle. The top of the rectangle is on the upper curve labeled “f(x*)” and the bottom of the rectangle is on the lower curve and labeled “g(x*)”. The second graph, labeled “(a)”, has two curves on the graph. The higher curve is labeled “f(x)” and the lower curve is labeled “g(x)”. There are two boundaries on the x-axis labeled a and b. There is shaded area between the two curves bounded by lines at x=a and x=b. The third graph, labeled “(b)” has two curves one over the other. The first curve is labeled “f(x*)” and the lower curve is labeled “g(x*)”. There is a shaded rectangle between the two. The width of the rectangle is labeled as “delta x”.

Figure 2. (a)We can approximate the area between the graphs of two functions, $f(x)$ and $g(x),$ with rectangles. (b) The area of a typical rectangle goes from one curve to the other.

The height of each individual rectangle is $f({x}_{i}^{*})-g({x}_{i}^{*})$ and the width of each rectangle is $\text{Δ}x.$ Adding the areas of all the rectangles, we see that the area between the curves is approximated by

A \approx \sum_{i = 1}^{n} [f (x_{i}^{*}) - g (x_{i}^{*})] Δ x .

$A\approx\displaystyle\sum_{i=1}^{n} \left[f({x}_{i}^{*})-g({x}_{i}^{*})\right]\text{Δ}x.$

This is a Riemann sum, so we take the limit as $n\to \infty$ and we get

A = lim_{n \to \infty} \sum_{i = 1}^{n} [f (x_{i}^{*}) - g (x_{i}^{*})] Δ x = \int_{a}^{b} [f (x) - g (x)] d x .

$A=\underset{n\to \infty }{\text{lim}}\displaystyle\sum_{i=1}^{n} \left[f({x}_{i}^{*})-g({x}_{i}^{*})\right]\text{Δ}x={\displaystyle\int }_{a}^{b}\left[f(x)-g(x)\right]dx.$

These findings are summarized in the following theorem.

Finding the Area between Two Curves

Let $f(x)$ and $g(x)$ be continuous functions such that $f(x)\ge g(x)$ over an interval $\left[a,b\right].$ Let $R$ denote the region bounded above by the graph of $f(x),$ below by the graph of $g(x),$ and on the left and right by the lines $x=a$ and $x=b,$ respectively. Then, the area of $R$ is given by

A = \int_{a}^{b} [f (x) - g (x)] d x

$A={\displaystyle\int }_{a}^{b}\left[f(x)-g(x)\right]dx$

We apply this theorem in the following example.

Example: Finding the Area of a Region between Two Curves 1

If R is the region bounded above by the graph of the function $f(x)=x+4$ and below by the graph of the function $g(x)=3-\frac{x}{2}$ over the interval $\left[1,4\right],$ find the area of region $R.$

Show Solution

The region is depicted in the following figure.

This figure is has two linear graphs in the first quadrant. They are the functions f(x) = x+4 and g(x)= 3-x/2. In between these lines is a shaded region, bounded above by f(x) and below by g(x). The shaded area is between x=1 and x=4.

Figure 3. A region between two curves is shown where one curve is always greater than the other.

We have

\begin{array}{cc} A & = \int_{a}^{b} [f (x) - g (x)] d x \\ = \int_{1}^{4} [(x + 4) - (3 - \frac{x}{2})] d x = \int_{1}^{4} [\frac{3 x}{2} + 1] d x \\ = {[\frac{3 x^{2}}{4} + x] |}_{1}^{4} = (16 - \frac{7}{4}) = \frac{57}{4} . \end{array}

$\begin{array}{cc}\hfill A& ={\displaystyle\int }_{a}^{b}\left[f(x)-g(x)\right]dx\hfill \\ & ={\displaystyle\int }_{1}^{4}\left[(x+4)-(3-\frac{x}{2})\right]dx={\displaystyle\int }_{1}^{4}\left[\frac{3x}{2}+1\right]dx\hfill \\ & ={\left[\frac{3{x}^{2}}{4}+x\right]|}_{1}^{4}=(16-\frac{7}{4})=\frac{57}{4}.\hfill \end{array}$

The area of the region is $\frac{57}{4}{\text{units}}^{2}.$

Try It

If $R$ is the region bounded by the graphs of the functions $f(x)=\frac{x}{2}+5$ and $g(x)=x+\frac{1}{2}$ over the interval $\left[1,5\right],$ find the area of region $R.$

Show Solution

In the last example, we defined the interval of interest as part of the problem statement. Quite often, though, we want to define our interval of interest based on where the graphs of the two functions intersect. This is illustrated in the following example.

Example: Finding the Area of a Region between Two Curves 2

If $R$ is the region bounded above by the graph of the function $f(x)=9-{(\frac{x}{2})}^{2}$ and below by the graph of the function $g(x)=6-x,$ find the area of region $R.$

Show Solution

The region is depicted in the following figure.

This figure is has two graphs in the first quadrant. They are the functions f(x) = 9-(x/2)^2 and g(x)= 6-x. In between these graphs, an upside down parabola and a line, is a shaded region, bounded above by f(x) and below by g(x).

Figure 4. This graph shows the region below the graph of $f(x)$ and above the graph of $g(x).$

We first need to compute where the graphs of the functions intersect. Setting $f(x)=g(x),$ we get

\begin{array}{ccc} f (x) & = & g (x) \\ 9 - {(\frac{x}{2})}^{2} & = & 6 - x \\ 9 - \frac{x^{2}}{4} & = & 6 - x \\ 36 - x^{2} & = & 24 - 4 x \\ x^{2} - 4 x - 12 & = & 0 \\ (x - 6) (x + 2) & = & 0. \end{array}

$\begin{array}{ccc}\hfill f(x)& =\hfill & g(x)\hfill \\ \\ \hfill 9-{(\frac{x}{2})}^{2}& =\hfill & 6-x\hfill \\ \hfill 9-\frac{{x}^{2}}{4}& =\hfill & 6-x\hfill \\ \hfill 36-{x}^{2}& =\hfill & 24-4x\hfill \\ \hfill {x}^{2}-4x-12& =\hfill & 0\hfill \\ \hfill (x-6)(x+2)& =\hfill & 0.\hfill \end{array}$

The graphs of the functions intersect when $x=6$ or $x=-2,$ so we want to integrate from -2 to 6. Since $f(x)\ge g(x)$ for $-2\le x\le 6,$ we obtain

\begin{array}{cc} A & = \int_{a}^{b} [f (x) - g (x)] d x \\ = \int_{- 2}^{6} [9 - {(\frac{x}{2})}^{2} - (6 - x)] d x = \int_{- 2}^{6} [3 - \frac{x^{2}}{4} + x] d x \\ = {[3 x - \frac{x^{3}}{12} + \frac{x^{2}}{2}] |}_{- 2}^{6} = \frac{64}{3} . \end{array}

$\begin{array}{cc}\hfill A& ={\displaystyle\int }_{a}^{b}\left[f(x)-g(x)\right]dx\hfill \\ & ={\displaystyle\int }_{-2}^{6}\left[9-{(\frac{x}{2})}^{2}-(6-x)\right]dx={\displaystyle\int }_{-2}^{6}\left[3-\frac{{x}^{2}}{4}+x\right]dx\hfill \\ & ={\left[3x-\frac{{x}^{3}}{12}+\frac{{x}^{2}}{2}\right]|}_{-2}^{6}=\frac{64}{3}.\hfill \end{array}$

The area of the region is $\frac{64}{3}$ units².

Try It

If R is the region bounded above by the graph of the function $f(x)=x$ and below by the graph of the function $g(x)={x}^{4},$ find the area of region $R.$

Show Solution

$\frac{3}{10}$ unit²

Hint

Use the process from the last example.

Watch the following video to see the worked solution to the above Try It.

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Try It

Areas of Compound Regions

So far, we have required $f(x)\ge g(x)$ over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.

Finding the Area of a Region between Curves That Cross

Let $f(x)$ and $g(x)$ be continuous functions over an interval $\left[a,b\right].$ Let $R$ denote the region between the graphs of $f(x)$ and $g(x),$ and be bounded on the left and right by the lines $x=a$ and $x=b,$ respectively. Then, the area of $R$ is given by

A = \int_{a}^{b} | f (x) - g (x) | d x

$A={\displaystyle\int }_{a}^{b}|f(x)-g(x)|dx$

In practice, applying this theorem requires us to break up the interval $\left[a,b\right]$ and evaluate several integrals, depending on which of the function values is greater over a given part of the interval. We study this process in the following example.

Example: Finding the Area of a Region Bounded by Functions That Cross

If R is the region between the graphs of the functions $f(x)= \sin x$ and $g(x)= \cos x$ over the interval $\left[0,\pi \right],$ find the area of region $R.$

Show Solution

The region is depicted in the following figure.

This figure is has two graphs. They are the functions f(x) = sinx and g(x)= cosx. They are both periodic functions that resemble waves. There are two shaded areas between the graphs. The first shaded area is labeled “R1” and has g(x) above f(x). This region begins at the y-axis and stops where the curves intersect. The second region is labeled “R2” and begins at the intersection with f(x) above g(x). The shaded region stops at x=pi.

Figure 5. The region between two curves can be broken into two sub-regions.

The graphs of the functions intersect at $x=\pi \text{/}4.$ For $x\in \left[0,\pi \text{/}4\right],$ $\cos x\ge \sin x,$ so

| f (x) - g (x) | = | \sin x - \cos x | = \cos x - \sin x

$|f(x)-g(x)|=| \sin x- \cos x|= \cos x- \sin x$

On the other hand, for $x\in \left[\pi \text{/}4,\pi \right],$ $\sin x\ge \cos x,$ so

| f (x) - g (x) | = | \sin x - \cos x | = \sin x - \cos x

$|f(x)-g(x)|=| \sin x- \cos x|= \sin x- \cos x$

Then

\begin{array}{cc} A & = \int_{a}^{b} | f (x) - g (x) | d x \\ = \int_{0}^{π} | \sin x - \cos x | d x = \int_{0}^{π / 4} (\cos x - \sin x) d x + \int_{π / 4}^{π} (\sin x - \cos x) d x \\ = {[\sin x + \cos x] |}_{0}^{π / 4} + {[− \cos x - \sin x] |}_{π / 4}^{π} \\ = (\sqrt{2} - 1) + (1 + \sqrt{2}) = 2 \sqrt{2} . \end{array}

$\begin{array}{cc}\hfill A& ={\displaystyle\int }_{a}^{b}|f(x)-g(x)|dx\hfill \\ & ={\displaystyle\int }_{0}^{\pi }| \sin x- \cos x|dx={\displaystyle\int }_{0}^{\pi \text{/}4}( \cos x- \sin x)dx+{\displaystyle\int }_{\pi \text{/}4}^{\pi }( \sin x- \cos x)dx\hfill \\ & ={\left[ \sin x+ \cos x\right]|}_{0}^{\pi \text{/}4}+{\left[\text{−} \cos x- \sin x\right]|}_{\pi \text{/}4}^{\pi }\hfill \\ & =(\sqrt{2}-1)+(1+\sqrt{2})=2\sqrt{2}.\hfill \end{array}$

The area of the region is $2\sqrt{2}$ units².

Try It

If R is the region between the graphs of the functions $f(x)= \sin x$ and $g(x)= \cos x$ over the interval $\left[\frac{\pi}{2},2\pi \right],$ find the area of region $R.$

Show Solution

$2+2\sqrt{2}$ units²

Hint

The two curves intersect at $x=\frac{(5\pi )}{4}.$

Watch the following video to see the worked solution to the above Try It.

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Example: Finding the Area of a Complex Region

Consider the region depicted in Figure 6. Find the area of $R.$

This figure is has two graphs in the first quadrant. They are the functions f(x) = x^2 and g(x)= 2-x. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The region is labeled R. The shaded area is between x=0 and x=2.

Figure 6. Two integrals are required to calculate the area of this region.

Show Solution

As with the last example, we need to divide the interval into two pieces. The graphs of the functions intersect at $x=1$ (set $f(x)=g(x)$ and solve for $x$ ), so we evaluate two separate integrals: one over the interval $\left[0,1\right]$ and one over the interval $\left[1,2\right].$

Over the interval $\left[0,1\right],$ the region is bounded above by $f(x)={x}^{2}$ and below by the $x$ -axis, so we have

A_{1} = \int_{0}^{1} x^{2} d x = {\frac{x^{3}}{3} |}_{0}^{1} = \frac{1}{3} .

${A}_{1}={\displaystyle\int }_{0}^{1}{x}^{2}dx={\frac{{x}^{3}}{3}|}_{0}^{1}=\frac{1}{3}.$

Over the interval $\left[1,2\right],$ the region is bounded above by $g(x)=2-x$ and below by the $x\text{-axis,}$ so we have

A_{2} = \int_{1}^{2} (2 - x) d x = {[2 x - \frac{x^{2}}{2}] |}_{1}^{2} = \frac{1}{2} .

${A}_{2}={\displaystyle\int }_{1}^{2}(2-x)dx={\left[2x-\frac{{x}^{2}}{2}\right]|}_{1}^{2}=\frac{1}{2}.$

Adding these areas together, we obtain

A = A_{1} + A_{2} = \frac{1}{3} + \frac{1}{2} = \frac{5}{6} .

$A={A}_{1}+{A}_{2}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}.$

The area of the region is $\frac{5}{6}$ units².

Try It

Consider the region depicted in the following figure. Find the area of $R.$

This figure is has two graphs in the first quadrant. They are the functions f(x) = squareroot of x and g(x)= 3/2 – x/2. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The shaded area is between x=0 and x=3.

Figure 7.

Show Solution

$\frac{5}{3}$ units²

Hint

The two curves intersect at $x=1.$

Watch the following video to see the worked solution to the above Try It.

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Module 6: Applications of Integration