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Area of a Region between Two Curves

Let \(f \left(\right. x \left.\right)\) and \(g \left(\right. x \left.\right)\) be continuous functions over an interval \(\left[\right. a , b \left]\right.\) such that \(f \left(\right. x \left.\right) \geq g \left(\right. x \left.\right)\) on \(\left[\right. a , b \left]\right. .\) 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 6.2 The area between the graphs of two functions, \(f \left(\right. x \left.\right)\) and \(g \left(\right. x \left.\right) ,\) on the interval \(\left[\right. a , b \left]\right. .\)

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 \(i = 0 , 1 , 2 ,\ldots , n ,\) let \(P = \left{\right. x_{i} \left.\right}\) be a regular partition of \(\left[\right. a , b \left]\right. .\) Then, for \(i = 1 , 2 ,\ldots , n ,\) choose a point \(x_{i}^{\star} \in \left[\right. x_{i - 1} , x_{i} \left]\right. ,\) and on each interval \(\left[\right. x_{i - 1} , x_{i} \left]\right.\) construct a rectangle that extends vertically from \(g \left(\right. x_{i}^{\star} \left.\right)\) to \(f \left(\right. x_{i}^{\star} \left.\right) .\) Figure 6.3(a) shows the rectangles when \(x_{i}^{\star}\) is selected to be the left endpoint of the interval and \(n = 10 .\) Figure 6.3(b) shows a representative rectangle in detail.

Media

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

The height of each individual rectangle is \(f \left(\right. x_{i}^{\star} \left.\right) - g \left(\right. x_{i}^{\star} \left.\right)\) and the width of each rectangle is \(\Delta x .\) Adding the areas of all the rectangles, we see that the area between the curves is approximated by

\[A \approx \sum_{i = 1}^{n} \left[\right. f \left(\right. x_{i}^{\star} \left.\right) - g \left(\right. x_{i}^{\star} \left.\right) \left]\right. \Delta x .\]

This is a Riemann sum, so we take the limit as \(n \rightarrow \infty\) and we get

\[A = \underset{n \rightarrow \infty}{\text{lim}} \sum_{i = 1}^{n} \left[\right. f \left(\right. x_{i}^{\star} \left.\right) - g \left(\right. x_{i}^{\star} \left.\right) \left]\right. \Delta x = \int_{a}^{b} \left[\right. f \left(\right. x \left.\right) - g \left(\right. x \left.\right) \left]\right. d x .\]

These findings are summarized in the following theorem.

Theorem 6.1

Finding the Area between Two Curves

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

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

We apply this theorem in the following example.

Example 6.1

Finding the Area of a Region between Two Curves 1

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

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

We have

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

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

Checkpoint 6.1

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

In Example 6.1, 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 6.2

Finding the Area of a Region between Two Curves 2

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

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 6.5 This graph shows the region below the graph of \(f \left(\right. x \left.\right)\) and above the graph of \(g \left(\right. x \left.\right) .\)

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

\[\begin{aligned} f \left(\right. x \left.\right) & = & g \left(\right. x \left.\right) \\ \\ 9 - \left(\left(\right. \frac{x}{2} \left.\right)\right)^{2} & = & 6 - x \\ 9 - \frac{x^{2}}{4} & = & 6 - x \\ 36 - x^{2} & = & 24 - 4 x \\ x^{2} - 4 x - 12 & = & 0 \\ \left(\right. x - 6 \left.\right) \left(\right. x + 2 \left.\right) & = & 0. \end{aligned}\]

The graphs of the functions intersect when \(x = 6\) or \(x = −2 ,\) so we want to integrate from \(−2\) to \(6 .\) Since \(f \left(\right. x \left.\right) \geq g \left(\right. x \left.\right)\) for \(−2 \leq x \leq 6 ,\) we obtain

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

The area of the region is \(64 / 3\) units2.

Checkpoint 6.2

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

This lesson is part of:

Applications of Integration

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