How do you find the area between two curves using integrals?

The area between two curves is found by integrating the difference between the upper curve and lower curve (or right curve minus left curve with respect to y) between their points of intersection: Area = integral from x1 to x2 of [f(x) - h(x)] dx.

Area Between Curves – Formulas, Examples & Solutions

Find the area between curves using definite integrals. Tutorials on the applications of integrals to calculate areas between curves, with examples and detailed solutions, are presented. This tutorial is a continuation of the tutorial on area under a curve. It starts from obvious examples to more challenging ones, which should be mastered for a thorough understanding of integral applications.

Formulas of Areas Between Curves

We can find the area between two curves by subtracting the area corresponding to the lower curve from the area of the upper curve as follows:

1) If \( f \) and \( h \) are functions of \( x \) such that \( f(x) \ge h(x) \) for all \( x \) in the interval \( [x_1, x_2] \), the area shown below (in blue) is given by:

\[ \text{Area} = \int_{x_1}^{x_2} \left(f(x) - h(x)\right) dx \quad (1) \]

finite area between two curves defined as functions of x — **Figure 1.** Finite area between two curves defined as functions of \( x \)

2) If \( Z \) and \( X \) are functions of \( y \) such that \( Z(y) \ge X(y) \) for all \( y \) in the interval \( [y_1, y_2] \), the area shown below (in blue) is given by:

\[ \text{Area} = \int_{y_1}^{y_2} \left(Z(y) - X(y)\right) dy \quad (2) \]

finite area between two curves defined as functions of y — **Figure 2.** Finite area between two curves defined as functions of \( y \)

Examples with Detailed Solutions

Example 1

Find the area of the region enclosed between the curves defined by the equations \( y = x^2 - 2x + 2 \) and \( y = -x^2 + 6 \).

Solution to Example 1

We first graph the two equations and examine the region enclosed between the curves.

area between curves, example 1 — **Figure 3.** Area between curves example 1.

The region whose area is in question is bounded above by the curve \( y = -x^2 + 6 \) and below by the curve \( y = x^2 - 2x + 2 \). The left and right endpoints of the region are the points of intersection of the curves, found by solving the system of equations:

\[ \begin{cases} y = x^2 - 2x + 2 \\ y = -x^2 + 6 \end{cases} \]

Substitute \( y \) to obtain one equation in one variable:

\[ x^2 - 2x + 2 = -x^2 + 6 \]

Rewrite in standard form:

\[ 2x^2 - 2x - 4 = 0 \implies 2(x + 1)(x - 2) = 0 \]

Which gives solutions:

\[ x = -1 \quad \text{and} \quad x = 2 \]

Between the points of intersection, \( -x^2 + 6 \ge x^2 - 2x + 2 \). Let \( f(x) = -x^2 + 6 \) and \( h(x) = x^2 - 2x + 2 \), and apply formula (1):

\[ \begin{aligned} A &= \int_{-1}^{2} \big(f(x) - h(x)\big)\, dx \\ &= \int_{-1}^{2} \big((-x^2 + 6) - (x^2 - 2x + 2)\big)\, dx \\ &= \int_{-1}^{2} \big(-2x^2 + 2x + 4\big)\, dx \\ &= \Big[-\tfrac{2}{3}x^3 + x^2 + 4x\Big]_{-1}^{2} = 9 \end{aligned} \]

Conclusion: The area of the region enclosed between the curves is equal to \( 9 \).

Example 2

Find the area of the region enclosed between the curves defined by the equations \( y = \sqrt{x + 2} \), \( y = x \), and \( y = 0 \).

Solution to Example 2

We examine the region using two methods.

area between curves, example 2 — **Figure 4.** Area between curves example 2.

Method 1: Integration with respect to \( x \)

The region from \( x = -2 \) to \( x = 0 \) is under the curve \( y = \sqrt{x + 2} \), so its area \( A_1 \) is:

\[ \begin{aligned} A_1 &= \int_{-2}^{0} \sqrt{x + 2}\, dx = \left[ \frac{2}{3} (x + 2)^{3/2} \right]_{-2}^{0} = \frac{2}{3} (2^{3/2}) \end{aligned} \]

The region from \( x = 0 \) to the intersection point of \( y = \sqrt{x + 2} \) and \( y = x \):

\[ \sqrt{x + 2} = x \implies x + 2 = x^2 \implies x^2 - x - 2 = 0 \implies x = 2 \]

The area \( A_2 \) from \( x = 0 \) to \( x = 2 \) is:

\[ \begin{aligned} A_2 &= \int_{0}^{2} \big(\sqrt{x + 2} - x\big)\, dx = \left[ \frac{2}{3}(x + 2)^{3/2} - \frac{x^2}{2} \right]_{0}^{2} = \frac{2}{3}(4)^{3/2} - 2 - \frac{2}{3}(2)^{3/2} \end{aligned} \]

Total area:

\[ A_1 + A_2 = \dfrac{10}{3} \; \text{unit}^2 \]

Method 2: Integration with respect to \( y \)

Express \( x \) in terms of \( y \): right curve is \( x = y \), left curve is \( x = y^2 - 2 \). Limits on \( y \) go from \( y = 0 \) to \( y = 2 \):

\[ \begin{aligned} \text{Area} &= \int_{0}^{2} \big( y - (y^2 - 2) \big)\, dy = \int_{0}^{2} \left( -y^2 + y + 2 \right)\, dy \\ &= \left[ -\frac{y^3}{3} + \frac{y^2}{2} + 2y \right]_{0}^{2} = \frac{10}{3} \; \text{unit}^2 \end{aligned} \]

Note that the second method is much faster.

Example 3

Find the area of the region enclosed by the curves \( y = \sin(x) \), \( y = \cos(x) \), \( x = 0 \), and \( x = 2\pi \).

Solution to Example 3

area between curves sin(x), cos(x), example 3 — **Figure 5.** Area between curves example 3.

Find points of intersection where \( \sin(x) = \cos(x) \implies \tan(x) = 1 \). Between \( 0 \) and \( 2\pi \), solutions are \( x = \pi/4 \) and \( x = 5\pi/4 \).

We divide into 3 regions:

Region 1 (\( 0 \) to \( \pi/4 \)): \( \cos x \ge \sin x \) \[ A_1 = \int_{0}^{\pi/4} (\cos x - \sin x)\, dx = \sqrt{2} - 1 \]
Region 2 (\( \pi/4 \) to \( 5\pi/4 \)): \( \sin x \ge \cos x \) \[ A_2 = \int_{\pi/4}^{5\pi/4} (\sin x - \cos x)\, dx = 2\sqrt{2} \]
Region 3 (\( 5\pi/4 \) to \( 2\pi \)): \( \cos x \ge \sin x \) \[ A_3 = \int_{5\pi/4}^{2\pi} (\cos x - \sin x)\, dx = \sqrt{2} + 1 \]

Total area \( A = A_1 + A_2 + A_3 = 4\sqrt{2} \; \text{unit}^2 \).

Example 4

Below are graphed \( y = 3x - x^2 \) and \( y = 0.5x \). Find the ratio of the area of region A to the area of region B.

Solution to Example 4

Find the intersection of \( y = 3x - x^2 \) and \( y = 0.5x \):

\[ 3x - x^2 = 0.5x \implies 2.5x - x^2 = 0 \implies x(2.5 - x) = 0 \implies x = 0, \; x = 2.5 \]

Area of region A:

\[ A = \int_{0}^{2.5} (3x - x^2 - 0.5x)\, dx = \int_{0}^{2.5} (2.5x - x^2)\, dx = \Big[\tfrac{2.5x^2}{2} - \tfrac{x^3}{3}\Big]_{0}^{2.5} = \dfrac{125}{48} \; \text{units}^2 \]

Total area under curve \( y = 3x - x^2 \) from \( x = 0 \) to \( x = 3 \):

\[ A_t = \int_{0}^{3} (3x - x^2)\, dx = \Big[\tfrac{3x^2}{2} - \tfrac{x^3}{3}\Big]_{0}^{3} = \dfrac{9}{2} \; \text{units}^2 \]

Area of region B is \( B = A_t - A \). The ratio \( r = \dfrac{A}{B} = \dfrac{A}{A_t - A} = \dfrac{125/48}{9/2 - 125/48} = \dfrac{125}{91} \).

Example 5

Find the area of the overlapping region of the circles with equations: \( x^2 + y^2 = 4 \) and \( x^2 + (y - 2)^2 = 1 \).

Solution to Example 5

area between enclosed by overlapping circles, example 5 — **Figure 7.** Area between overlapping circles, example 5.

Expand the second circle equation: \( x^2 + y^2 - 4y + 4 = 1 \). Subtracting from \( x^2 + y^2 = 4 \) yields \( y = 7/4 \). Substituting into \( x^2 + y^2 = 4 \) gives \( x = \pm\sqrt{15}/4 \).

Upper curve is \( y = \sqrt{4 - x^2} \) and lower curve is \( y = 2 - \sqrt{1 - x^2} \). By symmetry, half the area \( A' \) from \( x = 0 \) to \( x = \sqrt{15}/4 \) is:

\[ A' = \int_{0}^{\sqrt{15}/4} \left(\sqrt{4 - x^2} - 2 + \sqrt{1 - x^2}\right) dx \approx 0.70153 \]

Total overlapping area \( A = 2A' \approx 1.4 \; \text{unit}^2 \).

Exercises

Find the area of the region enclosed by \( y = (x - 1)^2 + 3 \) and \( y = 7 \).
Find the area of the region bounded by \( x = 0 \) on the left, \( x = 2 \) on the right, \( y = x^3 \) above, and \( y = -1 \) below.
Find the area enclosed by the curves \( y = \dfrac{1}{x^2} \), \( y = x \), and \( y = 3 \).

Solutions to the Above Exercises

\( \dfrac{32}{3} \)
\( 6 \)
\( 2(3 - \sqrt{3}) \)