Area Between Curves

A definite integral of a positive function gives the area under its graph. The area between two graphs comes from the same idea, with the height of each thin rectangle measured from one curve to the other.

Area with respect to x

Suppose

f(x) \ge g(x)

[a, b]

. Cut the region into vertical strips of width

\Delta x

. A strip at

x

is close to a rectangle of height

f(x) - g(x)

, so its area is about

\big(f(x) - g(x)\big)\Delta x

. Adding the strips gives a Riemann sum, and its limit is an integral.

The integrand is top minus bottom, and it stays correct when part of the region lies below the

x

-axis. When the region is enclosed by the two curves, the limits are the

x

-coordinates of the points where they meet.

Interactive: Area Between Two Curves

Below,

f(x) = \sqrt{x}

is on top and

g(x) = x^2

is on the bottom, and they meet at

x = 0

and

x = 1

. Sweeping from

0

1

builds up the area, which ends at

\int_0^1 \big(\sqrt{x} - x^2\big)\,dx = \frac{1}{3}

Example 1

Find the area of the region enclosed by the line

y = 7 - 2x

and the curve

y = \frac{3}{x}

The curves meet where

7 - 2x = \frac{3}{x}

, that is,

2x^2 - 7x + 3 = 0

, so

x = \frac{1}{2}

x = 3

. At

x = 1

, the line gives

5

and the curve gives

3

. The curves cross only at the ends, so the line is on top on all of

\left(\frac{1}{2}, 3\right)

The region between the line and the curve, with one vertical strip of height (7 − 2x) − 3/x.

\begin{aligned} A &= \int_{1/2}^{3} \left(7 - 2x - \frac{3}{x}\right)dx \\[4pt] &= \left(7x - x^2 - 3\ln x\right)\Big|_{1/2}^{3} \\[4pt] &= (12 - 3\ln 3) - \left(\tfrac{13}{4} + 3\ln 2\right) \\[4pt] &= \frac{35}{4} - 3\ln 6 \approx 3.375. \end{aligned}

Area with respect to y

Some regions are easier to cut into horizontal strips. A strip at height

y

has width from the left curve to the right curve, and thickness

\Delta y

. Write each boundary as

x

in terms of

y

The integrand is right minus left, and the limits are values of

y

. Choose the variable that gives one integral with a single formula for each boundary. If the top or bottom boundary changes formula partway across, the integral in

x

has to be split there.

Example 2

Find the area of the region enclosed by the curves

x = \frac{y^2}{4}

and

x = 5 - y^2

, first with respect to

y

, then with respect to

x

The curves meet where

\frac{y^2}{4} = 5 - y^2

, so

y^2 = 4

and

y = \pm 2

, at the points

(1, -2)

and

(1, 2)

. At

y = 0

the curves give

x = 0

and

x = 5

, so

x = 5 - y^2

is on the right.

A horizontal strip runs from x = y²/4 to x = 5 − y². Left of the dashed line x = 1 the region is bounded by one curve, right of it by the other.

With respect to $y$ . One integral covers the region:

\begin{aligned} A &= \int_{-2}^{2} \left(5 - y^2 - \frac{y^2}{4}\right)dy \\[4pt] &= \left(5y - \frac{5y^3}{12}\right)\Big|_{-2}^{2} \\[4pt] &= \frac{20}{3} - \left(-\frac{20}{3}\right) = \frac{40}{3}. \end{aligned}

With respect to $x$ . Solve each curve for

y

. The left curve gives

y = \pm 2\sqrt{x}

and the right curve gives

y = \pm\sqrt{5 - x}

. For

0 \le x \le 1

the region runs from

-2\sqrt{x}

up to

2\sqrt{x}

. For

1 \le x \le 5

it runs from

-\sqrt{5 - x}

up to

\sqrt{5 - x}

. So

\begin{aligned} A &= \int_{0}^{1} 4\sqrt{x}\,dx + \int_{1}^{5} 2\sqrt{5 - x}\,dx \\[4pt] &= \frac{8}{3}\,x^{3/2}\Big|_{0}^{1} - \frac{4}{3}(5 - x)^{3/2}\Big|_{1}^{5} \\[4pt] &= \frac{8}{3} + \frac{32}{3} = \frac{40}{3}. \end{aligned}

Both give

\frac{40}{3}

. The integral in

y

took one step, and the one in

x

took two integrals with square roots.

Intersections from a calculator

When the equation for the intersection points can't be solved by algebra, a calculator finds them. Store the values and use them as the limits, so no rounding builds up before the final answer.

Example 3

A calculator is allowed. Find the area of the region enclosed by the graphs of

y = 5 - x^2

and

y = 2e^{x/2}

, to three decimal places.

Let

h(x) = 5 - x^2 - 2e^{x/2}

. Since

h''(x) = -2 - \frac{1}{2}e^{x/2} < 0

h

is concave down and has at most two zeros. A calculator finds them at

x = a \approx -2.071

and

x = b \approx 1.180

. At

x = 0

the parabola gives 5 and the exponential gives 2, so the parabola is on top between them.

The region between the parabola and the exponential, from x ≈ −2.071 to x ≈ 1.180.

\begin{aligned} A &= \int_{a}^{b} \left(5 - x^2 - 2e^{x/2}\right)dx \\ &\approx 6.951. \end{aligned}