Finding the Area Bounded by a Curve

Area needs absolute value, not a signed integral

A definite integral gives signed area: positive above the x-axis, negative below it. To find the actual (unsigned) area enclosed by a curve that dips below the axis, integrate over each piece where the sign doesn't change, and add the absolute values together.

Worked example: a full period of cosine

Find the area bounded by $y=\cos x$ between $x=0$ and $x=2\pi$ .

Solution: $\cos x \ge 0$ on $[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},2\pi]$ and $\cos x \le 0$ on $[\frac{\pi}{2},\frac{3\pi}{2}]$ . By symmetry, the total area is 4 times the area of one quarter-period:

y = cos x from 0 to 2π. The two outer pieces lie above the axis and the middle piece below it. Each quarter-period encloses an area of 1.

4\int_0^{\pi/2}\cos x\,dx = 4\left[\sin x\right]_0^{\pi/2} = 4(1-0) = 4

Integrating straight across, $\displaystyle\int_0^{2\pi}\cos x\,dx = 0$ , because the area below the axis cancels the area above it. That is the signed-area trap the section above warns about.

Worked example: deriving the area of a circle

Find the area of the circle $x^2+y^2=a^2$ by integration.

Solution: The circle's upper half is $y=\sqrt{a^2-x^2}$ . By symmetry, the full circle's area is 4 times the area in the first quadrant:

4\int_0^a \sqrt{a^2-x^2}\,dx = 4\left[\dfrac{x}{2}\sqrt{a^2-x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}\right]_0^a

At $x=a$ the square-root term vanishes and $\sin^{-1}(1) = \frac{\pi}{2}$ , while both terms vanish at $x=0$ :

= 4\left(\dfrac{a^2}{2}\cdot\dfrac{\pi}{2}\right) = \pi a^2

Area between two curves

When a region is bounded by two curves instead of one curve and the axis, integrate the upper curve minus the lower curve between the points where they meet:

A = \int_a^b \left[f(x) - g(x)\right]dx, \qquad f(x) \ge g(x) \text{ on } [a, b]

Sketch both curves so you can see which one is on top.
Find the intersection points by solving $f(x) = g(x)$ . These are the limits $a$ and $b$ .
Integrate (upper minus lower) from $a$ to $b$ .
If the curves cross again inside the interval, split it there, since the upper and lower curves swap.

Worked example: a line and a parabola

Find the area enclosed between $y = x$ and $y = x^2$ .

Solution: They meet where $x^2 = x$ , so at $x = 0$ and $x = 1$ . Between them the line is on top (at $x = \dfrac{1}{2}$ , the line is at $\dfrac{1}{2}$ and the parabola at $\dfrac{1}{4}$ ):

A = \int_0^1 \left(x - x^2\right)dx = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}

The region between y = x (upper) and y = x² (lower) from x = 0 to x = 1. Its area is 1/6.

Common mistakes

Integrating straight across a sign change. Split the interval wherever the curve crosses the axis, or the pieces cancel, as $\cos x$ does over $[0, 2\pi]$ .
Subtracting the wrong way round. Lower minus upper gives a negative answer. An area is never negative, so a negative result means the order is reversed.
Guessing the limits. Solve for the intersection points. Don't read them off a rough sketch.