Area Between Two Curves, calculus lesson

A definite integral measures signed area, so pieces below the axis count as negative. Often you want plain area instead, such as the amount of paint needed to cover a region on a plan, or the gap between two cost curves. The idea is to slice the region into thin strips, measure each strip as top minus bottom, and add. The care goes into finding the right limits and the right top and bottom.

Key terms

Geometric area versus signed integral

Definition Let $f$ be continuous on $[a, b]$ . Its integral over $[a, b]$ is the signed area, where regions below the $x$ -axis count negatively. The geometric area of the region between the graph of $f$ and the $x$ -axis, for $a \le x \le b$ , is

A = \int_a^b \lvert f(x) \rvert \, dx.

The two agree exactly when $f \ge 0$ on $[a, b]$ .

In plain words The integral keeps a running balance with plus and minus signs. Area never goes negative, so you take absolute values first.

Example On $[0, 2\pi]$ ,

\int_0^{2\pi} \sin x \, dx = 0, \qquad \int_0^{2\pi} \lvert \sin x \rvert \, dx = \int_0^\pi \sin x \, dx - \int_\pi^{2\pi} \sin x \, dx = 2 + 2 = 4.

The arches above and below the axis cancel in the integral, but together they cover an area of $4$ .

Vertical slices

Definition Let $f$ and $g$ be continuous on $[a, b]$ with $f(x) \ge g(x)$ for all $x \in [a, b]$ . The area of the region $\{(x, y) \mid a \le x \le b, \ g(x) \le y \le f(x)\}$ is

A = \int_a^b \left(f(x) - g(x)\right) dx.

It is the limit, as the widths of the subintervals shrink to $0$ , of the Riemann sums

\sum_{i=1}^n \left(f(x_i^*) - g(x_i^*)\right) \Delta x_i,

where each term is the area of a thin vertical rectangle of height top minus bottom.

In plain words Cut the region into thin vertical strips. Each has height top curve minus bottom curve and width $dx$ . Integrate from the leftmost to the rightmost strip.

Example Between $y = x$ (top) and $y = x^2$ (bottom) on $[0, 1]$ , since $x \ge x^2$ there,

A = \int_0^1 (x - x^2) \, dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.

Intersections

Definition The curves $y = f(x)$ and $y = g(x)$ intersect at the $x$ values that solve $f(x) = g(x)$ . When a region is bounded only by the two curves, its left and right limits $a$ and $b$ are solutions of this equation. Between two consecutive solutions the continuous difference $f - g$ is never zero, so by the intermediate value theorem one function stays above the other there.

In plain words Set the two formulas equal and solve. The solutions are where the region starts and ends. Then test one point in between to see which curve is on top.

Example For $y = x + 2$ and $y = x^2$ , solve $x^2 = x + 2$ , so $(x - 2)(x + 1) = 0$ and $x = -1$ or $x = 2$ . At $x = 0$ the line gives $2$ and the parabola gives $0$ , so the line is on top. The area is

A = \int_{-1}^2 (x + 2 - x^2) \, dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^2 = \frac{10}{3} + \frac{7}{6} = \frac{9}{2}.

Horizontal slices

Definition Let $u$ and $v$ be continuous on $[c, d]$ with $u(y) \ge v(y)$ for all $y \in [c, d]$ . The area of the region $\{(x, y) \mid c \le y \le d, \ v(y) \le x \le u(y)\}$ is

A = \int_c^d \left(u(y) - v(y)\right) dy.

In plain words Cut the region into thin horizontal strips instead. Each has length right curve minus left curve. This helps when the region’s top or bottom changes formula partway across but its left and right sides do not.

Example Take the region bounded by $y = \sqrt{x}$ , $y = x - 2$ and the $x$ -axis. Vertically the bottom changes at $x = 2$ , so you would need two integrals. Horizontally, the left side is $x = y^2$ and the right side is $x = y + 2$ , which meet at $y = 2$ , so

A = \int_0^2 (y + 2 - y^2) \, dy = 2 + 4 - \frac{8}{3} = \frac{10}{3}.

The vertical way agrees, since

\int_0^4 \sqrt{x} \, dx - \int_2^4 (x - 2) \, dx = \frac{16}{3} - 2 = \frac{10}{3}.

Splitting at sign changes

Definition If $f$ and $g$ are continuous on $[a, b]$ , the area of the region between their graphs for $a \le x \le b$ is

A = \int_a^b \lvert f(x) - g(x) \rvert \, dx.

To evaluate it when $f - g$ has only finitely many zeros in $[a, b]$ , list those zeros together with $a$ and $b$ as $a = c_0 < c_1 < \dots < c_k = b$ . On each $[c_{i-1}, c_i]$ the difference $f - g$ keeps one sign, by the intermediate value theorem, so integrate whichever of $f - g$ or $g - f$ is nonnegative there and add the results.

In plain words When the curves cross, top and bottom swap. Split at every crossing so that each piece is integrated as top minus bottom.

Example Between $y = x^3$ and $y = x$ on $[-1, 1]$ , the difference $x^3 - x = x(x - 1)(x + 1)$ changes sign at $0$ . On $[-1, 0]$ the cubic is on top and on $[0, 1]$ the line is on top, so

A = \int_{-1}^0 (x^3 - x) \, dx + \int_0^1 (x - x^3) \, dx = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.

Without splitting you would get

\int_{-1}^1 (x - x^3) \, dx = 0,

which is not an area.

Area between a curve and an axis

Definition For $f$ continuous on $[a, b]$ , the area between the graph of $f$ and the $x$ -axis over $[a, b]$ is

A = \int_a^b \lvert f(x) \rvert \, dx,

the special case $g = 0$ of the formula above. For $h$ continuous on $[c, d]$ , the area between the curve $x = h(y)$ and the $y$ -axis over $[c, d]$ is

A = \int_c^d \lvert h(y) \rvert \, dy.

In plain words An axis is just a curve with formula $y = 0$ or $x = 0$ . Find where the graph crosses it and split there.

Example For $f(x) = x^2 - 4$ on $[0, 3]$ the graph crosses the axis at $x = 2$ . It lies below the axis before that and above it after, so

A = \int_0^2 (4 - x^2) \, dx + \int_2^3 (x^2 - 4) \, dx = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}.

The signed integral is quite different.

\int_0^3 (x^2 - 4) \, dx = -3

Check yourself

Find the area between $y = \sin x$ and $y = \cos x$ for $0 \le x \le \frac{\pi}{2}$ .

Show the answer

The curves cross at $x = \frac{\pi}{4}$ . Cosine is on top before that and sine after. Each piece gives $\sqrt{2} - 1$ , so the area is $2\sqrt{2} - 2 \approx 0.828$ .

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

Show the answer

They meet where $y^2 = y + 2$ , at $y = -1$ and $y = 2$ . The line is to the right, so

A = \int_{-1}^2 (y + 2 - y^2) \, dy = \frac{9}{2}.

What is the area between $y = x^3$ and the $x$ -axis on $[-2, 1]$ ?

Show the answer

Split at $0$ , where the cube changes sign.

\int_{-2}^0 (-x^3) \, dx + \int_0^1 x^3 \, dx = 4 + \frac{1}{4} = \frac{17}{4}

So the area is $17/4$ . The signed integral would be $-\frac{15}{4}$ .

Area Between Two Curves

Theory, part 125

Practice, part 126

Key terms

Geometric area versus signed integral

Vertical slices

Intersections

Horizontal slices

Splitting at sign changes

Area between a curve and an axis

Check yourself

How this lesson connects

Builds on