Area under a curve

The definite integral $\int_a^b f(x)\,dx$ gives the signed area between $y = f(x)$ and the $x$-axis on $[a, b]$. To find a true geometric area you must control the sign , splitting the interval where $f$ changes sign and integrating $|f|$ where needed.

Setup with $f \ge 0$

If $f(x) \ge 0$ on $[a, b]$ and $f$ is continuous, the region $\{(x, y) : a \le x \le b,\ 0 \le y \le f(x)\}$ has area $A = \int_a^b f(x)\,dx.$ This is the geometric interpretation of the integral. Each vertical "slice" at $x$ has width $dx$ and height $f(x)$, contributing area $f(x)\,dx$. Summing (integrating) gives the whole.

When $f$ goes below the axis

If $f(x) \le 0$ on $[a, b]$, the integral gives a negative number whose absolute value is the area. The region in question lies below the $x$-axis.

If $f$ takes both signs on $[a, b]$, find the zeros of $f$ inside the interval and split: $A = \int_a^b |f(x)|\,dx = \int_a^{c_1} |f| + \int_{c_1}^{c_2} |f| + \cdots.$ On each sub-interval $f$ has constant sign, so $|f|$ is either $f$ or $-f$ , easy to integrate.

Area to the left/right of the $y$-axis

If your curve is given as $x = g(y)$ on $[c, d]$ with $g \ge 0$, the area between the curve and the $y$-axis is $A = \int_c^d g(y)\,dy.$ Same idea, slices horizontal instead of vertical.

Worked examples

Example 1. Find the area under $y = x^2$ from $x = 0$ to $x = 3$.

$f(x) = x^2 \ge 0$, so $A = \int_0^3 x^2\,dx = \left[\dfrac{x^3}{3}\right]_0^3 = 9$.

Example 2. Find the area between $y = \sin x$ and the $x$-axis on $[0, 2\pi]$.

$\sin x$ changes sign at $\pi$. Split: $A = \int_0^\pi \sin x\,dx + \int_\pi^{2\pi} (-\sin x)\,dx = 2 + 2 = 4.$

Example 3. Find the area between $y = x^3$ and the $x$-axis from $x = -2$ to $x = 2$.

$x^3$ is odd; the signed integral is zero. But the area requires $|x^3|$. By symmetry, $A = 2\int_0^2 x^3\,dx = 2 \cdot \dfrac{16}{4} = 8$.

Example 4. Find the area between $y = \ln x$, the $x$-axis, and the line $x = e$.

The curve $y = \ln x$ is zero at $x = 1$ and positive on $[1, e]$. Below $x = 1$ it dips negative. The standard interpretation: area from $x = 1$ to $x = e$: $A = \int_1^e \ln x\,dx = [x\ln x - x]_1^e = (e - e) - (0 - 1) = 1.$

Example 5. Area enclosed by $y^2 = x$ and the line $x = 4$.

Solve simultaneously: $y^2 = 4$ gives $y = \pm 2$. Region is symmetric about the $x$-axis. Using horizontal slices: at height $y$, the region extends from $x = y^2$ to $x = 4$, with width $4 - y^2$. $A = \int_{-2}^2 (4 - y^2)\,dy = 2\int_0^2(4 - y^2)\,dy = 2\left[4y - \dfrac{y^3}{3}\right]_0^2 = 2 \cdot \dfrac{16}{3} = \dfrac{32}{3}.$

Example 6. Find the area bounded by $y = 4 - x^2$ and the $x$-axis.

Zeros at $x = \pm 2$, with $y \ge 0$ between them. By symmetry: $A = 2\int_0^2 (4 - x^2)\,dx = 2 \cdot (8 - 8/3) = \dfrac{32}{3}$.

Try it yourself

1. Area under $y = 2x + 1$ from $x = 0$ to $x = 4$.
2. Area under $y = e^x$ from $x = 0$ to $x = 1$.
3. Area between $y = \cos x$ and the $x$-axis on $[0, \pi]$.
4. Area enclosed by $y = x^2 - 4$ and the $x$-axis.
5. Area under $y = \sqrt{x}$ from $x = 0$ to $x = 9$.
6. Area between $y = x^2$ and the $x$-axis from $x = -1$ to $x = 2$.
7. Area enclosed by $y = 9 - x^2$ and the $x$-axis.
8. Area enclosed by $y = \sin 2x$ and the $x$-axis on $[0, \pi]$.
9. Area enclosed by $y^2 = 4x$ between $x = 0$ and $x = 1$ (in the upper half).
10. Area enclosed by the curve $x = y^2 - 1$ and the $y$-axis.
11. Area between $y = |x|$, the $x$-axis, and $x = \pm 3$.
12. Area between $y = 1/x$, the $x$-axis, $x = 1$, $x = e$.
13. Area enclosed by $y = e^{-x}$, the $x$-axis, and the lines $x = 0$, $x = 1$.
14. Area between $y = \tan x$ and the $x$-axis on $[0, \pi/4]$.

Pitfalls and tricks

• Sketch first. Without a picture you cannot identify the region or its limits.
• Mind signs. Geometric area uses $|f|$; the bare integral gives signed area.
• Find zeros of $f$ in the interval and split there if $f$ changes sign.
• Use symmetry. If the region is symmetric about the $x$-axis or $y$-axis, compute half and double.
• Choose slice direction. Vertical ($dx$) works when the region is "long horizontally"; horizontal ($dy$) when "tall vertically."