Reading the Integral Sign

A stretched S

The integral sign $\int$ is a letter S drawn tall and thin. The S stands for sum. Wherever it appears, it is an instruction to add something up, and the rest of the expression says what is being added and over what range.

One sliver: f(x) dx

What is added is slivers. Take the region under $y = x^{2}$ and cut out the thin strip from x = 2 to x = 2.3. Treated as a rectangle as tall as the curve at its left edge, it has height 4 and width 0.3, so area 1.2. The curve rises across the strip to $2.3^{2} = 5.29$ at the right edge, so the strip's true area lies between 1.2 and 5.29 × 0.3 = 1.587.

Make the strip thinner and the two bounds close in. From x = 2 to x = 2.01 they are 4 × 0.01 = 0.04 and 4.0401 × 0.01 = 0.040401, which differ by only 1%. The thinner the strip, the more exactly its area is its height times its width.

In $\int f\left(x\right) \,dx$, the part f(x) dx is one such sliver: a height f(x) times a width dx, where dx is a width shrunk toward nothing. The sign says: add up all the slivers.

The curve $y = x^{2}$, with the sliver from x = 2 to x = 2.3 shaded. Its area lies between 4 × 0.3 = 1.2 and 5.29 × 0.3 = 1.587.

One sliver under $y = x^{2}$ at x = 1.2, magnified. Its area is about $f\left(1.2\right) \times \Delta x = 1.44 \times 0.3 = 0.432$. Drag the width down toward 0.01: the sliver becomes a line of height 1.44, and $\Delta x$ becomes the dx of the integral.

Where the sweep starts and stops

Numbers written at the bottom and the top of the sign are the limits of integration. In $\int_{a}^{b} f\left(x\right) \,dx$ the slivers are added from x = a, the lower limit at the bottom, to x = b, the upper limit at the top. The lower limit is read first.

The simplest case is a constant. $\int_{1}^{5} 3 \,dx$ adds slivers of height 3 from x = 1 to x = 5. Together they fill a rectangle 3 tall and 5 − 1 = 4 wide, so $\int_{1}^{5} 3 \,dx = 12$.

The line y = 3, with the region under it from x = 1 to x = 5 shaded: a rectangle 3 tall and 4 wide, area 12.

Saying it aloud

Read $\int_{a}^{b} f\left(x\right) \,dx$ as "the integral from a to b of f of x, dx": the sign, then the limits, bottom first, then the function, then the dx. The function inside, f(x), is called the integrand.

So $\int_{0}^{3} x^{2} \,dx$ is "the integral from 0 to 3 of x squared, dx", with integrand $x^{2}$. And $\int_{0}^{2} x \,dx$ is "the integral from 0 to 2 of x, dx": its slivers fill the triangle under y = x from 0 to 2, with base 2 and height 2, so its value is ½ × 2 × 2 = 2.

What dx says

The dx names the variable the slivers are laid along. That matters when the integrand has more than one letter. In $\int x y \,dx$, x varies and y is held fixed, like a number, so the answer is $\frac{x^{2}y}{2} + C$. In $\int x y \,dy$, y varies and x is held fixed, and the answer is $\frac{xy^{2}}{2} + C$. Same integrand, different answers.

In a definite integral the letter itself makes no difference to the value. $\int_{0}^{2} x \,dx$ and $\int_{0}^{2} t \,dt$ add the same slivers over the same range, and both equal 2.

With limits and without

With no limits, $\int f\left(x\right) \,dx$ asks for an antiderivative, and the answer is a function with + C: $\int 3 \,dx = 3x + C$.

With limits, $\int_{a}^{b} f\left(x\right) \,dx$ is a single number. Where the curve is above the x-axis, it is the area of the region under the curve between x = a and x = b: $\int_{1}^{5} 3 \,dx = 12$.

The usual mistakes

Reading the limits top first. In $\int_{1}^{5} 3 \,dx$ the sweep starts at 1, at the bottom, and stops at 5.

Leaving out the dx. Without it the expression does not say which variable the slivers are laid along, and in $\int x y \,dx$ that decides the answer.

Taking the limits for heights or for the answer. They are positions on the x-axis; the heights come from the integrand, and the answer comes from adding the slivers.

Reading the sign as the letter S. It began as an S, but it is read "the integral".