What is a definite integral? The area under a graph, step by step

The idea: adding up many small pieces

The area of a rectangle is width times height. The trouble starts when the top edge of a region is curved. The idea of the integral is simple: cut the region into thin strips. Each strip is almost a rectangle, so its area is easy to work out. The total area of the strips approximates the whole area - and the thinner the strips, the better.

Lots of everyday quantities work the same way:

• distance is the area under a graph of speed - when the speed changes, you add up short stretches of time times the speed,
• the energy used is the area under a graph of power,
• a day’s takings are the sum of the takings in each minute.

From rectangles to the integral

Go through the animation step by step with the buttons or the arrow keys.

With very thin strips it no longer matters whether the height comes from the left end, the right end or the middle - the totals close in on the same number.

Try it yourself: rectangles under any graph

Pick a function and an interval, change the number of rectangles and how their heights are chosen, and compare the total with the integral. For $\sin x$ and $x^3 - 3x$ part of the graph is below the $x$ axis - there the rectangles count as negative.

What to notice:

• the more rectangles, the smaller the difference between their total and the integral,
• rectangles taken from the middle of each strip are usually far more accurate than those from the left or right end - with just a few rectangles the difference is already small,
• when part of the graph is above the axis and part below, the integral can come out close to zero even though the region is not small at all.

The definition

Cut the interval $[a, b]$ into $n$ strips of width $\Delta x = \frac{b - a}{n}$ and pick a point $x_i$ in each. The definite integral is the limit of the total area of the rectangles:

The $\int$ sign is a stretched letter S - for “sum” - and the $dx$ is a reminder of the width of the strips.

Working out the limit of a sum every time would be tedious. Fortunately there is a shortcut - the fundamental theorem of calculus. If $F$ is an antiderivative of $f$, that is $F'(x) = f(x)$, then

Why does this work? The derivative $f = F'$ says how fast $F$ grows. Over a thin strip $F$ therefore grows by about $f(x_i)\,\Delta x$ - exactly the area of the rectangle. Adding up these increases from $a$ to $b$ gives the whole increase of $F$, which is $F(b) - F(a)$. Integration and taking derivatives undo each other.

Example 1: the area under a parabola

An antiderivative of $x^2$ is $\frac{x^3}{3}$, because $\left(\frac{x^3}{3}\right)' = x^2$. So

That is exactly the number the rectangles were closing in on. And the area from the question at the start is $\int_0^1 x^2\,dx = \frac{1}{3}$.

Example 2: the area under a sine wave

An antiderivative of $\sin x$ is $-\cos x$, because $(-\cos x)' = \sin x$. So

Your turn

Work out $\int_1^3 (2x + 1)\,dx$, then check the answer geometrically: what shape is the region under this graph?

Common mistakes

• Area below the $x$ axis. The integral counts it as negative. $\int_0^{2\pi} \sin x\,dx = 0$, even though the graph and the axis enclose a region of area $4$. To get the area, integrate the parts above and below the axis separately and add up their absolute values.
• The wrong order. It is $F(b) - F(a)$, not $F(a) - F(b)$. Swapping them flips the sign of the answer.
• The constant $C$. An indefinite integral gets $+\,C$, because there are infinitely many antiderivatives. In a definite integral $C$ cancels: $(F(b) + C) - (F(a) + C) = F(b) - F(a)$.
• Mixing up definite and indefinite integrals. A definite integral, with limits $a$ and $b$, is a number. An indefinite integral is a family of functions, $F(x) + C$.

At university

The definite integral is one of the main topics of a first calculus course. In the open textbook Active Calculus, for example, the fundamental theorem of calculus is followed by integration by parts and by substitution, numerical integration and applications: areas between curves, arc length, volumes of solids of revolution, work and improper integrals. In physics an integral gives displacement from velocity and work from force.

Check yourself

Flashcards for review

Flip a card and say whether you knew the answer. The ones you did not know come back at the end of the deck. Why it pays to come back to them in a few days is explained in the guide Spaced repetition.