The Definite Integral

The definite integral of from to is the limit of Riemann sums as the number of rectangles goes to infinity:

It’s a single number: the net signed area between the graph of and the x-axis from to . Area above the axis counts as positive and area below as negative. In the Riemann sums lesson, the sums for on all closed in on ; in this notation, .

What Is a Definite Integral?

A Riemann sum estimates the area under a curve with rectangles of width . More rectangles give a better estimate, and the definite integral is the exact value those estimates approach.

When the limit exists, is called integrable on . Every continuous function is integrable, and for a continuous function the limit is the same whether the sums use left endpoints, right endpoints, midpoints or any other sample points.

Definite Integral Notation

• is the integral sign, a stretched S for “sum.”
• and are the lower and upper limits of integration. The interval is the interval of integration.
• is the integrand, the function whose area you’re measuring.
• names the variable of integration and stands for the width in the limit.

The variable is a placeholder. , and all mean the same number, so the variable of integration is called a dummy variable.

Definite vs Indefinite Integrals

The two share a symbol but answer different questions:

The fundamental theorem of calculus, in the next lesson, shows why they share a symbol: you can compute the definite integral from any antiderivative.

The Definite Integral as Net Signed Area

When on , the definite integral is the area of the region under the graph and above the x-axis. When dips below the axis, the rectangles there have negative heights, so that area is subtracted:

Example 1. Find .

The line crosses the axis at . From 0 to 2 it forms a triangle below the axis with base 2 and height 2, area 2. From 2 to 6 it forms a triangle above the axis with base 4 and height 4, area 8:

Net area vs total area. The total area between the graph and the axis counts every region as positive: here . To find it with integrals, split the interval where changes sign and add the absolute values:

If a problem asks for “the area of the region,” it means total area. If it asks for “the value of the integral,” it means net area.

How to Find a Definite Integral

There are three ways, and you’ll use all of them:

1. Geometry. If the region is made of triangles, rectangles, trapezoids or parts of circles, add their areas, subtracting any below the axis.
2. The limit definition. Write a Riemann sum with rectangles and take the limit as . This always works in principle but is slow.
3. The fundamental theorem of calculus. Find an antiderivative and compute . This is the method you’ll use most often, and it’s the subject of the next lesson.

Example 2 (a trapezoid). Find .

The region under from 0 to 4 is a trapezoid with parallel sides and and width 4:

Example 3 (a semicircle). Find .

The graph of is the upper half of the circle , with radius 3. The integral is the area of that semicircle:

None of the antiderivative rules so far can handle , but geometry gives the answer in one line.

Example 4 (the limit definition). Find as a limit of right Riemann sums.

With and , we get :

The sum of the first cubes is (see the summation formulas), so

As , the top and bottom both grow like , so by the rule for limits at infinity, :

Example 5 (a preview of the fundamental theorem). An antiderivative of is . Then

That’s the same answer as Example 4, with no sums and no limits. The next lesson shows why this always works.

Properties of Definite Integrals

These properties follow from the definition, because sums behave the same way. They let you break an integral into simpler pieces, and they answer many AP-style questions without any integration.

The comparison properties tell you about the size of an integral without computing it:

Additivity holds even when isn’t between and , as long as is integrable on all the intervals involved. The reversal rule makes the signs work out.

Example 6. Suppose , and . Find:

(a) . By additivity, , so .

(b) . Reversing the limits changes the sign: .

(c) . Split the integral and pull out the constants:

(d) . The interval has zero width, so the answer is 0.

Example 7 (estimating with bounds). Show that .

On , decreases from to . So , and :

The true value is about 0.747. This integral has no elementary antiderivative, so bounds and numerical sums are the practical tools for it.

Which Functions Are Integrable?

• Every continuous function on is integrable there.
• A function that is bounded with only finitely many jump discontinuities is also integrable. Each jump just separates two pieces of area.
• A function with a vertical asymptote inside the interval is not integrable in this sense. has no value as a definite integral, because the area near 0 is unbounded. Integrals like this are called improper integrals and are handled in Calculus 2.

Common Mistakes

• Adding to a definite integral. The answer is a number. The constants cancel in .
• Confusing net area with total area. The integral subtracts area below the axis. For total area, split at the zeros and add absolute values.
• Forgetting the sign when reversing limits. .
• Splitting a product. is not . Only sums and constant multiples split.
• Integrating across an asymptote. Check that is defined and bounded on the whole interval first.

For more examples of net area and the properties, see Section 5.2 of the free textbook OpenStax Calculus Volume 1.

Practice Problems

Try each one before opening the answer. The practice set has 12 more with instant checking.

Problem 1. Find .

Problem 2. Find using geometry.

Problem 3. Find .

Problem 4. Find .

Problem 5. If and , find .

Problem 6. If , find .

The definite integral turns area into a number. Next, the fundamental theorem of calculus shows how to compute it from an antiderivative, without any Riemann sums at all.