Evaluate a definite integral by integrating and substituting the upper and lower limits

What this dot point is asking

SEAB wants you to evaluate a definite integral, an integral with limits (an upper and a lower number). Unlike an indefinite integral, a definite integral gives a single number. You integrate as usual, then substitute the upper and lower limits and subtract. No constant of integration is needed, because it cancels in the subtraction. Definite integrals are the route to areas under curves.

The answer

What the limits mean

A definite integral is written with numbers at the top and bottom of the integral sign:

Here $a$ is the lower limit and $b$ is the upper limit. The result is a number that depends on $a$, $b$ and the function.

The evaluation rule

Integrate the function to get $F(x)$, write it in square brackets with the limits, then substitute the upper limit, substitute the lower limit, and subtract:

Why no constant of integration

If you kept the constant $c$, it would appear as $(F(b) + c) - (F(a) + c)$, and the two $c$ terms cancel. So you may safely drop the $+ c$ for a definite integral; including it does no harm but is unnecessary.

Working tidily

Keep the square-bracket line so the marker can see your integrated function, then show the substitution clearly. Evaluate the upper-limit value and the lower-limit value separately before subtracting, which makes sign errors easy to catch. Take care with negative limits: substituting a negative number into a power needs brackets.

Two useful properties

Two facts often save work. First, swapping the limits changes the sign of the result:

Second, an integral can be split at any point $k$ between the limits, which is exactly what you do when a curve crosses the $x$-axis:

Both follow directly from the rule $F(b) - F(a)$, and recognising them makes harder area questions much shorter.

Examples in context

Example 1. Distance travelled. Integrating a velocity function between two times gives the distance travelled in that interval as a single number. The definite integral is exactly how kinematics turns a velocity-time relationship into a total distance.

Example 2. Average value. Dividing a definite integral of a function by the width of the interval gives the average value of the function over that interval. The definite integral is the building block for averages of continuously varying quantities.

Try this

Q1. Evaluate $\displaystyle\int_0^1 2x\, dx$. [2 marks]

• Cue. $[x^2]_0^1 = 1 - 0 = 1$.

Q2. Evaluate $\displaystyle\int_1^2 3x^2\, dx$. [2 marks]

• Cue. $[x^3]_1^2 = 8 - 1 = 7$.

Q3. Evaluate $\displaystyle\int_0^2 (x + 1)\, dx$. [3 marks]

• Cue. $\left[\dfrac{x^2}{2} + x\right]_0^2 = (2 + 2) - 0 = 4$.