14.13 Definite Integrals and Areas

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Part 1: What is a Definite Integral?

You already know that when you integrate a function (find its integral), you get a + c at the end. For example:

$\int x^2 \, dx = \frac{1}{3}x^3 + c$

This is called an indefinite integral — "indefinite" means it has no fixed, single answer because c can be any number.

A definite integral is different. It is evaluated between two fixed values called limits. Because of these limits, the answer is always a single, definite number — no c involved.

The notation looks like this:

$\int_a^b f(x) \, dx$

• The lower limit is a (written at the bottom of the ∫ sign)
• The upper limit is b (written at the top of the ∫ sign)

Part 2: How to Evaluate a Definite Integral

Follow these three steps every time:

Step 1: Integrate the function normally (but leave out the + c).

Step 2: Write the result inside square brackets with the limits: $[F(x)]_a^b$

Step 3: Substitute — put in the upper limit first, then subtract the result when you put in the lower limit.

$\int_a^b f(x) \, dx = \left[F(x)\right]_a^b = F(b) - F(a)$

Part 3: Worked Examples — Evaluating Definite Integrals

Example 1 — Powers of x

Evaluate $\displaystyle\int_1^2 \frac{x^5 + 3}{x^2} \, dx$

Step 1: Simplify the fraction by dividing each term by $x^2$:

$\frac{x^5}{x^2} + \frac{3}{x^2} = x^3 + 3x^{-2}$

Step 2: Integrate:

$\int (x^3 + 3x^{-2}) \, dx = \frac{1}{4}x^4 - \frac{3}{x}$

Step 3: Apply the limits (upper limit 2, lower limit 1):

$= \left(\frac{1}{4}(2)^4 - \frac{3}{2}\right) - \left(\frac{1}{4}(1)^4 - \frac{3}{1}\right)$

$= \left(4 - \frac{3}{2}\right) - \left(\frac{1}{4} - 3\right) = \frac{5}{2} + \frac{11}{4} = 5\frac{1}{4}$