A definite integral is calculus’s answer to a simple-sounding question: how much area sits under this curve between here and there? Rectangles make it easy; curves don’t — until the Fundamental Theorem of Calculus turns the whole problem into subtraction. Find an antiderivative, evaluate it at both ends, subtract. That’s the entire game, and this guide walks you through it with fully worked examples.
The Core Idea: F(b) − F(a)
If F(x) is an antiderivative of f(x) (meaning F′(x) = f(x)), then:
∫[a,b] f(x) dx = F(b) – F(a)
The +C constant of integration cancels out completely — [F(b) + C] − [F(a) + C] = F(b) − F(a) — so you never write “+C” when evaluating definite integrals.
A definite integral computes net signed area: regions above the x-axis count positive, regions below count negative. That’s the single most misunderstood fact in this topic — read the “net vs. total” section before your exam.
The 4-Step Evaluation Framework
- Find the antiderivative F(x): use the reverse power rule — raise the exponent by 1, divide by the new exponent.
- Set up evaluation brackets: write [F(x)] from a to b.
- Substitute both limits: compute F(b) and F(a) separately.
- Subtract and simplify: F(b) − F(a), upper minus lower, always in that order.
Worked Example 1: Area Under a Parabola
Evaluate the area under the curve from x = 1 to x = 3:
∫[1,3] (3x^2 – 2x + 1) dx
Step 1: Find the antiderivative
- 3x² → 3(x³/3) = x³
- −2x → −2(x²/2) = −x²
- 1 → x
So F(x) = x³ − x² + x.
Steps 2–3: Evaluate at both limits
F(3) = (3)³ − (3)² + (3) = 27 − 9 + 3 = 21
F(1) = (1)³ − (1)² + (1) = 1 − 1 + 1 = 1
Step 4: Subtract
∫₁³ (3x² − 2x + 1) dx = F(3) − F(1) = 21 − 1 = 20 square units
Worked Example 2: A Curve That Crosses the Axis
Evaluate:
∫[0,2] (4x^3 – 6x) dx
Antiderivative: F(x) = x⁴ − 3x² (check: derivative of x⁴ is 4x³, of −3x² is −6x ✓)
F(2) = 16 − 12 = 4; F(0) = 0. So the integral equals 4.
But here’s the catch: f(x) = 4x³ − 6x = 2x(2x² − 3), which equals zero at x = √1.5 ≈ 1.225. The curve dips below the x-axis from 0 to 1.225, then rises above it. The answer 4 is the net signed area — the negative part partially canceled the positive part.
Net Signed Area vs. Total Geometric Area
The integral automatically computes (area above) − (area below). If a problem asks for the total physical area, you must:
- Find where the curve crosses the x-axis (set f(x) = 0).
- Split the integral at those points.
- Take the absolute value of each below-axis section.
- Add everything up.
For Example 2: F(1.225) = (1.5)² − 3(1.5) = 2.25 − 4.5 = −2.25, so the below-axis region has area 2.25, and the above-axis region has area 4 − (−2.25) = 6.25. Total geometric area = 2.25 + 6.25 = 8.5 square units — very different from the net value of 4.
Worked Example 3: Symmetric Limits
Evaluate:
∫[-2,2] x^2 dx
F(x) = x³/3. F(2) = 8/3; F(−2) = −8/3.
Integral = 8/3 − (−8/3) = 16/3 ≈ 5.33 square units
Since x² is an even function (symmetric about the y-axis), you could also double the integral from 0 to 2: 2 × (8/3) = 16/3. Spotting symmetry halves the work.
3 Mistakes to Avoid
- Subtracting in the wrong order: it’s F(b) − F(a), upper minus lower. Reversing it flips the sign of your entire answer.
- Ignoring below-axis regions: when total geometric area is asked, a single unbroken integral will cancel out real physical space. Split at the x-intercepts.
- Dragging “+C” into definite integrals: it always cancels, so writing it only invites arithmetic slips.
Before computing, glance at the graph. Knowing whether the function stays above the axis, crosses it, or is symmetric tells you which version of “area” the integral will give you — and whether the question wants something different.