A definite integral has limits a and b and gives an exact number — the area under the curve between x = a and x = b. No +c needed: it always cancels.
Upper and lower limitsAnswer is a numberNo + cF(b) − F(a)
📏 The Rules
F(b) − F(a)
Integrate, then subtract the value at the lower limit from the value at the upper limit.
no +c needed
The constant cancels out when you subtract F(a) from F(b).
answer is a number
Unlike indefinite integration, a definite integral gives an exact numerical value.
📖 Key Vocabulary
Definite integral — an integral evaluated between two limits, giving a numerical answer.
Limits — the upper (b) and lower (a) x-values between which the integral is evaluated.
F(x) — the integrated function, before the limits are substituted in.
∫ba f(x) dx = [ F(x) ]ab = F(b) − F(a)
F(x) is the anti-derivative of f(x). The +c cancels when you subtract, so you never write it.
🎯 Limit Explorer — drag a and b to see F(b)−F(a) update live
ℹ️ How to use: Drag the slider to explore. Watch how the values update live and connect to the definite integral idea below.
1.0
3.0
F(3.0) − F(1.0) = 9.000 − 2.333 = 6.667
📐
The Five-Step Method
Follow these steps for every definite integral. Show each step clearly.
📏 The Rules
1–2: integrate, write in brackets
Find F(x), then write [F(x)] with the limits a and b attached.
3–4: evaluate at b, then a
Substitute the upper limit, then the lower limit, into F(x) separately.
5: subtract
F(b) − F(a) gives the final numerical answer.
📖 Key Vocabulary
Upper limit (b) — the larger x-value bounding the integral.
Lower limit (a) — the smaller x-value bounding the integral.
Evaluate — substitute a specific x-value into a function to get a number.
∫ Definite Integral CalculatorFive-step method, fully worked
ℹ️ How to use: Type the coefficients of your polynomial and the upper/lower limits. The solver applies all five steps automatically — integrate, write in brackets, evaluate at b, evaluate at a, subtract.
Enter the coefficients and limits above.
👆
Try: x²−2x+3 from 1 to 3 ·
3x+1 from 0 to 2 ·
2x²−4 from -2 to 2
Five steps
1Integrate — find F(x)
2Write in square brackets: [ F(x) ]ab
3F(b) — substitute upper limit
4F(a) — substitute lower limit
5Subtract: F(b) − F(a)
⚠ Do NOT write + c for definite integrals — it always cancels.
Worked: ∫31 (x² − 2x + 3) dx
Integrate:F(x) = x3/3 − x2 + 3x
Write:[ x3/3 − x2 + 3x ]13
F(3):27/3 − 9 + 9 = 9
F(1):1/3 − 1 + 3 = 7/3
Answer:9 − 7/3 = 20/3 = 6.667 ✓
✏️
Worked Examples
📏 The Rules
write out every step
Examiners award marks for each stage — integrate, brackets, F(b), F(a), subtract.
keep fractions exact
Don't round mid-calculation — simplify fractions only at the final answer.
📖 Key Vocabulary
Anti-derivative — another name for the integrated function F(x).
Substitute — replace x with a specific number to evaluate the function.
Example 1 — ∫20 (3x + 1) dx
Integrate:F(x) = 3x2/2 + x
Write:[ 3x2/2 + x ]02
F(2):3(4)/2 + 2 = 8
F(0):0
Answer:8 ✓
Example 2 — ∫41 √x dx = ∫41 x½ dx
Integrate:F(x) = 2x3/2/3
F(4):2(8)/3 = 16/3
F(1):2/3
Answer:16/3 − 2/3 = 14/3 ≈ 4.67 ✓
Example 3 — ∫π0 sin x dx
Integrate:F(x) = −cos x
F(π):−cos π = −(−1) = 1
F(0):−cos 0 = −1
Answer:1 − (−1) = 2 ✓ (one arch of sine)
⚠
Negative Areas — a Common Trap
When the curve is below the x-axis, the integral is negative. For actual area, use the modulus. If the curve crosses, split the integral.
📏 The Rules
below the axis → negative integral
The signed value of the integral is negative when the curve dips below y = 0.
true area = |integral|
For actual physical area, take the modulus (positive value) of a negative result.
curve crosses? split it
Integrate each section separately, take the modulus of any negative part, then add.
📖 Key Vocabulary
Signed area — the raw result of integration, which can be negative below the x-axis.
Modulus — the positive (absolute) value of a number, written |x|.
Root — a point where the curve crosses the x-axis, where y = 0.
⚠ Negative Area CheckerSpot when the curve dips below the axis
ℹ️ How to use: Type the coefficients of a quadratic and the limits. The solver checks whether the curve is entirely above, entirely below, or crosses the x-axis within those limits, and calculates both the signed integral AND the true (always positive) area.
Enter the coefficients and limits above.
👆
Try: x²−4 from -2 to 2 (below axis) ·
x² from -1 to 1 (above axis) ·
x²−x−2 (crosses axis!)
⚠ The integral can be negative but area is always positive. Always check if the curve dips below the x-axis.
Curve below x-axis
Area: y = x² − 4, x = −2 to 2
Note:Curve is BELOW axis for x ∈ (−2, 2)
Integrate:F(x) = x3/3 − 4x
F(2)−F(−2):(8/3−8)−(−8/3+8) = −32/3
Area:|−32/3| = 32/3 ≈ 10.67 ✓
Curve crosses axis — split it
Total area: y = x3, x = −1 to 1
Root at x=0:Split [−1,0] and [0,1]
∫0-1 x3 dx:[x4/4]−10 = −1/4
∫10 x3 dx:[x4/4]01 = +1/4
Total area:|−1/4| + |1/4| = 1/2 ✓
Area = |∫ra f dx| + |∫br f dx|
🎮
Definite Integral Game
Evaluate the definite integral and choose the correct value.
🔢 LIMIT CHALLENGE — 8 Questions
ℹ️ How to play: A definite integral question appears. Click the correct answer. Score and streak update after each — click Next to continue.
Score: 0Q 1/8
What is the value of this definite integral?
∫₀² 2x dx
Score: 0/8
⚙️
Applications
Where this topic is used in engineering, manufacturing, maintenance and daily life.
📖 Key Vocabulary
Mean value — the average value of a function over an interval, found using a definite integral.
Work done — energy transferred by a variable force, found by integrating force over distance.
⚡Electrical:q=∫₀ᵀ i dt for a time-varying current.
🔧Mechanical:Variable force work W=∫F(x)dx.
📊Analysis:Mean value f̄=(1/(b−a))∫f dx.
🏭Manufacturing:Integrate stress profile for total force on cross-section.
📝
Practice — Definite Integrals
Show F(x), then [F(x)] with limits, then F(b) − F(a).
🟢 Tier 1
ℹ️ How to use: Work through each question and write down your answer. When ready, click Show Answer for the full step-by-step solution.
