A definite integral gives the exact area between the curve and the x-axis from x = a to x = b. Explore different functions to see the area.
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The definite integral: \intab f(x) dx = [F(x)]ab = F(b) - F(a) where F(x) is any antiderivative of f(x). This equals the signed area between the curve and the x-axis from x = a to x = b. Note: NO +C needed for definite integrals — it cancels out.
Example 1: Basic definite integral
Step by step
- Evaluate \int13 (x2 + 1) dx.
- Find antiderivative: F(x) = (x3)/(3) + x.
- Evaluate: F(3) - F(1).
- F(3) = (27)/(3) + 3 = 9 + 3 = 12.
- F(1) = (1)/(3) + 1 = (4)/(3).
- 12 - (4)/(3) = (36)/(3) - (4)/(3) = (32)/(3).
Final answer
\int13 (x2 + 1) dx = (32)/(3)
Signed vs unsigned area: If the curve goes BELOW the x-axis (f(x) < 0) in the interval, the integral gives a NEGATIVE value. This is called signed area. For geometric area (always positive), you must split the integral at x-intercepts and take absolute value: Area = \intab |f(x)| dx
Example 2: Area below x-axis
Step by step
- Find the area enclosed between f(x) = x2 - 4 and the x-axis from x = −2 to x = 2.
- Root: x² − 4 = 0 → x = ±2.
- Check f(0) = −4 < 0. Curve is BELOW the x-axis on [−2, 2].
- \int-22(x2 - 4) dx = [(x3)/(3) - 4x]-22.
- = ((8)/(3) - 8) - ((-8)/(3) + 8) = -(16)/(3) - (16)/(3) = -(32)/(3).
- Area = |-(32)/(3)| = (32)/(3).
Final answer
Area = (32)/(3)
Area vs integral in IB exams: IB questions are specific: • 'Evaluate the integral' → can be negative, keep the sign • 'Find the area' → always positive, take absolute value if below x-axi • 'Find the area enclosed between the curve and the x-axis' → find roots first, integrate between them, make positive
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Area under a curve in context
Area under a curve (above the x-axis): For a curve y = f(x) that lies above the x-axis between x = a and x = b, the area of the region between the curve and the x-axis is the definite integral ∫ₐᵇ f(x) dx.
Example: area under a parabola
Find the area between y = 4x − x² and the x-axis from x = 0 to x = 4.
(The curve is above the x-axis on this interval.)
Step by step
- Integrate f(x) between the limits.
- Evaluate at the limits.
Final answer
Area = 32/3 ≈ 10.7 square units.
Example: area under y = 3x²
Find the area between y = 3x² and the x-axis from x = 1 to x = 3.
Step by step
- Integrate between the limits.
- Evaluate.
Final answer
Area = 26 square units.
Use the GDC for the area: On both papers you may compute the area directly with the GDC's definite-integral function — enter f(x) and the limits a and b.
AI SL only asks for the area between a curve and the x-axis, not between two curves.
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Definite integrals on the GDC
Technology is allowed: AI SL lets you evaluate definite integrals using your GDC.
Use the calculator's definite-integral (∫) tool: enter the function, the variable, and the lower and upper limits.
Example: by hand and by GDC
Evaluate ∫₀² (x³ + 1) dx.
Step by step
- By hand: find the antiderivative and evaluate.
- By GDC: enter ∫ from 0 to 2 of (x³ + 1) → 6. Same answer.
Final answer
∫₀² (x³ + 1) dx = 6.
Example: a messier function (GDC)
Find the area between y = 0.5x³ − 2x + 4 and the x-axis from x = 0 to x = 3 (the curve is above the axis).
Step by step
- Enter the definite integral on the GDC with limits 0 and 3.
- Read the value from the GDC.
Final answer
Area ≈ 13.1 square units (from the GDC).
When to use the GDC: For a quick polynomial you can integrate by hand; for anything messy, use the GDC definite-integral tool.
There is no + C for a definite integral — it cancels in F(b) − F(a).
Increase in a quantity = definite integral of its rate: If you know the RATE OF CHANGE of a quantity, the increase over an interval [a, b] i: \text(Increase) = \intab (dQ)/(dx) dx = Q(b) - Q(a) This appears in IB a: 'dP/dx is given.
Find the increase in profit when x goes from a to b.'
IB-style: Increase in profit (2025 style)
Step by step
- The rate of change of profit is (dP)/(dx) = -10x + 460, where P is in MUR and x is kg produced.
- She makes a profit of 3300 MUR when producing 10 kg.
- (a) Find P(x). [5 marks]
- Integrate: P(x) = -5x2 + 460x + C.
- P(10) = −500 + 4600 + C = 3300 → C = −800.
- So P(x) = −5x² + 460x − 800.
- (b) Find the increase in profit when production increases from 25 to 50 kg. [2 marks]
- \int2550(-10x + 460) dx = [-5x2 + 460x]2550.
- At x = 50: −12500 + 23000 = 10500. At x = 25: −3125 + 11500 = 8375.
- Increase = 10500 − 8375 = 2125 MUR.
Final answer
Increase in profit = 2125 MUR
Two ways to find the increase: Method 1 (direct): Integrate the RATE from a to b — ∫[a→b] dP/dx dx Method 2 (after finding P): Calculate P(b) − P(a) Both give the same answer.
You might use the GDC for method 1 directly.
IB-style: Rate of change from derivative (2021 style)
Step by step
- A company's profit rate is (dP)/(dx) = -1.6x + 48. The company produces 30 to 50 cars per month.
- (a) Find P(x) given P(15) = 260.
- Integrate: P(x) = -0.8x2 + 48x + C.
- P(15) = −0.8(225) + 48(15) + C = −180 + 720 + C = 260 → C = −280.
- P(x) = −0.8x² + 48x − 280.
- (b) Describe how profit changes for 30 < x ≤ 50.
- dP/dx at x = 30: −1.6(30) + 48 = −48 + 48 = 0 (x = 30 is the maximum).
- For x > 30: dP/dx = −1.6x + 48 < 0. Profit is DECREASING.
Final answer
Profit is decreasing for x > 30 (rate of change is negative)
'Describe how profit changes' — use f'(x) sign: If asked to 'describe how a quantity changes' over an interval: • Find dQ/dx in that interval • If dQ/dx > 0 → increasing • If dQ/dx < 0 → decreasing • Give a specific reason: 'because dP/dx < 0 for 30 < x ≤ 50'
IB-style question — area between two curves
Find the area enclosed between the curves y = x² and y = 2x.
Step by step
- Find where they meet (the limits) by setting them equal.
- Between 0 and 2 the line y = 2x is on top, so integrate (top − bottom).
Final answer
Area = 4⁄3 ≈ 1.33.
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