IB Maths AA SLTopic 5 — CalculusPaper 1 & 2~8 min read
Definite Integrals
A definite integral has limits — it gives a number, not a function. The recipe is simple: integrate as usual (no “+ c”), plug in the upper limit, plug in the lower limit, subtract. There are six handy properties that can save you a ton of work, especially on Paper 1 — knowing them turns long calculations into one-line answers.
📘 What you need to know
Definite integral: ∫ab f(x) dx — has a lower limit a and upper limit b.
Fundamental Theorem of Calculus: ∫ab f(x) dx = F(b) − F(a), where F is any antiderivative of f.
No “+ c” — it cancels in F(b) − F(a).
Six properties let you split, swap, scale, and shift integrals without redoing the calculation.
On Paper 2, your GDC evaluates definite integrals directly. Use it to check, even when working is required.
What is a definite integral?
A definite integral is the integral with two numbers — the limits — attached:
Definite integral notation
∫ab f(x) dx
The lower limit a sits at the bottom, the upper limit b at the top. The result is a single number — usually representing an area, a total, or a net change.
The Fundamental Theorem of Calculus
Fundamental Theorem of Calculus
∫ab f(x) dx = [F(x)]ab = F(b) − F(a)
✓ in formula booklet
F(b) − F(a) — the antiderivative tells you the area
The “+ c” disappears because it appears in both F(b) and F(a) and cancels when you subtract. So in definite integrals, just leave it out from the start.
Three-step method
Definite integral recipe
Name the integral — call it I. If needed, expand brackets first to get an integrable form.
Integrate with limits in square brackets — [F(x)]ab. No “+ c”.
Substitute and subtract — F(b) − F(a). Be careful with signs.
📍
Paper 2: GDC shortcut
On the calculator paper, your GDC evaluates definite integrals directly. Even when you must show working manually, use the GDC to check your final answer — a quick safety net.
The six properties
These shortcuts let you transform integrals without recomputing them. Memorising the look of each one is enough — you’ll spot when to use them.
The six properties of definite integrals
1 · Constant out
∫ab kf(x) dx = k ∫ab f(x) dx
pull constants outside
2 · Term by term
∫ab [f(x) ± g(x)] dx = ∫ f ± ∫ g
split sums and differences
3 · Equal limits
∫aa f(x) dx = 0
no width = no area
4 · Swap limits
∫ba f(x) dx = − ∫ab f(x) dx
flipping limits negates
5 · Split interval
∫ab = ∫ac + ∫cb
break long intervals at c
6 · Horizontal shift
∫ab f(x) dx = ∫a−kb−k f(x+k) dx
shift function and limits together
🧠
“Same limits = 0 · Swap = negate · Split = add”
The three most useful properties to spot in exam questions. If two limits are equal — answer is zero. If they’re flipped — flip a minus sign. If a “middle” value is given — split into two integrals.
Worked examples
WE 1
Manual evaluation — show that integral equals 144
Show that ∫24 3x(x² − 2) dx = 144.
step 1 — expandI = ∫(2 to 4) (3x³ − 6x) dxstep 2 — integrateI = [¾x⁴ − 3x²] from 2 to 4step 3 — substitute and subtractat x = 4: ¾(256) − 3(16) = 192 − 48 = 144at x = 2: ¾(16) − 3(4) = 12 − 12 = 0I = 144 − 0 = 144 ✓∫(2 to 4) 3x(x² − 2) dx = 144always expand brackets BEFORE integrating — never integrate a product directly!
WE 2
GDC evaluation — Paper 2 only
Use your GDC to evaluate ∫01 3ex² sin x dx, giving your answer to 3 significant figures.
use GDCtype ∫ from 0 to 1 of 3e^(x² sin x) dxread outputGDC: 3.872 957…I = 3.87 (3 sf)this integrand has no nice antiderivative — GDC is the only way!
WE 3
Using properties — equal limits & swap
f(x) is continuous on 5 ≤ x ≤ 15. It’s known that ∫510 f(x) dx = 12 and ∫1015 f(x) dx = 5. Write down the values of:
(a) ∫77 f(x) dx (b) ∫105 f(x) dx
part (a) — equal limits∫(a to a) f = 0 (zero width)∫(7 to 7) f(x) dx = 0part (b) — swap limits∫(b to a) f = − ∫(a to b) f∫(10 to 5) f = − ∫(5 to 10) f = −12∫(10 to 5) f(x) dx = −12two of the easiest properties — spot them and write the answer instantly!
WE 4
Using properties — split, factor & shift
Same setup as WE 3. Find the values of:
(a) ∫515 f(x) dx (b) ∫510 6f(x + 5) dx
part (a) — split interval∫(5 to 15) f = ∫(5 to 10) f + ∫(10 to 15) f= 12 + 5 = 17∫(5 to 15) f(x) dx = 17part (b) — factor + shiftpull the 6 out (constant property):I = 6 ∫(5 to 10) f(x + 5) dxshift property with k = 5:∫(5 to 10) f(x + 5) dx = ∫(10 to 15) f(x) dx = 5so I = 6 × 5 = 30∫(5 to 10) 6f(x + 5) dx = 30shift property: limits move by k AND the inside changes by +k. They cancel out!
💡 Top tips
No “+ c” on definite integrals. Don’t even write it.
Expand brackets first. ∫3x(x² − 2) dx = ∫(3x³ − 6x) dx — never integrate a product directly.
Use square bracket notation with limits at the closing bracket: [F(x)]ab.
Check on GDC on Paper 2 — quick safety net for sign errors.
Spot the property. Same limits → 0. Flipped limits → negative. Middle value given → split.
Shift property: when both limits and the function shift by the same k, they cancel — areas are preserved.
⚠ Common mistakes
Adding “+ c” to a definite integral. Pointless — it cancels.
Sign errors in F(b) − F(a). Especially with negative numbers in the second bracket — use parentheses!
Forgetting to expand a product before integrating.
Mixing up upper and lower limits. Lower at the bottom of ∫, upper at the top — F(upper) − F(lower).
Not spotting properties. If you find yourself integrating something that’s already given to you in the question, you’ve missed a shortcut.
Bad shift property. The function shifts by +k but the limits shift by −k. Keep them straight.
Next up: Negative Integrals — what happens when the curve dips below the x-axis. Spoiler: a definite integral can come out negative, but areas can’t. The fix involves the modulus function.
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