IB Maths AI HL
Further Integration
Paper 1 & 2
~6 min read
Definite Integrals
A definite integral has limits — it evaluates to a single number, not a family of functions. Integrate as usual, then compute F(b) − F(a): the upper limit minus the lower. The “+ c” cancels, and a handful of properties let you flip, split, and simplify integrals before you even evaluate.
📘 What you need to know
- Definition: ∫ab f(x) dx = F(b) − F(a), where F is an antiderivative of f.
- No “+ c“: the constant appears in both F(b) and F(a) and cancels.
- Square-bracket notation: write [F(x)]ab, then substitute top minus bottom.
- Swap limits, change sign: ∫ab = −∫ba.
- Zero width: ∫aa f(x) dx = 0.
- Split & scale: you can break the interval at a midpoint and pull constants out.
Evaluating with the FTC
Fundamental Theorem of Calculus
∫ab f(x) dx = [F(x)]ab = F(b) − F(a)
✓ the FTC link is standard
🤔 Why does the “+ c” disappear?
Take any antiderivative F(x) + c. Evaluating top minus bottom gives [F(b) + c] − [F(a) + c] — the two c‘s subtract to zero. So whichever constant you pick, the definite integral is the same number. That’s why you simply omit it.
🧠 “Top minus bottom”
Integrate, drop the “+ c“, put it in square brackets with the limits, then substitute the upper limit and subtract the lower: F(b) − F(a). Order matters — upper first.
Properties of definite integrals
🧭 The properties you can use
- Swap limits: ∫ab f = −∫ba f.
- Equal limits: ∫aa f = 0.
- Constant multiple: ∫ab k f = k ∫ab f.
- Sum/difference: ∫ab (f ± g) = ∫ab f ± ∫ab g.
- Split the interval: ∫ab f = ∫ac f + ∫cb f.
Why split? If a question gives you ∫05 and ∫03, you can find ∫35 by subtraction — no need to integrate again.
Worked examples
WE 1Evaluate ∫13 (2x + 1) dx
Integrate, then substitute top minus bottom.
[x² + x]13
= (9 + 3) − (1 + 1)
= 12 − 2
= 10
WE 2Evaluate ∫02 (3x2 − 4x + 1) dx
Integrate term by term; the lower limit 0 makes the bottom substitution vanish.
[x³ − 2x² + x]02
= (8 − 8 + 2) − (0)
= 2
WE 3Evaluate ∫14 1x dx, giving an exact answer
A special-function integral: 1x → ln|x|.
[ln|x|]14 = ln 4 − ln 1
ln 1 = 0
ln 4 (≈ 1.39)
WE 4Given ∫02 f(x) dx = 2, write down ∫20 f(x) dx
Swapping the limits flips the sign — no re-integration needed.
∫20 f = − ∫02 f
= −(2)
= −2
WE 5Evaluate ∫0π/2 cos x dx
cos → sin; work in radians.
[sin x]0π/2
= sin π2 − sin 0
= 1 − 0
= 1
💡 Top tips
- Square brackets — write [F(x)]ab before substituting, so you don’t lose track.
- Upper minus lower, in that order.
- No “+ c“ in a definite integral.
- Use the properties to flip or split before grinding through algebra.
- Radians for trig integrals.
- GDC backup — Paper 2 lets you evaluate the integral directly to check.
⚠ Common mistakes
- Subtracting the wrong way — it’s F(b) − F(a), not F(a) − F(b).
- Adding “+ c“ to a definite integral.
- Sign slips when the lower limit is negative.
- Forgetting to flip the sign when limits are swapped.
- Degrees mode on the GDC for a trig integral.
Next up — Areas Between Curves. You can now evaluate a definite integral and use its properties to flip and split. The next topic turns that number into geometry: the integral of f − g between intersection points gives the area enclosed between two curves, with the definite integral doing the measuring.
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