ZIMSEC A Level · Pure Mathematics · Form 6
Standard results
∫xⁿ dx = xⁿ⁺¹/(n + 1) + c (n ≠ −1); ∫1/x dx = ln|x| + c; ∫eᵃˣ dx = eᵃˣ/a + c; ∫sin ax dx = −cos ax / a + c; ∫cos ax dx = sin ax / a + c; ∫sec²x dx = tan x + c.
Techniques
- Recognition: ∫f′(x)/f(x) dx = ln|f(x)| + c.
- Substitution: change the variable, the dx, and the limits.
- Integration by parts: ∫u dv = uv − ∫v du (choose u to become simpler when differentiated).
- Partial fractions for rational functions.
- Trigonometric identities, e.g. sin²x = ½(1 − cos 2x).
Worked example: Find ∫x eˣ dx.
u = x, dv = eˣ dx → du = dx, v = eˣ
∫x eˣ dx = x eˣ − ∫eˣ dx = eˣ(x − 1) + c
Worked example: Evaluate ∫ from 0 to 1 of 2x/(x² + 1) dx.
= [ln(x² + 1)] from 0 to 1 = ln 2 − ln 1 = ln 2 ≈ 0.693
Applications
- Area under a curve between x = a and x = b: ∫y dx (areas below the axis come out negative).
- Area between curves: ∫(upper − lower) dx.
- Volume of revolution about the x-axis: V = π∫y² dx.
Exam tip: For definite integrals by substitution, change the limits as well – then you need not substitute back.
Check yourself
- Find ∫(3x² − 4/x) dx.
- Evaluate ∫ from 0 to π/2 of cos x dx.
- Find the volume when y = √x from x = 0 to x = 4 is rotated about the x-axis.
Show answers
- x³ − 4 ln|x| + c
- [sin x] = 1 − 0 = 1
- π∫x dx from 0 to 4 = π[x²/2] = 8π
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