Topic 8 covers integration as the reverse of differentiation and as the limit of a sum: standard integrals, definite integrals and areas (including areas between curves and below the axis), integration by substitution, by parts and using partial fractions, differential equations solved by separating the variables, and the trapezium rule.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.
A definite integral needs no + C: evaluate the antiderivative at the limits and subtract.
Antiderivative: x³ + 2x.
At x = 3: 27 + 6 = 33. At x = 1: 1 + 2 = 3.
Integral = 33 − 3 = 30.
Exam trap: the n ≠ −1 exclusion is not decoration — ∫x^(−1) dx is ln|x| + C, because x⁰/0 is nonsense. And the modulus signs in ln|x| matter for negative limits.
[x³ + 2x] from 1 to 3 = (27 + 6) − (1 + 2) = 33 − 3 = 30.
Substitution reverses the chain rule. Choose u to be the awkward inner function; the rest of the integrand should be (a multiple of) du/dx.
Let u = x² + 1 ⇒ du/dx = 2x ⇒ du = 2x dx. The 2x dx in the integral is exactly du.
New limits: x = 0 ⇒ u = 1; x = 1 ⇒ u = 2.
Integral = ∫ (from 1 to 2) u³ du = [u⁴/4] from 1 to 2 = 16/4 − 1/4 = 15/4 = 3.75.
Exam trap: if you change the variable you must change the limits (or convert back to x before substituting). Putting x-limits into a u-expression is the single most common error in this topic.
∫ (from 1 to 2) u³ du = [u⁴/4] = 2⁴/4 − 1⁴/4 = 4 − 0.25 = 3.75.
(= 15/4.)
By parts reverses the product rule.
Priority for u: ln x, then powers of x, then trig/exponential. (ln x has no simple integral, so it must be the u.)
u = x ⇒ du/dx = 1. dv/dx = e^(2x) ⇒ v = ½e^(2x).
= [x · ½e^(2x)] (0 to 1) − ∫ (0 to 1) ½e^(2x) dx
= ½e² − [¼e^(2x)] (0 to 1) = ½e² − (¼e² − ¼) = ¼e² + ¼
= (7.389056)/4 + 0.25 = 1.847264 + 0.25 = 2.097 (3 d.p.)
Exam trap: for ∫ ln x dx take u = ln x and dv/dx = 1, so v = x. You get x ln x − ∫ 1 dx = x ln x − x + C. Check by differentiating: ln x + 1 − 1 = ln x ✓
= [½x e^(2x)] − ∫ ½e^(2x) dx = ½e² − [¼e^(2x)] from 0 to 1
= ½e² − ¼e² + ¼ = ¼e² + ¼ = 1.847264 + 0.25 = 2.09726…
= 2.097 (3 d.p.)
Tap an integral, then the method you would use.
A rational function with a factorised denominator splits into simple logs (Topic 2 pays off here).
5 ≡ A(x + 3) + B(x − 2). x = 2 ⇒ 5 = 5A ⇒ A = 1. x = −3 ⇒ 5 = −5B ⇒ B = −1.
Integral = ∫ [1/(x − 2) − 1/(x + 3)] dx = [ln|x − 2| − ln|x + 3|] = [ln |(x − 2)/(x + 3)|]
At x = 4: ln(2/7). At x = 3: ln(1/6).
Value = ln(2/7) − ln(1/6) = ln[(2/7) × 6] = ln(12/7) = 0.539 (3 d.p.)
Exam trap: combine the logs before substituting the limits — it is faster and avoids the classic error ln a − ln b ≠ ln(a − b).
[ln|(x − 2)/(x + 3)|] from 3 to 4 = ln(2/7) − ln(1/6)
= ln[(2/7) ÷ (1/6)] = ln(12/7) = ln(1.714286) = 0.5390
= 0.539 (3 d.p.)
Separate the variables: get all the y terms on one side with dy, all the x terms with dx, then integrate both sides.
ln|y| = x² + C. At x = 0, y = 3: ln 3 = 0 + C ⇒ C = ln 3.
ln y = x² + ln 3 ⇒ y = 3e^(x²).
At x = 1: y = 3e = 3 × 2.718282 = 8.15 (3 s.f.)
Exam trap: apply the boundary condition immediately after integrating, while the constant is still additive. Trying to fit C after exponentiating leads to errors. And a “rate proportional to amount” model is always dP/dt = kP, which separates to P = P₀e^(kt).
∫(1/y) dy = ∫2x dx ⇒ ln y = x² + C. y(0) = 3 ⇒ C = ln 3.
y = 3e^(x²). At x = 1: y = 3e = 8.1548… = 8.15 (2 d.p.)
A definite integral is the signed area between the curve and the x-axis. Area below the axis comes out negative — split the integral at the roots if you want the total physical area.
When there is no analytic antiderivative, estimate numerically:
h = (5 − 1)/4 = 1. Ordinates: y(1) = 1, y(2) = 0.5, y(3) = 0.333333, y(4) = 0.25, y(5) = 0.2.
≈ 1/2 [1 + 0.2 + 2(0.5 + 0.333333 + 0.25)] = 0.5 [1.2 + 2(1.083333)] = 0.5 × 3.366667 = 1.683 (3 d.p.)
Exact value: ln 5 = 1.609. The estimate is an overestimate because 1/x is convex — the chords lie above the curve.
Exam trap: n strips means n + 1 ordinates. And to improve the estimate, increase the number of strips (reduce h) — not the number of decimal places.
≈ ½ [ (1 + 0.2) + 2(0.5 + 0.33333 + 0.25) ]
= ½ [1.2 + 2.16667] = ½ × 3.36667 = 1.6833
= 1.683 (3 d.p.), an overestimate of the true ln 5 = 1.609.
Tap an integral on the left, then its result.
Always check by differentiating. It takes ten seconds and catches sign errors, missing factors and forgotten chain-rule constants.
Standard: ∫xⁿ = x^(n+1)/(n+1) + C (n ≠ −1) · ∫1/x = ln|x| + C · ∫eᵏˣ = eᵏˣ/k + C
Substitution: reverses the chain rule — change the limits with the variable
By parts: ∫u(dv/dx) = uv − ∫v(du/dx); pick u = ln x or the polynomial
Partial fractions: split, integrate to logs, combine before substituting limits
Differential equations: separate the variables, integrate, apply the boundary condition at once
Trapezium rule: h/2[ends + 2 × middles]; convex ⇒ overestimate
Topic 8 rewards method choice above all — decide the technique before writing a single line. Press Finish to see your score.
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