This mini-lesson covers AQA section H — Integration: integration as the reverse of differentiation, definite integrals and areas, integration by substitution, by parts, and using partial fractions, the trapezium rule (and whether it over- or under-estimates), and differential equations solved by separating the variables.
Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect ⭐ stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.
Integration undoes differentiation. Every indefinite integral needs the constant of integration, since differentiating a constant gives zero.
∫₁³ 3x² dx = [x³]₁³ = 27 − 1 = 26.
For definite integrals the constant c cancels, so you can omit it.
The definite integral gives the signed area between the curve and the x-axis.
∫₀² (x² − 4) dx = [x³/3 − 4x]₀² = (8/3 − 8) − 0 = −16/3.
The curve is below the axis throughout, so the area is 16/3 — the sign is telling you the position, not the size.
Substitution is the chain rule in reverse. Look for a function and (a multiple of) its derivative both present.
Evaluate ∫₀² 2x(x² + 1)³ dx.
Let u = x² + 1, so du/dx = 2x, i.e. du = 2x dx — exactly the 2x dx already sitting in the integral.
Change the limits: x = 0 ⇒ u = 1; x = 2 ⇒ u = 5.
∫₁⁵ u³ du = [u⁴/4]₁⁵ = (625 − 1)/4 = 624/4 = 156.
For a definite integral, change the limits to the new variable — or convert back to x before substituting. Applying the old x-limits to a u-expression is a guaranteed loss of marks.
Tap an integral, then tap the method that will crack it.
By parts is the product rule in reverse. Use it for a product of two unrelated functions, such as x × ex or x × ln x.
Choose u = x (since du/dx = 1 — simpler) and dv/dx = ex (so v = ex).
∫ x ex dx = x ex − ∫ ex dx = x ex − ex + c.
Evaluate: [x ex − ex]₀¹ = (e − e) − (0 − 1) = 0 − (−1) = 1.
The exception: ∫ ln x dx. Here you must take u = ln x and dv/dx = 1, because ln x has no elementary integral to start from. That gives ∫ln x dx = x ln x − x + c.
When the integrand is an algebraic fraction with a factorisable denominator, split it first, then integrate each piece as a log.
∫ (5x + 1)/((x − 1)(x + 2)) dx.
Partial fractions: (5x + 1)/((x − 1)(x + 2)) = 2/(x − 1) + 3/(x + 2).
Integrate term by term: ∫ = 2 ln|x − 1| + 3 ln|x + 2| + c.
Two habits worth building: keep the modulus signs inside the logarithm, and remember the 1/a factor when the denominator is ax + b rather than x + b.
When a function cannot be integrated exactly, estimate the area with trapezia.
Estimate ∫₀² 2x dx with n = 4, so h = 0.5.
Ordinates: y(0) = 1, y(0.5) = 1.4142, y(1) = 2, y(1.5) = 2.8284, y(2) = 4.
T = 0.25 × [1 + 2(1.4142 + 2 + 2.8284) + 4] = 0.25 × [1 + 12.4853 + 4] = 0.25 × 17.4853 = 4.371 (3 d.p.).
The exact value is 3/ln2 = 4.328, so the estimate is an overestimate.
Over or under? If the curve is convex (bending upwards, f″ > 0) each chord lies above the curve, so the trapezium rule overestimates. For a concave curve it underestimates. Sketch it — you should be able to justify which.
A first-order differential equation of the form dy/dx = f(x)g(y) is solved by getting all the y-terms on one side and all the x-terms on the other.
dN/dt = −kN (the rate of decay is proportional to the amount present).
Separate: ∫ (1/N) dN = ∫ −k dt ⇒ ln N = −kt + c.
Exponentiate: N = e−kt + c = A e−kt, where A = ec is the initial amount N₀.
So the exponential model of the last topic is a consequence of the differential equation.
General vs particular: the general solution contains c; use the given initial condition (for example N = 200 when t = 0) to find c and produce the particular solution. A question asking to "solve" usually wants the particular one.
Tap an item on the left, then its partner on the right.
The hardest part of an integration question is deciding what to do. Work through this order:
Always check by differentiating. Integration is the only topic where you can verify your answer completely — differentiate it and see whether you get the integrand back.
Standard: ∫xⁿ = x^{n+1}/(n+1) + c (n ≠ −1) · ∫1/x = ln|x| + c · ∫e^{kx} = (1/k)e^{kx} + c
Area: signed — split at roots and take the modulus of each part; between curves, integrate (upper − lower)
Substitution: a function and its derivative both present — and change the limits
By parts: ∫u dv/dx = uv − ∫v du/dx — choose u to be the part that simplifies
Partial fractions: split, then integrate each term to a log
Trapezium rule: h/2[y₀ + 2(middles) + yₙ] — convex curve ⇒ overestimate
Differential equations: separate the variables, integrate both sides, then use the initial condition
That is AQA section H — Integration. Press Finish to see your score.
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