amazon.co.ukStandard results, reverse chain rule, substitution, integration by parts, areas between curves, volumes of revolution and differential equations - A-Level pure maths calculus (Edexcel 9MA0 Topic 8, AQA 7357 Section H) revised with worked examples and mark-scheme logic.
Integration is the reverse of differentiation plus a constant of integration (Edexcel 9MA0 8.1; AQA 7357 H1; OCR A H640 12.01). The power rule: the integral of x^n dx = x^(n+1)/(n+1) + C for any n except -1 - raise the power by one and divide by the new power. The two exceptions and extras are recall material: the integral of 1/x dx = ln|x| + C (note the modulus); the integral of e^x dx = e^x + C; and the trigonometric results the integral of sin x dx = -cos x + C, the integral of cos x dx = sin x + C, and the integral of sec^2 x dx = tan x + C. The sign on the sine integral is the classic slip - check it by differentiating your answer, since d/dx(-cos x) = sin x confirms it.
A definite integral has limits and evaluates to a number: the integral from a to b of f(x) dx = F(b) - F(a), with no constant needed. It measures signed area: regions below the axis contribute negatively, so area questions split the integral where the curve crosses the axis. Polynomials integrate term by term exactly as they differentiate, so fluency here is mechanical - the marks are lost in the exceptions (the modulus in ln|x|, the sign on -cos x) and in the techniques that follow. Writing the antiderivative line explicitly before substituting limits earns the method mark - jumping straight to a number is where marks are lost.
Evaluate the integral from 1 to 3 of (2x + 1/x) dx.
Line 1: Antiderivative: x^2 + ln|x| (M1: power rule plus ln result).
Line 2: Substitute limits: (9 + ln 3) - (1 + ln 1) = 8 + ln 3 (A1).
Line 3: = 8 + 1.0986 = 9.10 (A1: 3 s.f. if decimals requested; 8 + ln 3 is the exact form).
The reverse chain rule handles integrals where a derivative sits alongside a function of itself: the integral of f'(x)[f(x)]^n dx = [f(x)]^(n+1)/(n+1) + C, and the integral of f'(x)e^(f(x)) dx = e^(f(x)) + C. Spot the pattern first - if the derivative of the inner function (up to a constant) multiplies the outer function, the reverse chain rule is the fast route, no substitution needed.
Substitution (Edexcel 9MA0 8.5; AQA 7357 H2; OCR A H640 17.01) is the general version: let u = f(x), compute du/dx, rewrite dx = du/(du/dx), and replace every x in the integrand. For definite integrals, change the limits from x-values to u-values - then the integral is entirely in u and no back-substitution is needed. Forgetting the limit change is the most common error, and it silently produces a wrong answer if the integrand is also back-substituted inconsistently.
Evaluate the integral from 0 to 2 of x sqrt(4 - x^2) dx using the substitution u = 4 - x^2.
Line 1: du/dx = -2x, so x dx = -du/2 (M1: substitution set up).
Line 2: Limits: x = 0 gives u = 4; x = 2 gives u = 0 (M1: limits changed).
Line 3: Integral becomes the integral from 4 to 0 of u^(1/2) (-1/2) du = (1/2) times the integral from 0 to 4 of u^(1/2) du (A1: flipped and simplified).
Line 4: = (1/2)[(2/3)u^(3/2)] from 0 to 4 = (1/3)(8) = 8/3 (A1: exact answer).
Evaluate the integral from 0 to 1 of x/(1 + x^2)^2 dx.
Line 1: The derivative of 1 + x^2 is 2x, which matches the numerator up to a constant (M1: reverse chain spot).
Line 2: Antiderivative: -1/(2(1 + x^2)) (A1).
Line 3: Limits: [-1/(2(1 + x^2))] from 0 to 1 = -1/4 - (-1/2) = 1/4 (A1).
Integration by parts (Edexcel 9MA0 8.5; AQA 7357 H3; OCR A H640 17.01) reverses the product rule: the integral of u (dv/dx) dx = uv - the integral of v (du/dx) dx. Choosing u follows the LIATE order - Logs, Inverse trig, Algebraic, Trig, Exponential - pick u as the earliest category present, because u must simplify when differentiated and dv/dx must survive integration. Logs go first: ln x differentiates to the clean 1/x; exponentials go last: e^x integrates to itself without simplifying.
Use integration by parts to find the integral of x ln x dx.
Line 1: u = ln x, dv/dx = x, so du/dx = 1/x and v = x^2/2 (M1: LIATE choice of u).
Line 2: Integral = (x^2/2) ln x - the integral of (x^2/2)(1/x) dx (M1: formula applied).
Line 3: = (x^2/2) ln x - the integral of x/2 dx = (x^2/2) ln x - x^2/4 + C (A1).
Find the integral of x cos x dx.
Line 1: u = x (algebraic), dv/dx = cos x, so du/dx = 1 and v = sin x (M1).
Line 2: Integral = x sin x - the integral of sin x dx (M1).
Line 3: = x sin x + cos x + C (A1: sign on the sine integral).
Some integrals need by parts twice: the integral of x^2 e^x dx = x^2 e^x - the integral of 2x e^x dx, and the remaining integral needs by parts again, giving x^2 e^x - 2x e^x + 2 e^x + C = e^x(x^2 - 2x + 2) + C. The classic cyclic case is the integral of e^x sin x dx, where applying by parts twice returns the original integral, which is then solved algebraically. Keep the same u-type on the second application, or the algebra cycles without progress.
The definite integral measures signed area, so area questions need care where the curve crosses the axis (Edexcel 9MA0 8.3; AQA 7357 H4; OCR A H640 12.01/17.01). Find where f(x) = 0, split the integral at those points, and add the absolute values. The area between two curves is the integral of (top - bottom) dx over the interval between their intersections - find the intersections first, and check which curve is on top in that interval.
The region R is bounded by y = 4 - x^2 and the x-axis. Find the exact area of R.
Line 1: Intersections: 4 - x^2 = 0, so x = -2 and x = 2 (M1: limits found).
Line 2: The curve is above the axis throughout (-2, 2), so area = the integral from -2 to 2 of (4 - x^2) dx (M1).
Line 3: Antiderivative: 4x - x^3/3; at 2: 8 - 8/3 = 16/3; at -2: -8 + 8/3 = -16/3 (A1).
Line 4: Area = 16/3 - (-16/3) = 32/3 (A1: exact).
The curve y = sqrt(x) from x = 0 to x = 4 is rotated a full turn about the x-axis. Find the exact volume formed.
Line 1: V = (pi) times the integral from 0 to 4 of y^2 dx = (pi) times the integral from 0 to 4 of x dx (M1: formula, y^2 = x).
Line 2: = (pi)[x^2/2] from 0 to 4 = (pi)(8) = 8(pi) (A1: exact volume, units cubed).
For rotation about the y-axis, use V = (pi) times the integral of x^2 dy with the y-limits, rewriting the curve as x = f(y) first. Volumes always carry the (pi) factor and units cubed - dropping the (pi) or integrating y instead of y^2 are the two standard errors. For a region between two curves rotated about the axis, the washer method applies: V = (pi) times the integral of (R^2 - r^2) dx, where R is the outer radius (the top curve) and r the inner radius (the bottom curve) - integrate the difference of the squares, not the square of the difference.
A differential equation involving dy/dx is solved by separation of variables (Edexcel 9MA0 8.6; AQA 7357 H5; OCR A H640 17.01): move every y term to the dy side and every x term to the dx side, integrate both sides, and use an initial condition to find C. A general solution keeps C arbitrary; a particular solution fixes it from the condition.
Solve dy/dx = 2xy, given that y = 3 when x = 0.
Line 1: Separate: (1/y) dy = 2x dx (M1: variables separated).
Line 2: Integrate: ln|y| = x^2 + C (A1: both sides integrated).
Line 3: Exponentiate: y = A e^(x^2), absorbing the constant (M1: general solution).
Line 4: Initial condition: y(0) = 3 gives A = 3, so y = 3e^(x^2) (A1: particular solution).
Exam technique across the topic: (1) write the antiderivative line explicitly before substituting limits - the method mark lives there; (2) change limits when substituting in a definite integral, or back-substitute consistently - mixing the two produces silent errors; (3) check every indefinite integral by differentiating the answer (the -cos x check); (4) in area questions, split at axis crossings and add absolute values; (5) in volume questions, keep the (pi) and integrate y^2, not y; (6) in differential equations, separate cleanly, keep ln|y| with the modulus, and state the particular solution in the form asked. Board coverage: Edexcel 9MA0 Topic 8, AQA 7357 H1-H5, OCR A H640 12.01 and 17.01, WJEC/Eduqas AS/A2 Units, CCEA AS1/A21 - identical techniques, differing wording.