Integration — Edexcel A-Level Mathematics
Test yourself on Integration with PEARSON EDEXCEL A-Level practice questions.
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Integration explained
The Fundamental Theorem of Calculus links differentiation and integration as inverse processes.
Read the full explanation
If F is an antiderivative of f, so F′(x) = f(x), then the definite integral of f from a to b equals F(b) − F(a). This lets you evaluate a definite integral by finding any antiderivative, substituting the upper limit, substituting the lower limit and subtracting. For example, ∫ from 1 to 3 of 2x dx = [x²] from 1 to 3 = 3² − 1² = 9 − 1 = 8. The theorem also shows that differentiating a definite integral with a variable upper limit returns the integrand, so d/dx of ∫ from a to x of f(t) dt = f(x). Use it to connect rates of change with accumulated change.
8.2 Integrate xⁿ (excluding n = −1) and related sums, differences and constant multiples. Integrate e^(kx), 1/x, sin kx, cos kx and related sums, differences and constant multiples.
Standard integration reverses standard differentiation. For powers, ∫ xⁿ dx = x^(n+1)/(n+1) + c, valid for all n except n = −1; for example ∫ x³ dx = x⁴/4 + c. The exceptional case is ∫ 1/x dx = ln|x| + c, where the modulus keeps the logarithm defined for negative x. Exponential functions integrate to themselves with a scale factor: ∫ e^(kx) dx = (1/k)e^(kx) + c. Trigonometric functions give ∫ sin kx dx = −(1/k)cos kx + c and ∫ cos kx dx = (1/k)sin kx + c. Sums, differences and constant multiples are integrated term by term, so ∫ (3x² − 2 sin 2x + 5/x) dx = x³ + cos 2x + 5 ln|x| + c.
8.3 Evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves.
A definite integral is evaluated by finding an antiderivative and subtracting its value at the lower limit from its value at the upper limit. Geometrically, the definite integral of a positive function from a to b gives the area under the curve y = f(x) between those limits and the x-axis. If the curve dips below the axis, that part contributes a negative value, so split the integral at the roots and add the absolute areas. For the area between two curves, integrate the upper curve minus the lower curve over the interval between their intersection points; for example, between y = x and y = x² from x = 0 to x = 1, the area is ∫ from 0 to 1 of (x − x²) dx = [x²/2 − x³/3] from 0 to 1 = 1/2 − 1/3 = 1/6.
8.4 Understand and use integration as the limit of a sum.
Integration as the limit of a sum means the definite integral is defined by splitting the interval [a, b] into n strips of width Δx = (b − a)/n, forming the sum Σ f(xᵢ)Δx, and taking the limit as n → ∞ and Δx → 0. For example, for f(x) = x² on [0, 1], the right-endpoint sum is Σ (i/n)²(1/n) = (1/n³)Σ i² = (1/n³)·n(n + 1)(2n + 1)/6, which tends to 1/3, so ∫₀¹ x² dx = 1/3. This limit definition explains why integration gives signed area and why the definite integral is a number, while the indefinite integral is a family of functions. It also links to the fundamental theorem: differentiation and integration are inverse processes.
8.5 Carry out simple cases of integration by substitution and integration by parts; understand these methods as the inverse processes of the chain and product rules respectively (Integration by substitution includes finding a suitable substitution and is limited to cases where one substitution will lead to a function which can be integrated; integration by parts includes more than one application of the method but excludes reduction formulae).
Integration by substitution reverses the chain rule: choose u = g(x), so du = g′(x) dx, and rewrite ∫ f(g(x))g′(x) dx as ∫ f(u) du. For example, ∫ 2x cos(x²) dx uses u = x², du = 2x dx, giving ∫ cos u du = sin u + C = sin(x²) + C. Integration by parts reverses the product rule: ∫ u dv = uv − ∫ v du, often written ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx. For example, ∫ x eˣ dx with u = x and dv = eˣ dx gives x eˣ − ∫ eˣ dx = x eˣ − eˣ + C. Some integrals need more than one application, such as ∫ x² eˣ dx, but reduction formulae are excluded. Substitution is limited to cases where one substitution produces an integrable function.
8.6 Integrate using partial fractions that are linear in the denominator.
Integrating using partial fractions with linear denominators means decomposing a rational function into simpler fractions whose denominators are linear factors, then integrating each term. For example, (3x + 2)/((x + 1)(x + 2)) can be written as A/(x + 1) + B/(x + 2). Multiplying out gives 3x + 2 = A(x + 2) + B(x + 1). Setting x = −1 gives A = −1; setting x = −2 gives B = 4. So the integral is ∫ (−1/(x + 1) + 4/(x + 2)) dx = −ln|x + 1| + 4 ln|x + 2| + C. The method applies when the denominator factors into distinct linear factors; repeated or quadratic factors are not required by this statement. The absolute value is needed because ln is defined only for positive arguments.
8.7 Evaluate the analytical solution of simple first order differential equations with separable variables, including finding particular solutions (Separation of variables may require factorisation involving a common factor).
A first order differential equation with separable variables can be written as dy/dx = f(x)g(y). Separate the variables to get (1/g(y)) dy = f(x) dx, then integrate both sides. For example, dy/dx = xy² separates to y⁻² dy = x dx, giving −y⁻¹ = x²/2 + C. If y = 1 when x = 0, then −1 = 0 + C, so C = −1 and −y⁻¹ = x²/2 − 1, which rearranges to y = 2/(2 − x²). Some equations require factorisation first to reveal a common factor, such as dy/dx = xy + x. Taking out the common factor x gives dy/dx = x(y + 1), which separates to (1/(y + 1)) dy = x dx. The analytical solution is ln|y + 1| = x²/2 + C. A particular solution uses a given initial condition to find the constant C.
8.8 Interpret the solution of a differential equation in the context of solving a problem, including identifying limitations of the solution; includes links to kinematics.
A differential equation models how a quantity changes; its general solution contains an arbitrary constant, and an initial or boundary condition fixes that constant to give the particular solution. Interpreting means translating the algebraic solution back into the problem: state what each variable and constant represents, with units, and check the solution is valid only where the model holds. In kinematics, for example, integrating acceleration a = dv/dt gives velocity v = ∫a dt + C, and integrating again gives displacement; the constants follow from given values of v and s at t = 0. Limitations include restricted domains, unrealistic long-term behaviour, and assumptions such as constant acceleration or negligible resistance.
Your focus
- State the Fundamental Theorem of Calculus and explain the inverse relationship it describes.
- Evaluate definite integrals accurately using upper limit minus lower limit.
- Apply the theorem to problems involving variable upper limits or accumulated change.
Show all 24 objectives
- Integrate powers of x correctly, including the exception for n = −1.
- Integrate e^(kx), 1/x, sin kx and cos kx with correct coefficients and signs.
- Combine standard integrals to integrate sums, differences and constant multiples.
- Evaluate definite integrals accurately using limits and an antiderivative.
- Calculate the area under a curve, including cases where the curve crosses the x-axis.
- Determine the area between two curves using their intersection points as limits.
- Define a Riemann sum for a function on a closed interval and explain each part of the expression.
- Evaluate a simple definite integral by taking the limit of a sum as n → ∞.
- Explain the connection between the limit of a sum and the fundamental theorem of calculus.
- Perform integration by substitution for simple cases where one substitution leads to an integrable function.
- Apply integration by parts, including more than one application, to evaluate integrals.
- Explain how substitution and integration by parts reverse the chain rule and product rule respectively.
- Decompose a rational function with linear denominator factors into partial fractions.
- Integrate each resulting partial fraction term to obtain logarithmic expressions.
- Evaluate definite integrals of rational functions using partial fractions.
- Separate the variables in a first order differential equation and integrate both sides.
- Find a particular solution using a given initial condition.
- Factorise expressions to reveal a separable form where a common factor is present.
- Solve a first-order differential equation and apply a given condition to obtain the particular solution.
- Explain the meaning of the solution and its constants in the original problem context, including units.
- State limitations of the solution, such as restricted domains or unrealistic long-term behaviour.
Integration exam tips
Marking Points
- States that differentiation and integration are inverse operations, with F′(x) = f(x) when F is an antiderivative of f.
- Evaluates a definite integral as F(b) − F(a) after finding an antiderivative F.
- Substitutes the upper limit first and the lower limit second, then subtracts in the correct order.
- Applies the theorem to a variable upper limit, recognising that d/dx of ∫ from a to x of f(t) dt = f(x).
- Uses the theorem to interpret a definite integral as the accumulated change in an antiderivative over an interval.
- Handles constants of integration correctly by omitting them in definite evaluation because they cancel in the subtraction.
- Applies the power rule ∫ xⁿ dx = x^(n+1)/(n+1) + c correctly for positive, negative and fractional n except n = −1.
- Recognises the exceptional case and integrates 1/x as ln|x| + c rather than using the power rule.
- Integrates e^(kx) as (1/k)e^(kx) + c, including the reciprocal scale factor.
- Integrates sin kx as −(1/k)cos kx + c and cos kx as (1/k)sin kx + c with correct signs.
- Integrates sums, differences and constant multiples term by term, carrying the constant of integration c.
- Simplifies coefficients and exponents correctly, for example writing x^(5/2)/(5/2) as (2/5)x^(5/2).
- Evaluates a definite integral by substituting the upper and lower limits into an antiderivative and subtracting correctly.
- Identifies the correct limits from the context, including roots or intersection points found by solving equations.
- Finds the area under a curve above the x-axis by integrating the function between the given limits.
- Splits the integral at roots when the curve crosses the x-axis and adds the absolute values of each part.
- Finds the area between two curves by integrating upper curve minus lower curve between their intersection points.
- Interprets a negative integral value correctly as area below the axis rather than treating it as a positive area.
- Defines a partition of [a, b] into n subintervals of equal width Δx = (b − a)/n and identifies a sample point xᵢ in each subinterval.
- Forms the Riemann sum Σ f(xᵢ)Δx and explains that it approximates the signed area under y = f(x) between x = a and x = b.
- Takes the limit as n → ∞ (equivalently Δx → 0) and states that the definite integral is the value of this limit when it exists.
- Uses the limit definition to evaluate a simple integral, such as ∫₀¹ x² dx = 1/3, by summing squares and letting n → ∞.
- Interprets the definite integral as a signed area, distinguishing regions below the x-axis from those above.
- Connects the limit-of-a-sum definition to the fundamental theorem of calculus, showing that integration reverses differentiation.
- For substitution, selects a suitable u = g(x), computes du = g′(x) dx, and rewrites the integral entirely in terms of u and du.
- For substitution, integrates the transformed expression and substitutes back to the original variable, including the constant of integration for indefinite integrals.
- For definite integrals by substitution, changes the limits to u-values or substitutes back before evaluating at the original limits.
- For integration by parts, applies the formula ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx, choosing u and dv/dx appropriately.
- For integration by parts, applies the method more than once where necessary, for example for ∫ x² eˣ dx, without using reduction formulae.
- Explains that substitution reverses the chain rule and integration by parts reverses the product rule.
- Factorises the denominator into linear factors and sets up the correct partial fraction form with unknown constants.
- Determines the constants by substituting suitable values of x or by comparing coefficients.
- Integrates each term of the form A/(ax + b) as (A/a) ln|ax + b| + C, including the constant of integration for indefinite integrals.
- Handles definite integrals by evaluating the antiderivative at the limits after finding the partial fractions.
- Uses the absolute value inside the logarithm to keep the argument positive where necessary.
- Separates the variables correctly, writing the equation in the form (1/g(y)) dy = f(x) dx.
- Factorises expressions where necessary, including taking out a common factor, to expose the separable form.
- Integrates both sides with respect to their variables, including the constant of integration.
- Finds a particular solution by substituting the given initial condition and solving for the constant.
- Rearranges the general or particular solution into the required form, such as y as a function of x, where possible.
- Checks the solution by differentiating and substituting back into the original differential equation.
- Form the differential equation from the problem context, defining each variable and its units before solving.
- Separate variables or integrate directly, remembering the arbitrary constant in the general solution.
- Apply the given initial or boundary condition to determine the constant and obtain the particular solution.
- Interpret the particular solution in context, stating what the expression predicts for the modelled quantity.
- Identify limitations: domain restrictions, times or values for which the model is invalid, and unrealistic limiting behaviour.
- For kinematics, link displacement, velocity and acceleration by differentiating or integrating with respect to time and using initial conditions.
Examiner Tips
- 💡Write the antiderivative in square brackets with the limits attached before substituting.
- 💡Show the substitution of each limit on a separate line so the subtraction order is clear.
- 💡Check by differentiating your antiderivative mentally to confirm it returns the integrand.
- 💡Write each term on its own line before combining, so signs and coefficients are easy to check.
- 💡Differentiate your answer mentally to confirm it returns the original integrand.
- 💡Remember the modulus in ln|x| and the constant c in every indefinite integral.
- 💡Sketch the curve or region first to see where it crosses the axis and which curve is on top.
- 💡Write the integral with limits clearly before evaluating, and keep the subtraction order visible.
- 💡State area as a positive value and include units if the context supplies them.
- 💡When asked to explain integration as a limit, define Δx, write the sum, and state the limit in symbols before giving a numerical example.
- 💡Use a simple function such as f(x) = x² on [0, 1] to demonstrate the method; show the algebra for Σ i² and the limiting value.
- 💡If a question asks for the limit of a sum, check whether the interval is [a, b] and adjust Δx and the sample points accordingly.
- 💡Link the limit definition to the fundamental theorem when explaining why ∫ₐᵇ f(x) dx can be evaluated using an antiderivative.
- 💡For substitution, state u and du clearly, then show the integral fully in terms of u before integrating.
- 💡For integration by parts, write down u, dv/dx, du/dx and v in a small table to avoid sign and substitution errors.
- 💡If an integral requires more than one application of integration by parts, show each stage clearly and simplify between stages.
- 💡Check whether a substitution is suitable by confirming that the derivative of the chosen u appears in the integrand, up to a constant factor.
- 💡Always factorise the denominator completely before writing the partial fraction form.
- 💡Use substitution of convenient x-values to find constants quickly, but check by comparing coefficients if unsure.
- 💡For definite integrals, either integrate first and then substitute the limits, or change limits if using a substitution; do not mix the two.
- 💡Write the final answer with ln|ax + b| and include + C for indefinite integrals.
- 💡Write dy/dx as a fraction and move all y-terms to one side and all x-terms to the other before integrating.
- 💡If the equation does not look separable, try factorising the right-hand side to reveal a common factor.
- 💡Write down the meaning and units of every symbol before integrating, so the final interpretation is ready to state.
- 💡After finding the particular solution, substitute the initial condition back in to check the constant.
- 💡Finish with a sentence interpreting the result and naming at least one limitation of the model, since interpretation is explicitly required.
Common Mistakes
- Subtracting in the wrong order by doing F(a) − F(b); correct this by always taking upper limit value minus lower limit value.
- Including a constant of integration in a definite integral; correct this by omitting it, since it cancels.
- Confusing the variable of integration with the limit variable; correct this by using a dummy variable such as t inside the integral when the upper limit is x.
- Applying the power rule to 1/x and obtaining x⁰/0; correct this by using ln|x| + c instead.
- Forgetting the reciprocal factor when integrating e^(kx) or trigonometric functions; correct this by dividing by k.
- Omitting the constant of integration in an indefinite integral; correct this by adding c to every indefinite result.
- Subtracting lower limit value from upper limit value in the wrong order; correct this by always doing upper minus lower.
- Integrating across a root without splitting and reporting a net signed value as the total area; correct this by splitting at the root and adding absolute areas.
- Subtracting the curves in the wrong order and obtaining a negative area; correct this by identifying which curve is above and integrating upper minus lower.
- Treating Δx as a fixed finite width rather than a quantity that tends to zero; correct by stating Δx = (b − a)/n and taking the limit as n → ∞.
- Omitting the factor Δx when forming the sum; correct by writing each term as f(xᵢ)Δx before summing.
- Assuming every function has a limit of sums; correct by noting the definition applies when the limit exists, for example for continuous functions on a closed interval.
- Confusing the definite integral with the indefinite integral; correct by stating that the limit of a sum produces a number, whereas the indefinite integral produces a family of functions.
- Forgetting to replace dx with du in a substitution; correct by computing du = g′(x) dx and substituting both the integrand and the differential.
- Choosing u and dv/dx poorly in integration by parts, leading to a more complicated integral; correct by choosing u to simplify on differentiation and dv/dx to be integrable.
- Omitting the constant of integration for indefinite integrals; correct by adding + C after integrating.
- For definite integrals by substitution, failing to change the limits; correct by converting the limits to u-values or by substituting back before evaluating.
- Attempting a reduction formula when the specification excludes it; correct by applying integration by parts repeatedly for the required number of steps.
- Omitting the absolute value in ln|ax + b|; correct by writing ln|ax + b| because the logarithm requires a positive argument.
- Forgetting to divide by the coefficient of x when integrating A/(ax + b); correct by writing (A/a) ln|ax + b| + C.
- Setting up an incorrect partial fraction form for repeated linear factors; correct by using separate terms for each power, such as A/(x + 1) + B/(x + 1)², when the denominator has a repeated factor.
- Making sign errors when solving for the constants; correct by substituting values carefully and checking by expanding.
- Forgetting the constant of integration when integrating both sides; correct by adding + C immediately after integrating.
- Dividing by g(y) without considering g(y) = 0; correct by noting any constant solutions separately if required.
- Failing to factorise before separating; correct by taking out the common factor, for example writing xy + x as x(y + 1) rather than treating it as non-separable.
- Substituting the initial condition before integrating; correct by integrating first, then using the condition to find C.
- Omitting the constant of integration, which loses the family of solutions; the correction is to add the arbitrary constant immediately after integrating and evaluate it from the condition.
- Treating the constant as always zero because the problem starts at the origin; the correction is to substitute the actual initial values of the modelled quantity, which need not be zero.
- Quoting a solution outside its valid domain, such as a negative time or a population exceeding a carrying capacity; the correction is to state the interval on which the model applies and reject values outside it.