Integration is the second half of calculus, and it is one of the most 'profitable' topics on the A-level Pure papers — technical, predictable, and full of 'easy' marks that students lose over a minus sign or a forgotten constant. In this guide we build the topic from the ground up: what an antiderivative is, how to apply the power rule for integration, the difference between an indefinite and a definite integral, and how to calculate the area under a curve and between two curves — including the mistakes that lose the most marks and a practice plan that will take you to full command.
If differentiation is the first half of calculus, integration is the second — and, to many people's surprise, it is the most 'profitable' topic on the A-level Pure papers. The reason is simple: while optimisation problems demand verbal analysis and thought, calculating an integral and an area is technical and predictable. Anyone who knows the power rule for integration, understands the formula F(b) − F(a), and watches the sign — collects those marks almost automatically. And that is exactly where the tragedy lies: thousands of students lose marks every year on the integration question not because they did not understand the idea, but because they forgot the constant C, got confused by the sign of an area below the x-axis, or substituted the limits the wrong way round. This guide is built so that it does not happen to you: we start with what an antiderivative is, go through the power rule and the definite integral, learn to calculate the area under a curve and between two curves, map all the common mistakes, and finish with a focused practice plan.
What is an integral? — the antiderivative
Integration is the reverse operation to differentiation. In differentiation you start from f(x) and calculate the derivative f'(x). In integration you go backwards: you are given the rate of change, and you look for the original function whose derivative it is. The function you find is called the antiderivative (or 'the reverse derivative').
Example: given f(x) = 2x. Which function do you differentiate to get 2x? The answer is x², because the derivative of x² is exactly 2x. So x² is an antiderivative of 2x. But it is not the only one: x² + 5, x² − 3, and x² + 100 — all of them differentiate to 2x, because the derivative of any constant is zero. So the general antiderivative is written with a constant of integration: x² + C.
The notation for an indefinite integral is ∫f(x) dx = F(x) + C. The sign ∫ is a 'stretched S' (from the word sum), the dx says 'with respect to the variable x', and F(x) is some antiderivative. So ∫2x dx = x² + C. The C is an integral part of the answer to an indefinite integral — without it the answer is incomplete and you lose a mark in the exam.
The power rule for integration
The central rule to know by heart is the power rule for integration: ∫xⁿ dx = x^(n+1)/(n+1) + C. In words: raise the power by 1, and divide by the new power. This is exactly the reverse of the power rule for differentiation (there you reduce the power by 1 and multiply by the old power).
Direct examples: ∫x³ dx = x⁴/4 + C. ∫x⁵ dx = x⁶/6 + C. ∫x dx = x²/2 + C (because x is x¹). A special case: ∫1 dx = ∫x⁰ dx = x + C — the integral of the constant 1 is simply x. In general ∫k dx = kx + C for any constant k.
Exactly as in differentiation, the integral of a sum is the sum of the integrals, and a constant multiplier comes outside. That is ∫(af(x) + bg(x)) dx = a∫f(x) dx + b∫g(x) dx. You work term by term.
A combined example: ∫(3x² − 4x + 5) dx. Integrate each term: ∫3x² dx = 3·(x³/3) = x³. ∫(−4x) dx = −4·(x²/2) = −2x². ∫5 dx = 5x. Add them up and add a single constant for the whole expression: x³ − 2x² + 5x + C.
| Function f(x) | Integral ∫f(x) dx | Short explanation |
|---|---|---|
| k (constant) | kx + C | The integral of a constant = the constant times x |
| x | x²/2 + C | Power rule with n = 1 |
| x² | x³/3 + C | Raise the power, divide by 3 |
| xⁿ | x^(n+1)/(n+1) + C | The general power rule (n ≠ −1) |
| 3x² − 4x + 5 | x³ − 2x² + 5x + C | Term by term + one C |
The definite integral — the formula F(b) − F(a)
So far we have talked about the indefinite integral — its result is a family of functions (with C). The definite integral, by contrast, gives a single number. Its notation is ∫[a to b] f(x) dx, where a and b are the limits of integration (a the lower, b the upper).
The fundamental theorem of calculus states: ∫[a to b] f(x) dx = F(b) − F(a), where F is any antiderivative of f. That is: find an antiderivative, substitute the upper limit b into it, substitute the lower limit a, and subtract. The order matters — always upper minus lower.
Example: calculate ∫[1 to 3] x² dx. The antiderivative is F(x) = x³/3. Substitute: F(3) − F(1) = 3³/3 − 1³/3 = 27/3 − 1/3 = 26/3. Note — there is no need to write the constant C in a definite integral, because it appears twice (in F(b) and in F(a)) and cancels: (F(b) + C) − (F(a) + C) = F(b) − F(a).
Calculating the area under a curve and between curves
The geometric meaning of the definite integral is the area enclosed between the graph of f(x) and the x-axis, on the interval [a, b]. That is the essence of most integration questions in the exam: 'Calculate the area enclosed between the curve, the x-axis and the lines x = a and x = b'.
When the function is entirely positive on the interval (the curve is above the x-axis), the area is simply ∫[a to b] f(x) dx. Example: the area under y = x² between x = 1 and x = 3 is 26/3 square units, as we calculated.
But be careful with the sign: if the curve lies below the x-axis on part of the interval, the integral there will come out negative. 'Area' is always a positive quantity, so on an interval where the function is negative you take the absolute value of the integral (or flip the sign). If the function crosses the x-axis inside the interval, split the integral at the crossing point, calculate each part separately, and add the absolute values.
The area between two curves: to calculate the area enclosed between two curves f(x) and g(x), you use the formula ∫[a to b] (f(x) − g(x)) dx, where f is the upper (larger) function and g is the lower one on the interval. The limits a and b are the intersection points of the two curves, which you find by solving the equation f(x) = g(x).
A full example — the area between two curves: find the area enclosed between the parabola f(x) = 6 − x² and the line g(x) = x. Step 1 — intersection points: 6 − x² = x, that is x² + x − 6 = 0, giving (x + 3)(x − 2) = 0, so x = −3 and x = 2. These are the limits. Step 2 — which is on top? Substitute a point in between, say x = 0: f(0) = 6, g(0) = 0, so the parabola is on top. Step 3 — the integral: ∫[−3 to 2] ((6 − x²) − x) dx = ∫[−3 to 2] (6 − x − x²) dx. The antiderivative: 6x − x²/2 − x³/3. Substitution: at 2: 12 − 2 − 8/3 = 10 − 8/3 = 22/3. At −3: −18 − 9/2 + 9 = −9 − 9/2 = −27/2. The result: 22/3 − (−27/2) = 22/3 + 27/2 = 44/6 + 81/6 = 125/6 square units.
Common mistakes that lose marks
Mistake 1 — forgetting the constant C. In an indefinite integral, ∫f(x) dx must end in + C. This is the most common mistake, and it costs a mark almost every time. Rule of thumb: an integral without limits → must have C. An integral with limits (definite) → no need for C because it cancels.
Mistake 2 — confusing the differentiation rule with the integration rule. In differentiation you reduce the power; in integration you raise it. Tired students sometimes differentiate instead of integrating. A quick check: differentiate your answer — if you do not get back to the original function, you went wrong.
Mistake 3 — the sign of an area below the x-axis. If you calculated an integral and got a negative number when asked for an area — you must take the absolute value. You must not report a negative area. If the function crosses the axis, split into intervals.
Mistake 4 — substituting the limits the wrong way round. In the formula F(b) − F(a), b is the upper limit and a the lower. Reversing the substitution flips the sign of the whole answer. In the area between curves, make sure you have also identified correctly which is the upper function.
Mistake 5 — forgetting to solve f(x) = g(x) in the area between curves. Without the intersection points there are no limits of integration. Do not assume the limits from the graph without calculating them algebraically — the exam requires the calculation.
A focused practice plan
Integration is a technical topic, so repeated practice is the key. The following plan assumes about an hour a day and is built to take you from the basics to exam level in about two to three weeks.
- Days 1-2: the power rule for integration. Thirty indefinite integrals of polynomials, until adding C becomes a reflex. For every exercise differentiate back to check.
- Days 3-4: definite integrals. Twenty F(b) − F(a) exercises. Get used to writing the antiderivative in square brackets before substituting.
- Days 5-6: the area under a curve, including cases where the function dips below the x-axis. Practise absolute values and splitting into intervals.
- Days 7-8: the area between two curves. Ten complete exercises: finding intersection points, identifying the upper function, and calculating.
- Days 9-10: full past-paper questions under timed conditions. At this stage the remaining weaknesses will only be technical — sign, C, limits.
Summary
Integration is the topic where the investment pays off fast: it is technical, predictable, and full of 'safe' marks for anyone who practises. The four components you need to command are the antiderivative (with C), the power rule for integration, the definite integral F(b) − F(a), and calculating area — under a curve and between two curves. If you always remember to add C in an indefinite integral, to check the sign of an area, and to substitute the limits in the right order — the integration question in the exam will become one of your easy questions. The way there is through repeated practice.
Frequently asked questions
What is the difference between a definite and an indefinite integral?
An indefinite integral, ∫f(x) dx, gives a family of antiderivatives and its result is an expression with a constant C (for example x² + C). A definite integral, ∫[a to b] f(x) dx, gives a single number — it is calculated as F(b) − F(a), and in it the constant C cancels so there is no need to write it. The definite integral represents an area; the indefinite one represents a function.
Why do you have to add + C in an integral?
Because every function has infinitely many antiderivatives that differ from each other only by a constant. The derivative of x² + 5 and the derivative of x² − 100 are both 2x, because the derivative of a constant is zero. The C in x² + C represents that whole family. In an indefinite integral forgetting the C loses a mark. In a definite integral the C cancels automatically, so it is not required there.
How do you calculate an area when the curve is below the x-axis?
When the function is negative on the interval, the integral ∫[a to b] f(x) dx comes out as a negative number. Area is always positive, so you take the absolute value of the result. If the function crosses the x-axis inside the interval, split the integral at the crossing point, calculate each part separately, and add the absolute values of the parts.
How do you find the limits of integration for the area between two curves?
The limits of integration are the intersection points of the two curves. Solve the equation f(x) = g(x); the solutions are the x-values of the intersection points, and they are the limits a and b. Then check at a point in between which is the upper function, and calculate ∫[a to b] (upper − lower) dx.
What is the most common mistake in integration questions?
Forgetting the constant C in an indefinite integral, and right behind it — a sign error for an area below the x-axis and substituting the limits in the wrong order in F(b) − F(a). These are technical mistakes that lose marks even for students who understand the material. The way to prevent them is repeated practice and self-checking: differentiate the answer to the integral and make sure you get back to the original function.
A graded set of exercises — from antiderivatives to the area between two curves in exam style
Practise integration and finding areas ←