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Integration

The second half of calculus: reverse differentiation, areas under curves, and the applications that close out the chain.

Topic Importance

Integration is the second half of calculus. It has appeared in 26 questions over the last 12 years — and like differentiation, it is always there. As we saw in the Paper One overview, students tend to find it a bit trickier, which makes it well worth focusing on.

Integration appeared in 26 questions over the last 12 years, usually sitting inside a differentiation question

It rarely stands alone: integration is usually joined with differentiation in some way. Sometimes there is a separate question that is mostly integration, but often it is mixed into a differentiation question — so if you plan to take on a differentiation question and aim for top marks, you will most likely need integration to finish it.

What the Topic Covers

The topic starts with the concept: integration as the reverse of differentiation. One idea deserves emphasis from the very beginning — the constant of integration. When you differentiate, any constant disappears, so functions that differ only by a constant share the same derivative. Reversing that, the integral cannot know which constant you started with — that is what the \(+\,c\) is for.

Integration and differentiation are inverses: differentiating the integral of f(x) gives back f(x), and integrating f prime of x gives back f(x)

From there come the rules — factoring out constants, sums and differences — which combine into polynomial integration, mirroring the differentiation case. Then the expression types broaden to trig functions and exponentials and logarithms, each reversing the derivatives you already know.

That covers the theory; then come the applications. The definite integral finds the area under a curve between two values — and with start and end values in play, the \(c\)s cancel out.

A curve with the region between x equals a and x equals b shaded: the area under the curve equals the definite integral of f from a to b

The area idea then extends in three directions: the average value of a function, the area a curve encloses beside the y-axis (found by integrating the inverse function), and areas in different quadrants, where absolute values are needed. Finally, kinematics runs the differentiation applications in reverse.

Example question

Find the total area enclosed between \(y = -x\) and the x-axis from \(x = 0\) to \(x = 4\).

Exam Correlations

Integration appears 26 times over 12 years, and its exam life is dominated by one partner: differentiation, with 18 shared appearances. Algebra (11) and functions (7) follow.

Questions shared with integration over 12 years: differentiation 18, algebra 11, functions 7, exponentials and logs 4, trigonometry 3, and smaller counts for the remaining topics

That differentiation number is the story of the topic. The two halves of calculus are usually examined together — a part (a) that differentiates, a part (b) or (c) that integrates, or a check of one using the other. The usual caveat about co-occurrence applies, but here the mathematical link genuinely is deep: they are reverse operations.

As noted in the differentiation articles: for top marks on a differentiation question, you will most likely need some integration to finish it — and vice versa.

Concept Connections

The lead-in to integration is Differentiation 3 — and that is mostly it. The other feeders you might expect (polynomials, trigonometry, exponentials and logs) all lead into differentiation first, so by the time you arrive here they are assumed.

Comes from The idea it gives you Where you will use it in Integration

Differentiation 3 \(\dfrac{d}{dx}\, f(x)\) Every differentiation rule, reversed — integration undoes differentiation, so each rule you mastered becomes an integration fact.

Functions \(f^{-1}(x)\) Inverse functions — finding the area a curve encloses beside the y-axis means integrating the inverse of the function.

And outward? Nothing. Integration is the final topic of calculus — no other topic depends on it. Once it is done, the whole calculus chain, from algebra through differentiation, is complete.

Study Order

There are ten sub-topics in Integration: five of theory, then five of applications that stack on top. Click each step to see how they build.

Integration as Reverse Differentiation

Start with the concept: the integral undoes the derivative. And because differentiating a constant gives zero, functions differing only by a constant have the same derivative — so every indefinite integral carries a \(+\,c\) to stand for the constant you cannot recover.

\(\displaystyle\int f'(x)\, dx = f(x) + c\)

Constant and Sum Rules

The combining rules, mirroring differentiation: constants factor out, and a sum or difference can be integrated term by term.

\(\displaystyle\int n\, f(x)\, dx = n \int f(x)\, dx\)

\(\displaystyle\int \left(f(x) + g(x)\right) dx = \int f(x)\, dx + \int g(x)\, dx\)

Polynomial Integration

Combine the rules and you can integrate any polynomial, just as combining the differentiation rules unlocked polynomial differentiation.

\(\displaystyle\int \left(8x + 3\right) dx = 4x^2 + 3x + c\)

Trigonometric Functions

Reverse the trig derivatives you learned in Differentiation 1.

\(\displaystyle\int \cos x \, dx = \sin x + c\)

Exponentials and Logarithms

The same again for exponential and log forms — the last of the theory steps.

\(\displaystyle\int e^x \, dx = e^x + c\)

Areas Under Curves

The first application: integrating between start and end values gives the area under the curve. With limits in play, the \(c\)s cancel out — which is why definite integrals do not carry one.

\(\text{Area} = \displaystyle\int_a^b f(x)\, dx\)

Average Value of a Function

Builds directly on areas: divide the area by the width of the interval and you get the average height — which is the average value of the function.

\(\text{Average} = \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\, dx\)

Areas Beside the y-axis

Instead of the area above the x-axis, the area the curve encloses with the y-axis. You find it by taking the inverse of the function — from the Functions topic — and integrating that.

\(\displaystyle\int f^{-1}(y)\, dy\)

Areas in Different Quadrants

So far the picture has been the upper-right quadrant. Below the x-axis, the integral comes out negative — integrate \(y = -x\) from \(0\) to \(4\) and you get \(-8\), a nonsense value for an area. Where a function cuts the x-axis, split the integral and sum the absolute values of each piece.

\(\displaystyle\int_{0}^{4} (-x)\, dx = -8 \;\Rightarrow\; \text{Area} = \left|-8\right| = 8\)

Kinematics

The reverse of the differentiation applications: integrate acceleration to get velocity, and velocity to get displacement.

\(v = \displaystyle\int a\, dt, \quad s = \displaystyle\int v\, dt\)

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