Integration and area under a curve
Indefinite and definite integration, and its use for areas.
Review these first
Learning objectives
What you need to be able to do
Teacher-mapped phrasing — check against the official Cambridge syllabus for exact wording.
- 4.1.1Integrate powers of x and evaluate definite integrals to find areas.
6 minute read
Integration and area under a curve
Integration is the reverse of differentiation.
The basic rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + c (for n ≠ −1): increase the power by one, then divide by the new power. It is exactly differentiation run backwards.
∫x³ dx = x⁴/4 + c∫6x dx = 3x² + c∫5 dx = 5x + c
The constant of integration
The + c is essential for indefinite integrals. Differentiating x² + 7 and x² − 3 both give 2x, so reversing cannot recover which constant was there. Omitting c is a guaranteed lost mark. If extra information is given — a point the curve passes through — substitute it to find c.
Definite integration
A definite integral has limits and produces a number, so no + c is needed (it cancels): ∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a) Always top limit minus bottom limit, in that order.
Area under a curve
The definite integral between a and b gives the area between the curve and the x-axis. Two cautions:
- Area below the x-axis comes out negative. If a question asks for total area and the curve crosses the axis, split the integral at the crossing point and add the magnitudes.
- To find the area between two curves, integrate the difference:
∫(upper − lower) dx.
Reversing kinematics
Since differentiating displacement gives velocity and differentiating velocity gives acceleration, integration goes the other way: integrate acceleration for velocity, and velocity for displacement — using initial conditions to find each constant.
Think of it like this
Integration is reassembling something you took apart: differentiation discarded the constant term, so putting it back requires either extra information or an honest "+ c" admitting you cannot know it.
Worked examples
Method, step by step
Evaluate the definite integral of 3x² from x = 1 to x = 3.
- 1Integrate: ∫3x² dx = x³.
- 2Apply the limits: [x³]₁³ = 3³ − 1³.
- 3= 27 − 1 = 26.
26
A curve has dy/dx = 4x + 3 and passes through (1, 6). Find the equation of the curve.
- 1Integrate: y = 2x² + 3x + c.
- 2Substitute the point (1, 6): 6 = 2(1)² + 3(1) + c.
- 36 = 5 + c, so c = 1.
y = 2x² + 3x + 1
Common misconceptions
- Omitting + c from an indefinite integral, which is an automatic lost mark.
- Reporting a negative area. A negative definite integral means the region lies below the x-axis; area is its magnitude.
- Subtracting the limits the wrong way round. It is always F(b) − F(a), top minus bottom.
In the exam
- If a question gives "the curve passes through (1, 5)", it expects you to find c — that is why the point was supplied.
- Sketch the curve before computing an area. It reveals whether the region dips below the axis and needs splitting.