MathematicsCore22 min read

Integration

Adding up infinitely many infinitely thin slices

This topic appears in:

2nd Year Mathematics · ch.3A Level Mathematics (9709) · ch.1A Level Mathematics (9709) · ch.2A Level Mathematics (9709) · ch.3
01

Two questions, one answer

Integration answers two questions that look unrelated. "What function has this derivative?" — the indefinite integral. And "What is the area under this curve?" — the definite integral.

That these turn out to be the same operation is the great surprise of calculus, and it has a name: the Fundamental Theorem. Until you meet it, treat them as two skills that happen to share notation.

∫ f(x) dx = F(x) + Cwhere F′(x) = f(x)the +C is not optional — differentiating kills constants, so integrating must restore an unknown one
02

Area as a sum of rectangles

To find the area under a curve, slice it into thin vertical rectangles, add their areas, then make the slices thinner and thinner. The sum converges on the true area, and that limit is the definite integral.

The notation is a picture of the process: ∫ is an elongated S for "sum", f(x) is the height of each rectangle, and dx is its vanishing width.

Start at n = 4 and watch the rectangles overshoot and undershoot badly. Drag n upward and watch the error readout collapse. The exact area is what the sum approaches as n runs to infinity — you can see the convergence happen.

03

The rules worth knowing by heart

Almost every integral you meet at school level is one of these, or becomes one after a substitution. Notice that the power rule has an exception at n = −1, because dividing by n + 1 would mean dividing by zero — that case gives a logarithm instead.

∫ f(x) dx where f(x) =gives
xⁿ (n ≠ −1)xⁿ⁺¹ / (n + 1) + C
1/xln|x| + C
eˣeˣ + C
sin x−cos x + C
cos xsin x + C
sec² xtan x + C

The two techniques that unlock the rest

Substitution reverses the chain rule: spot an inner function whose derivative also appears, and let u be that inner function. Integration by parts reverses the product rule: ∫u dv = uv − ∫v du, and the art is choosing which factor to call u. LIATE — logs, inverse trig, algebraic, trig, exponential — is a reliable ordering for that choice.

04

The Fundamental Theorem ties it together

To find the area from a to b, you do not need rectangles at all. Find any antiderivative F, evaluate it at both ends, and subtract. That is it.

This is why integration is worth learning as an algebraic skill rather than a numerical one: a problem that would need thousands of rectangles to approximate is solved exactly in two lines.

∫ₐᵇ f(x) dx = F(b) − F(a)the constant C cancels in the subtraction, which is why definite integrals never carry one
05

Integration by substitution

The chain rule run backwards. When an integrand contains a function and something close to its derivative, substituting for the inner function turns an impossible integral into a standard one.

The routine: choose u to be the awkward inner expression, differentiate to get du = u′ dx, replace every x — including the dx — and integrate in u. For a definite integral, change the limits to u values as well and there is no need to substitute back.

∫ f(g(x)) g′(x) dx = ∫ f(u) duwhere u = g(x)∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C(n ≠ −1)∫ 1/x dx = ln|x| + C∫ eˣ dx = eˣ + C∫ sin x dx = −cos x + C∫ cos x dx = sin x + Cthe n ≠ −1 exclusion is why 1/x needs its own rule — the power rule would divide by zero
Worked example

Evaluate ∫ 2x(x² + 1)⁵ dx.

  1. Let u = x² + 1, the expression inside the bracket.Choose the inner function of the composition — it is nearly always the right choice.
  2. du/dx = 2x, so du = 2x dx.The 2x already sitting in the integrand is exactly what is needed, which is why this substitution works.
  3. The integral becomes ∫ u⁵ du = u⁶/6 + C.Every trace of x must be gone before integrating, including the dx.
  4. Substitute back: (x² + 1)⁶/6 + C.Check by differentiating: 6(x²+1)⁵(2x)/6 = 2x(x²+1)⁵ ✓

(x² + 1)⁶ / 6 + C

06

Definite integrals and the area between curves

A definite integral has limits and produces a number rather than a family of functions, so the constant of integration cancels and is left out. Geometrically it is a signed area: regions below the x-axis contribute negatively.

That signing matters. A question asking for the total area enclosed by a curve that crosses the axis must be split at the crossing points and the pieces added as positive quantities — otherwise the parts cancel and the answer comes out too small, or even zero.

∫ₐᵇ f(x) dx = F(b) − F(a)where F′ = farea between two curves = ∫ₐᵇ [upper − lower] dx∫ₐᵇ f(x) dx = −∫ᵦᵃ f(x) dxfor the area between curves, find the intersections first — they are the limits
Worked example

Find the area enclosed between y = x² and y = 2x.

  1. Find the intersections: x² = 2x gives x(x − 2) = 0, so x = 0 and x = 2.The intersections are the limits of integration; without them there is nothing to integrate between.
  2. Between 0 and 2, test x = 1: the line gives 2 and the curve gives 1, so the line is the upper curve.Deciding which is upper by testing one interior point is faster and safer than sketching.
  3. Area = ∫₀² (2x − x²) dx = [x² − x³/3]₀².Upper minus lower, always in that order.
  4. = (4 − 8/3) − 0 = 4/3.A positive answer, as an area must be. A negative result means the curves were subtracted the wrong way round.

4/3 square units

Split the integral where the curve crosses the axis

∫₋₁¹ x³ dx = 0, but the region between the curve and the axis is certainly not empty. The part below the axis has cancelled the part above. For a genuine area, integrate each side of the crossing separately and add the absolute values — here 2 × ¼ = ½.

07

The standard integrals beyond powers of x

Integration is reverse differentiation, so every derivative learned becomes an integral read backwards. The only genuinely new feature is the linear inside: integrating f(ax + b) divides by a, because differentiating would have multiplied by it.

∫ e^(ax+b) dx = (1/a)e^(ax+b) + c∫ 1/x dx= ln|x| + c∫ 1/(ax+b) dx = (1/a)ln|ax + b| + c∫ sin(ax+b) dx = −(1/a)cos(ax+b) + c∫ cos(ax+b) dx = (1/a)sin(ax+b) + c∫ sec²(ax+b) dx = (1/a)tan(ax+b) + ca linear inside costs a division by a; anything more complicated needs substitution
1/a
the correction factoronly valid when the inside is linear
|x|
the modulus in the logarithmneeded because ln of a negative does not exist
+ c
the arbitrary constantomitted only for definite integrals

The 1/a trick works for linear insides only

∫ cos(3x) dx = (1/3)sin(3x) + c is correct. ∫ cos(x²) dx is not (1/2x)sin(x²) — that reasoning fails completely once the inside is not linear, and no elementary answer exists for that particular integral at all. Check that the inside is of the form ax + b before dividing.

08

Substitution, and integration by parts

Two techniques cover the rest of the syllabus. Substitution reverses the chain rule and is signalled by a function appearing alongside its own derivative. Integration by parts reverses the product rule and is signalled by a product of two unrelated functions.

One important special case is worth recognising instantly: an integrand of the form f′(x)/f(x) integrates to ln|f(x)|, because the numerator is exactly the derivative of the denominator. Spotting this saves a full substitution.

substitution: ∫ f(g(x))g′(x) dx = ∫ f(u) du, u = g(x)special case: ∫ f′(x)/f(x) dx = ln|f(x)| + cby parts:∫ u(dv/dx) dx = uv − ∫ v(du/dx) dxby parts trades one integral for another — choose u so the new one is easier
u
the part you differentiatepick the one that simplifies when differentiated
dv/dx
the part you integratemust be something you can actually integrate
du
the substitution differentialchange the limits too, for a definite integral
Worked example

Find ∫ x·ln x dx.

  1. This is a product of unrelated functions, so use by parts. Let u = ln x and dv/dx = x.ln x is chosen as u because differentiating it gives the much simpler 1/x, while integrating it would be harder.
  2. Then du/dx = 1/x and v = x²/2.Both pieces are needed before applying the formula.
  3. ∫ x ln x dx = (x²/2)ln x − ∫ (x²/2)(1/x) dx.Direct substitution into uv − ∫v(du/dx)dx.
  4. = (x²/2)ln x − ∫ (x/2) dx = (x²/2)ln x − x²/4 + c.The new integral is elementary, which confirms u was chosen correctly.

(x²/2)ln x − x²/4 + c

Choosing the method

  1. A function and its derivative both present → substitution.
  2. Numerator is the derivative of the denominator → ln|denominator| immediately.
  3. Product of unrelated functions → by parts, differentiating the one that simplifies.
  4. A rational function with a factorisable denominator → partial fractions first.
  5. An even power of sine or cosine → use a double-angle formula to lower the power first.
  6. For definite integrals by substitution, change the limits to the new variable rather than substituting back.

Practice questions

6 questions · 26 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 7 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Find ∫ (6x² − 4x + 3) dx.
Model answer

2x³ − 2x² + 3x + c

Examiner tip. The constant of integration is a mark on an indefinite integral. Losing it is the cheapest mark in the paper to throw away.

SQ2[3 marks]
Evaluate ∫₁³ (2x + 1) dx.
Model answer

[x² + x]₁³ = (9 + 3) − (1 + 1) = 10

Examiner tip. Square brackets, upper limit minus lower limit. No constant is needed on a definite integral because it cancels.

SQ3[2 marks]
Explain what a definite integral represents geometrically.
Model answer

The area between the curve and the x axis, between the two limits. Area below the axis counts as negative.

Examiner tip. The sign convention is the second mark, and it is the one that makes area questions work.

Long questions

2 · 13 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
The curve y = x² − 4x crosses the x axis at two points.
  1. Find the two points of intersection. [2]
  2. Find the area enclosed between the curve and the x axis. [4]
Mark scheme
  1. Sets y = 0: x(x − 4) = 0[1]
  2. x = 0 and x = 4[1]
  3. Integrates: ∫(x² − 4x) dx = x³/3 − 2x²[1]
  4. Evaluates: (64/3 − 32) − 0 = −32/3[1]
  5. Recognises the curve is below the axis on this interval[1]
  6. Area = 32/3 ≈ 10.7 square unitsthe magnitude is taken[1]

(a) x = 0 and x = 4 (b) 32/3 square units

Examiner tip. A negative integral is not a wrong answer — it tells you the region lies below the axis. Take the magnitude for area, and say why.

LQ2[7 marks]
A curve passes through the point (2, 9) and has gradient dy/dx = 3x² − 2x.
  1. Find the equation of the curve. [4]
  2. Find the area under the curve between x = 0 and x = 2. [3]
Mark scheme
  1. Integrates: y = x³ − x² + c[1]
  2. Substitutes the point: 9 = 8 − 4 + c[1]
  3. c = 5[1]
  4. y = x³ − x² + 5[1]
  5. Integrates again: ∫₀² (x³ − x² + 5) dx = [x⁴/4 − x³/3 + 5x]₀²[1]
  6. = 4 − 8/3 + 10[1]
  7. = 34/3 ≈ 11.3 square units[1]

(a) y = x³ − x² + 5 (b) 34/3 square units

Examiner tip. Part (a) is what the constant of integration is for. Being given a point on the curve is the examiner telling you to find c.

Exam questions

1 · 6 marks

Multi-part questions with a full mark scheme.

Q1[6 marks]
The region bounded by y = x + 2 and y = x² is to be found.
  1. Find the x coordinates of the points of intersection. [2]
  2. Write down the integral that gives the enclosed area. [2]
  3. Evaluate it. [2]
Mark scheme
  1. Sets x + 2 = x² → x² − x − 2 = 0[1]
  2. (x − 2)(x + 1) = 0, so x = −1 and x = 2[1]
  3. Recognises the line is above the curve between them[1]
  4. Area = ∫₋₁² [(x + 2) − x²] dx[1]
  5. = [x²/2 + 2x − x³/3]₋₁²[1]
  6. = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 9/2[1]

x = −1 and 2; area = 9/2 square units

Examiner tip. Always upper curve minus lower curve. Test with a value between the limits — at x = 0 the line gives 2 and the parabola gives 0, so the line is on top.