Integration

— Pure Mathematics 1 · AS & A Level (9709) · Chapter 9 —

★ Exam Tip: Integration is the 2nd big tool of calculus. In Paper 1 it appears as find a curve from its gradient · area under / between curves · volume of revolution — often all in one question, mixed with tangents & normals. Golden habit: sketch → set up the integral → evaluate → check sign/units.

1. Integration = Reverse of Differentiation

Idea: differentiation takes y → dy/dx. Integration goes back: dy/dx → y. But many functions share one derivative: x³ − 7, x³ + 4, x³ + ½ all give 3x². So every answer carries an arbitrary constant + c.

same gradient 3x², curves only shifted up/down ⇒ the “+ c”

Fun fact: Newton & Leibniz (17th c.): finding area under a graph ≡ reversing differentiation (Fundamental Theorem of Calculus). So we just say “integrate” for both.

Notation: ∫x³ dx = ¼x⁴ + c is the indefinite integral of x³ w.r.t. x — “indefinite” because it has infinitely many answers (one per c).

Key rule — powers∫ xⁿ dx = 1n + 1 xⁿ⁺¹ + c (n ≠ −1)

In words: raise the power by 1 → divide by the NEW power → add c.

Function form: if f′(x) = xⁿ then f(x) = 1n+1xⁿ⁺¹ + c. (n = −1 would mean ÷ 0 — that case is not met in this chapter.)

Constant multiples, sums & differences∫ k f(x) dx = k ∫ f(x) dx (k constant)
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx

Index form: √x = x^½ 1/xⁿ = x⁻ⁿ ³√x = x^⅓
Convert FIRST, integrate after.

Worked example (find y):

dy/dx = x³ , 1/x² , x√x — rewrite as powers, then apply the rule.

a) dy/dx = x³y = 13+1 x³⁺¹ + cy = ¼x⁴ + c

b) dy/dx = 1/x²= x⁻²y = 1−2+1 x⁻¹ + cy = −1/x + c

c) x√x = x · x^½ = x^3/2y = 13/2 + 1 x^5/2 + c = 15/2 x^5/2 + cy = ⅖ x^5/2 + c

Tidy-up first: split fractions, expand brackets

f′(x) = 4x³ − 2/x² + 4 = 4x³ − 2x⁻² + 4x⁰f(x) = x⁴ − 2x⁻¹−1 + 4x + cf(x) = x⁴ + 2/x + 4x + c

f′(x) = 8x² − 12x⁴ + 2x = 8x² − ½x⁻⁴ + 2xf(x) = 83x³ − ½x⁻³−3 + x² + cf(x) = 8/3 x³ + 1/(6x³) + x² + c

f′(x) = (x + 3)(x − 1)√x = x² + 2x − 3x^½= x^3/2 + 2x^½ − 3x^−½f(x) = ⅖x^5/2 + 2·⅔x^3/2 − 3·2x^½ + cf(x) = ⅖x^5/2 + ⁴⁄₃x^3/2 − 6√x + c

Warning: There is NO product/quotient rule for integrals. Expand brackets, split fractions, turn roots into powers — then integrate term by term.
e.g. ∫x(2x−1)(2x+3)dx → expand to 4x³+4x²−3x → x⁴ + ⁴⁄₃x³ − 3⁄₂x² + c. And ∫(4x² − 3√x)/x dx → ∫(4x − 3x^−½)dx = 2x² − 6√x + c.

routine for every “find y / find f(x)”

2. Finding the Constant of Integration

Use: if you know the gradient function and one point on the curve, you can find the exact curve.

a known point on the curve picks out ONE value of c

Integrate (don’t forget + c).
Substitute the known point (x, y) into your answer.
Solve for c, then write the full equation.

If k is unknown too: use the extra clue (gradient of tangent/normal, a stationary point…) to find k first, then c.

Ex 1gradient + point

dy/dx = 6x⁵ − 18x³ , curve passes through (1, 6).

dy/dx = 6x² − 18x⁻³y = 2x³ + 9x⁻² + cx = 1, y = 6 : 6 = 2 + 9 + cc = −5

y = 2x³ + 9/x² − 5

Ex 2function notation

f′(x) = 15x⁴ − 6x and f(−1) = 1. Find f(x).

f(x) = 3x⁵ − 3x² + c1 = 3(−1)⁵ − 3(−1)² + c1 = −3 − 3 + cc = 7

f(x) = 3x⁵ − 3x² + 7

Ex 3unknown k + normal

dy/dx = 6x + k. Normal at (1, −3) has gradient ½. Find the curve.

y = 3x² + kx + cpoint (1, −3): −3 = 3 + k + c ⇒ c + k = −6 …(1)normal ½ ⇒ tangent gradient = −2 (perpendicular: m₁m₂ = −1)dy/dx at x=1 : 6 + k = −2 ⇒ k = −8(1): c = −6 + 8 = 2

y = 3x² − 8x + 2

Second derivative given? Integrate twice: d²y/dx² → dy/dx + c₁ (use gradient / stationary-point info: dy/dx = 0) → y + c₂ (use a point on the curve). Two integrations ⇒ two constants, so you need two pieces of information.

3. Integrating (ax + b)ⁿ

Key rule — powers of a LINEAR bracket∫ (ax + b)ⁿ dx = 1a(n + 1) (ax + b)ⁿ⁺¹ + c (n ≠ −1, a ≠ 0)

Why ÷ a: Differentiating (ax+b)ⁿ⁺¹ produces an extra factor a (chain rule). So when integrating we divide by a to cancel it.

Check: d/dx [ 13×7(3x − 1)⁷ ] = (3x − 1)⁶ ⇒ ∫(3x − 1)⁶ dx = 13×7(3x − 1)⁷ + c.

Common mistake: This works only for a linear bracket. ∫(ax² + b)⁶ dx is not 13a(ax² + b)³… (differentiate it to see why not).

∫(2x − 3)⁴ dx = 12(4+1)(2x − 3)⁴⁺¹ + c= (2x − 3)⁵/10 + c

∫20(1 − 4x)⁶dx = 20∫(1 − 4x)⁻⁶ dx= 20(−4)(−6+1)(1 − 4x)⁻⁵ + c= (1 − 4x)⁻⁵ + c= 1/(1 − 4x)⁵ + c

∫5√(2x+7)dx = 5∫(2x + 7)^−½ dx= 52(−½+1)(2x + 7)^½ + c= 5√(2x + 7) + c

Trick: a = coefficient of x inside the bracket. Divide by it, e.g. 1/(−4) for (1 − 4x).

4. Further Indefinite Integration

Key idea — “differentiate, then reverse”If ddx[F(x)] = f(x) then ∫ f(x) dx = F(x) + c

Differentiate the expression suggested (usually with the chain rule).
Compare with what you have to integrate — it will differ by a constant factor.
Divide/multiply by that factor, then add c.

Worked example:

(a) Show d/dx[(3x² − 4)⁸] = 48x(3x² − 4)⁷. (b) Hence find ∫6x(3x² − 4)⁷ dx.

(a) y = (3x² − 4)⁸dy/dx = (6x)(8)(3x² − 4)⁷ chain rule= 48x(3x² − 4)⁷ shown(b) ∫6x(3x² − 4)⁷ dx = ⅛∫48x(3x² − 4)⁷ dx= ⅛(3x² − 4)⁸ + c

Another: d/dx[(x²+1)⁻¹] = −2x(x²+1)⁻²⇒ ∫x(x²+1)²dx = −½ × 1x²+1 + c= −1/(2(x²+1)) + c

Pattern: the “inside” differentiates to (a multiple of) the other factor — here 3x²−4 → 6x, and 6x is sitting outside. Spot that and you can integrate.

5. Definite Integration

Integrate between two limits x = a and x = b. No “+ c” needed — it cancels.

Key rule — evaluating∫_a^b f(x) dx = [F(x)]_a^b = F(b) − F(a)

Tip: Limits go to the RIGHT of ∫ (or directly above/below it) — never to the left.

∫₂⁴ x³ dx = [ ¼x⁴ ]₂⁴= (¼·4⁴) − (¼·2⁴)= 64 − 4= 60

Rules for definite integrals∫_a^b k f dx = k∫_a^b f dx
∫_a^b [f ± g] dx = ∫_a^b f dx ± ∫_a^b g dx
∫_a^b f dx = −∫_b^a f dx (swap limits ⇒ sign flips)
∫_a^b f dx + ∫_b^c f dx = ∫_a^c f dx (join intervals)

Worked examples

(a) ∫₁²6x⁴ − 1x²dx = ∫₁²(6x² − x⁻²)dx= [ 2x³ + x⁻¹ ]₁²= (16 + ½) − (2 + 1)= 13½

(b) ∫₀³√(5x+1) dx = ∫₀³(5x+1)^½dx= [ 15·(3/2)(5x+1)^3/2 ]₀³ = [ 215(5x+1)^3/2 ]= 215(16^3/2) − 215(1)= 12815 − 215= 8⅖

Careful: Put brackets round negative limits/values when substituting: (−2) not −2 — signs are the #1 slip.

6. Area Under a Curve

Idea: area bounded by y = x², the x-axis, x = 2 and x = 5. Cut into thin rectangles of width δx and height y: A ≈ Σ y δx. As δx → 0 the sum becomes an integral.

A ≈ Σ y δx (sum of strips)

strips get infinitely thin ⇒ exact area = integral

Key rule — area under y = f(x) between x = a and x = bA = ∫_a^b y dx (curve above the x-axis, y ≥ 0)

Area under y = 3x² from x = 4 to 6A = ∫₄⁶3x² dx = [ x³ ]₄⁶= 216 − 64= 152 units²

Region BELOW the x-axis

region under the x-axis ⇒ integral negative, area = |integral|

curve crosses the axis ⇒ split at 2, find each area, ADD the positives: 6⅔ + 42⅔ = 49⅓

If the region lies below the axis, all y-values are negative so the integral is negative: area = |integral|.

y = x² − 6x from 0 to 6∫ = [ ⅓x³ − 3x² ]₀⁶ = (72 − 108) − (0)= −36Area = 36 units²

Warning: If the curve crosses the x-axis inside the region, integrating straight through lets + and − cancel and gives the WRONG area. Split at each root, integrate each part, make each part positive, then add.

y = x(x−2)(x−6) = x³ − 8x² + 12x, roots 0, 2, 6∫₀² = [ ¼x⁴ − ⁸⁄₃x³ + 6x² ]₀² = 6⅔ (above axis)∫₂⁶ = (−36) − (6⅔) = −42⅔ (below axis)Total area = 6⅔ + 42⅔ = 49⅓ units²

Area between a curve and the y-axis

Remember: Slices are horizontal: width = x, thickness δy. Limits are y-values. If given y = f(x), rearrange to x = …

Key rule — area against the y-axisA = ∫_a^b x dy (x ≥ 0; between y = a and y = b)

slice horizontally: width = x, thickness δy, y-limits are 0 and 4

x = y(4 − y), y from 0 to 4A = ∫₀⁴(4y − y²)dy = [ 2y² − ⅓y³ ]₀⁴= (32 − 64/3) − 0= 10⅔ units²

7. Area Between a Curve and a Line / Two Curves

Sketch both graphs.
Equate the equations → x-values of intersection (a and b).
Decide which graph is on top over [a, b].
Area = ∫ (top − bottom) dx.

area = ∫(top − bottom) between the two intersection points

Method 1 — subtract a shape: curve y = −x² + 8x − 5 and line y = x + 1 meet at (1, 2) and (6, 7).

A = (area under curve) − (area of trapezium)= ∫₁⁶(−x² + 8x − 5)dx − ½(2 + 7)(5)= [ −⅓x³ + 4x² − 5x ]₁⁶ − 22½= 42 − (−4/3) − 22½= 20⅚ units²

Method 2 — key rule (f on top, g below)A = ∫_a^b f(x) dx − ∫_a^b g(x) dx or A = ∫_a^b [ f(x) − g(x) ] dx

f − g = (−x² + 8x − 5) − (x + 1) = −x² + 7x − 6A = ∫₁⁶(−x² + 7x − 6)dx = [ −⅓x³ + 7⁄₂x² − 6x ]₁⁶= 18 − (−17⁄₆)= 20⅚ units²

here the LINE is on top: ∫(line − curve) dx = 36

Tip: Method 2 is easier (and the only sensible one) when the region dips below the axis or there is no neat trapezium.

Curve y = x² − 6x − 2, line y = 2x − 9equate: x² − 8x + 7 = 0 ⇒ x = 1, 7A = ∫₁⁷[(2x − 9) − (x² − 6x − 2)]dx = ∫₁⁷(−x² + 8x − 7)dx= [ −⅓x³ + 4x² − 7x ]₁⁷ = 32⅔ − (−3⅓)= 36 units²

Exam link: Subtract the right way round: top − bottom (check by a quick sketch or a test x-value). Tangent / normal questions: differentiate for the gradient → equation of line → then area.

8. Improper Integrals

Improper = part of the integral “blows up”. Two types. In both we replace the bad limit by a letter X, integrate normally, then let X tend to the bad value. If the limit is finite, the integral exists; if not, it is undefined.

Type 1 — an infinite limit

Key rule (provided the limit exists)∫_a^∞ f dx : replace ∞ by X, take X → ∞ | ∫_−∞^b f dx : replace −∞ by X, take X → −∞

infinite width, but FINITE area (limit exists)

∫₁^∞ 1/x² dx → ∫₁^Xx⁻²dx = [ −x⁻¹ ]₁^X= −1/X + 1as X → ∞ : 1/X → 0value = 1

Curve y = 1/(1 − x)³, area from x = p to 0, p → −∞A = ∫_p⁰(1 − x)⁻³dx = [ 12(1 − x)² ]_p⁰= ½ − 12(1 − p)²p → −∞ : second term → 0A → ½

Type 2 — function undefined at an end point

e.g. ∫₀¹ 1/x² dx: 1/x² → ∞ as x → 0, but is fine elsewhere in [0, 1]. (∫₋₁¹ 1/x² dx is not even allowed — bad point is inside; here we only treat bad end-points.)

Key rules (provided the limit exists)f undefined at x = a : replace a by X, take X → a
f undefined at x = b : replace b by X, take X → b

curve shoots up at x = 2 but area still tends to 6√2

∫₀² 5/x² dx — f undefined at x = 0 ⇒ lower limit X∫_X² 5x⁻²dx = [ −5/x ]_X² = −5/2 + 5/XX → 0 : 5/X → ∞does NOT exist (undefined)

Curve y = 3/√(2 − x), area from 0 to p, p → 2A = ∫₀^p3(2 − x)^−½dx = [ −6√(2 − x) ]₀^p= 6√2 − 6√(2 − p)p → 2 : √(2−p) → 0A → 6√2

Quick check (power test): ∫₁^∞ x^−kdx finite ⇔ k > 1 (1/x² ✓, 1/√x ✗). ∫₀¹ x^−kdx finite ⇔ k < 1 (1/√x ✓, 1/x² ✗). Still show the limit working in the exam.

9. Volumes of Revolution

Spin an area 360° about an axis ⇒ solid of revolution; its volume = volume of revolution. Slice the solid into thin cylinders (discs): radius = y, thickness δx ⇒ volume of one disc = π y² δx; add them up → integral.

spin about the x-axis ⇒ stack of discs: V = ∫ π y² dx

Key rule — rotate about the x-axisV = ∫_a^b π y² dx (about the x-axis, from x = a to x = b)

Key rule — rotate about the y-axisV = ∫_a^b π x² dy (about the y-axis, from y = a to y = b)

Watch out: Square y first, then integrate — π y² means π(f(x))², not π(∫y)². Expand brackets / use index form before integrating. For dy: write x² in terms of y and use y-limits.

y = 9/(3x + 2), x from 1 to 2, about x-axisV = π∫₁²81(3x+2)²dx = π∫₁²81(3x + 2)⁻²dx= π[ 813(−1)(3x + 2)⁻¹ ]₁² = π[ −273x+2 ]₁²= π( −27/8 + 27/5 )V = 81π/40 units³

about the y-axis: discs are horizontal, radius x, limits are y-values

y = x², region from y = 2 to y = 5, about the y-axisV = π∫₂⁵x² dy = π∫₂⁵y dy (using x² = y)= π[ y²/2 ]₂⁵ = π(25/2 − 2)V = 21π/2 units³

When the region leaves a hole

region sits ABOVE y = 1 ⇒ rotating leaves a hole (r = 1, length 2)

Region lies above y = 1 under 2x² + y² = 9, x from 0 to 2, rotated about the x-axis ⇒ the solid has a cylindrical hole (radius 1, length 2).

V = π∫₀²y² dx − (volume of cylinder)= π∫₀²(9 − 2x²)dx − π(1)²(2)= π[ 9x − ⅔x³ ]₀² − 2π = π(18 − 16/3) − 2πV = 32π/3 units³

Proof — volume of a sphere (radius r)

Semicircle y = √(r² − x²) spun about the x-axis, x from −r to rV = π∫_−r^r(r² − x²)dx = π[ r²x − ⅓x³ ]_−r^r= π[ (r³ − ⅓r³) − (−r³ + ⅓r³) ]V = 4⁄3 πr³

rotate the semicircle ⇒ sphere

Quick Recap — Chapter Cheat-Sheet

✓ Indefinite integration

∫xⁿ dx = xⁿ⁺¹/(n+1) + c (n ≠ −1) — power +1, ÷ new power, + c
∫k f dx = k∫f dx ; ∫(f ± g)dx = ∫f ± ∫g
∫(ax+b)ⁿ dx = (ax+b)ⁿ⁺¹ / [a(n+1)] + c — linear bracket only
If d/dx F = f then ∫f dx = F + c (reverse chain rule: differentiate, compare, adjust factor)
Expand / split / index form BEFORE integrating. Find c from a known point; find k first if needed.

✓ Definite integrals & area

∫ₐᵇ f dx = F(b) − F(a) ; swap limits ⇒ sign flips ; ∫ₐᵇ + ∫ᵦᶜ = ∫ₐᶜ
Area under curve: A = ∫ₐᵇ y dx (y ≥ 0). Below axis ⇒ negative ⇒ take |·|. Split at roots, add areas.
Area against y-axis: A = ∫ₐᵇ x dy (x ≥ 0) — y-limits
Between two graphs: A = ∫ₐᵇ [f(x) − g(x)] dx, f on top, a & b = intersection x-values (or: area under top − area under bottom / trapezium)

✓ Improper integrals

Infinite limit: replace ∞ (or −∞) by X, integrate, let X → ±∞. Finite limit ⇒ exists, else undefined.
Function undefined at an end-point: replace that limit by X, let X → the bad value.
Examples: ∫₁^∞ 1/x² = 1 ; ∫₀² 5/x² does not exist ; ∫₀² 3/√(2−x) = 6√2

✓ Volumes of revolution (360°)

About x-axis: V = ∫ₐᵇ π y² dx (square y first!)
About y-axis: V = ∫ₐᵇ π x² dy (x² in terms of y, y-limits)
Region not touching the axis ⇒ hole ⇒ V = (outer solid) − (inner solid)
Sphere: V = 4⁄3 πr³ (from semicircle y = √(r²−x²), −r to r)

Before you hand in — checklist: ✓ “+ c” on every indefinite integral · ✓ brackets round negative numbers · ✓ areas positive · ✓ units² / units³ · ✓ exact values (π, fractions, surds) unless asked for decimals · ✓ a quick sketch for every area/volume question.