IB DP Maths AA · Topic 5 Calculus · SL 5.5, 5.9–5.11 · AHL 5.15–5.17 · ~9 min read

Integration and kinematics

Antiderivatives and the constant of integration, definite integrals and areas, area between curves, integrating eˣ, 1/x, sin and cos, reverse chain rule, kinematics with displacement and distance, and for HL: substitution, integration by parts, arctan and arcsin integrals, and volumes of revolution.

Indefinite integralsBoundary conditionsDefinite integralsArea under a curveArea between curveseˣ, 1/x, sin, cosReverse chain ruleKinematics

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No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

Integration undoes differentiation. ∫xⁿ dx = x^{n + 1}n + 1 + C (n ≠ −1). Never forget + C on an indefinite integral.
Definite integral ∫_a^b f(x) dx = F(b) − F(a) is the signed area: regions below the x-axis count as negative.
Area between curves = ∫(top − bottom) dx between the intersection points.
Kinematics: v = ds/dt and a = dv/dt, so s = ∫v dt and v = ∫a dt. Displacement = ∫v dt; distance = ∫ǀvǀ dt.
HL: substitution, integration by parts, ∫1a² + x² and ∫1√(a² − x²), and volumes of revolution V = π∫y² dx.

By the end you can

Find indefinite integrals and use a boundary condition to find C.
Evaluate definite integrals and find areas under a curve and between curves.
Integrate eˣ, 1/x, sin x, cos x and composites of the form f(ax + b), and use the reverse chain rule.
Use calculus for displacement, velocity, acceleration and total distance.
Integrate by substitution and by parts, including repeated parts. HL
Use the standard arctan and arcsin integrals. HL
Find volumes of revolution about the x-axis and y-axis. HL

1Indefinite integrals

∫xⁿ dx = x^{n + 1}n + 1 + C, n ≠ −1 FORMULA BOOKLET

add 1 to the power, divide by the new power

Worked example

Find ∫(6x² − 4x + 3) dx.

= 2x³ − 2x² + 3x + C A1A1

Worked example

f′(x) = 3x² − 2 and f(1) = 4. Find f(x).

f(x) = x³ − 2x + C A1
1 − 2 + C = 4 ⇒ C = 5 M1
f(x) = x³ − 2x + 5 A1

Rewrite first: 1x³ = x⁻³, √x = x^½, and split fractions: x² + 1x = x + 1x. You can't integrate a product or quotient term by term.

Common mistakeIntegrating (x + 1)(x − 2) as "(x²/2 + x)(x²/2 − 2x)". Expand the brackets first: x² − x − 2.

2Standard integrals

f(x)	∫ f(x) dx
eˣ	eˣ + C
1x	ln ǀxǀ + C
sin x	−cos x + C
cos x	sin x + C
f(ax + b)	1a F(ax + b) + C

All of these are in the formula booklet. The last row is the "divide by a" rule: ∫e^3x dx = ⅓e^3x + C, ∫cos 2x dx = ½ sin 2x + C, ∫12x + 1 dx = ½ ln ǀ2x + 1ǀ + C.
Reverse chain rule: if you can see a function and (a multiple of) its derivative, guess the answer and differentiate to check.

Worked example

Find ∫x(x² + 1)⁴ dx.

the derivative of x² + 1 is 2x, so try (x² + 1)⁵ M1
d/dx (x² + 1)⁵ = 10x(x² + 1)⁴, so divide by 10 M1
(x² + 1)⁵10 + C A1

Exam tip∫f′(x)f(x) dx = ln ǀf(x)ǀ + C is the most useful reverse-chain pattern: ∫2xx² + 5 dx = ln(x² + 5) + C.

3Definite integrals and area

The definite integral from 1 to 3 is the area under the curve between those x-values.

∫_a^b f(x) dx = F(b) − F(a)

no + C needed

Worked example

Find the area under y = x² + 1 between x = 1 and x = 3.

[x³/3 + x]₁³ M1
= (9 + 3) − (13 + 1) M1
= 323 A1

An integral adds signed areas. To find a total area, split at the roots.

Common mistakeIntegrating straight across a root when asked for the area. Find where the curve crosses the axis, integrate each part separately, and add the absolute values.

Shade the region, decide which curve is on top, then integrate the difference.

Worked example

Find the area enclosed by y = x² and y = x + 2.

x² = x + 2 ⇒ x = −1 or 2 M1A1
∫₋₁² (x + 2 − x²) dx = [x²/2 + 2x − x³/3]₋₁² M1
= 103 − (−76) = 4.5 A1

GDC (calculator) methodPaper 2: use the GDC's ∫ function (fnInt) for definite integrals. For an area, integrate ǀf(x)ǀ directly; it does the splitting for you.

4Kinematics

v = dsdt, a = dvdt = d²sdt²

differentiate going down

displacement = ∫_{t_{1}}^t₂ v dt, distance = ∫_{t_{1}}^t₂ ǀvǀ dt FORMULA BOOKLET

integrate going up

Area above the axis is forward motion, below is backward. Displacement adds them with signs; distance adds their sizes.

Worked example

A particle has velocity v = t² − 4t + 3 m s⁻¹ for 0 ≤ t ≤ 4. Find its displacement and the total distance travelled.

displacement = ∫₀⁴ v dt = 43 m M1A1
v = 0 at t = 1 and t = 3, so split there R1
distance = 43 + 43 + 43 = 4 m A1

Speed is ǀvǀ. The particle changes direction where v changes sign, not just where v = 0 (a double root of v doesn't count).
Speeding up or slowing down? Speeding up when v and a have the same sign; slowing down when they have opposite signs.

5Integration by substitution HL

Method: choose u (usually the inside of a bracket or root), find du/dx, replace every x and dx, integrate in u, then substitute back. For a definite integral, change the limits to u-values instead.

Worked example

Find ∫x√(x + 1) dx.

let u = x + 1, so x = u − 1 and dx = du M1
∫(u − 1)u^½ du = ∫(u^3/2 − u^½) du M1
= 25(x + 1)^5/2 − 23(x + 1)^3/2 + C A1A1

6Integration by parts HL

∫u dvdx dx = uv − ∫v dudx dx FORMULA BOOKLET

choose u to be the part that gets simpler when differentiated

Worked example

Find ∫xeˣ dx.

u = x, dv/dx = eˣ ⇒ v = eˣ M1
= xeˣ − ∫eˣ dx A1
= xeˣ − eˣ + C A1

Worked example

Find ∫x ln x dx.

u = ln x (ln is easier to differentiate than to integrate), dv/dx = x ⇒ v = x²2 M1
= x²2 ln x − ∫x2 dx A1
= x²2 ln x − x²4 + C A1

Memory trickChoosing u: LIATE, which stands for Logs, Inverse trig, Algebraic, Trig, Exponential. Whichever comes first is u.

Repeated parts: ∫x²eˣ dx needs parts twice. For ∫eˣ sin x dx, use parts twice until the original integral reappears, then solve for it.

7arctan and arcsin integrals HL

∫1a² + x² dx = 1a arctan(xa) + C FORMULA BOOKLET

complete the square first if needed

∫1√(a² − x²) dx = arcsin(xa) + C, ǀxǀ < a FORMULA BOOKLET

note: no 1/a in front this time

Worked example

Find ∫₀³ 1x² + 9 dx.

= [⅓ arctan(x/3)]₀³ M1
= 13 × π4 = π12 A1

Completing the square turns ∫1x² + 4x + 13 dx into ∫1(x + 2)² + 9 dx = 13 arctan(x + 23) + C.

8Volumes of revolution HL

Each thin slice is a disc of radius y and thickness dx.

V = ∫_a^b πy² dx (about the x-axis), V = ∫_a^b πx² dy (about the y-axis) FORMULA BOOKLET

square the radius before integrating

Worked example

The region under y = √x from x = 0 to x = 4 is rotated 360° about the x-axis. Find the volume.

V = π∫₀⁴ x dx M1
= π[x²/2]₀⁴ = 8π A1

Worked example

The region between y = x², the y-axis and the line y = 4 (x ≥ 0) is rotated about the y-axis. Find the volume.

x² = y, so V = π∫₀⁴ y dy M1
= 8π A1

Common mistakeWriting π∫y dx instead of π∫y² dx. The disc area is πr²: square y first.

Linking questionIntegration is the reverse of Differentiation, and differential equations (Differential equations and Maclaurin series) are solved by integrating. Areas under a pdf give probabilities (Random variables, binomial and normal).

Idea for your IA / explorationEstimate the area of an irregular shape (a lake, a leaf) by fitting functions to its boundary and integrating, or find the volume of a real object (a vase, a bottle) as a volume of revolution and compare with a measured volume of water.

Test yourself — cover the answers first

Quick check

Find ∫x⁻² dx.
Find ∫4x dx.
Find ∫₀¹ eˣ dx.
Find ∫sin 3x dx.
What is the difference between distance and displacement?
HL In ∫x cos x dx, which part should be u?

Exam-style

A particle starts at the origin with velocity 2 m s⁻¹, and its acceleration is a = 6t m s⁻². Find s when t = 2.
HL Find ∫ln x dx.

Show answers

−1/x + C
4 ln ǀxǀ + C
e − 1
−⅓cos 3x + C
distance uses ǀvǀ; displacement can cancel
u = x
v = 3t² + 2
s = t³ + 2t (s = 0 at t = 0)
s(2) = 12 m
write it as ∫1 × ln x dx with u = ln x, v = x
= x ln x − ∫x × 1x dx
= x ln x − x + C

Practise this chapter

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