DP Maths AA SL - Calculus - Integration and Its Applications

DP Maths AA SL - Calculus - Integration and Its Applications - Full notes

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DP Maths AA SL | Integration and Its Applications

Integration and Its Applications

This topic contains 1 note:

5.5 Integration

DP MATHEMATICS AA · TOPIC 5: CALCULUS

5.5

Integration

Full notes for SL · Syllabus SL 5.5, 5.10, 5.11

Contents

1	Antidifferentiation
2	Standard integrals
3	Finding the constant of integration
4	Reverse chain rule and substitution
5	Definite integrals
6	Area under a curve
7	Area between two curves
8	Kinematics with integration
9	Indefinite integrals: more practice
10	Integrals of linear compositions
11	Integration by inspection (reverse chain rule)
12	Definite integrals with exact answers
13	Properties of definite integrals
14	Areas under curves: more practice
15	Areas between curves: more practice
16	Kinematics with integration
17	Accumulation problems
18	Trapezoidal-style estimates and technology
19	Mixed review: worked problems
20	Additional exam-style practice
21	Integration and differentiation together
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
understand integration as the reverse of differentiation
integrate xn (n ≠ −1), 1/x, ex, sin x and cos x, and linear compositions of these
find the constant of integration from a boundary condition
integrate by inspection (reverse chain rule) and by simple substitution
evaluate definite integrals exactly and with technology
find areas under curves and between curves, including regions below the x-axis
find displacement and distance travelled from velocity

This booklet covers SL 5.5, 5.10, 5.11, which is common to SL and HL.

Guiding questions

How does reversing differentiation lead to areas? How can we recover position from velocity?

How this topic is examined

Integration appears on every paper: indefinite integrals (including reverse chain rule), finding f from f′ and a point, exact definite integrals on Paper 1, areas between curves with technology on Paper 2, and kinematics (displacement and total distance).

Always include + c for indefinite integrals, and split areas at the x-axis when the curve crosses it.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Antidifferentiation

Undoing the derivative.

Think about it

If the gradient of a curve is 2x everywhere, what could the curve be? Is there only one possible answer?

Integration reverses differentiation. If F′(x) = f(x), then F is an antiderivative of f, and we write the indefinite integral:

The constant of integration c appears because the derivative of a constant is zero: x2, x2 + 5 and x2 − 3 all have derivative 2x. The antiderivatives of a function form a family of curves, vertical translations of each other.

Figure 1. The family of antiderivatives of 2x.

2 Standard integrals

Reverse each derivative.

f(x)	∫f(x) dx
xn (n ≠ −1)	xn+1/(n + 1) + c
1/x	ln\|x\| + c
ex	ex + c
sin x	−cos x + c
cos x	sin x + c
(ax + b)n	(ax + b)n+1/(a(n + 1)) + c
eax+b	eax+b/a + c
cos(ax + b)	sin(ax + b)/a + c

To integrate xn, add 1 to the power and divide by the new power. Rewrite roots and fractions as powers first, and expand products. For a linear inside function ax + b, integrate as usual and divide by a.

WORKED EXAMPLE 1 Powers (basic)

Find ∫(6x2 − 4/x3 + 3√x) dx.

Solution

= 6x3/3 − 4x−2/(−2) + 3x3/2/(3/2)M1

= 2x3 + 2/x2 + 2x3/2 + cA1 A1

WORKED EXAMPLE 2 Linear compositions (standard)

Find (a) ∫e3x−1 dx; (b) ∫sin(2x) dx; (c) ∫1/(2x + 5) dx.

Solution

(a) e3x−1/3 + cA1

(b) −cos(2x)/2 + cA1

Common mistake

Integrating 1/x as x0/0. The power rule fails for n = −1; the integral is ln|x|.

Practice exercise 1

1. Find ∫(x3 − 2x + 7) dx.

2. Find ∫(x + 1)2/x2 dx (expand first).

3. Find ∫(4e2x − cos 3x) dx.

4. Find ∫(3x − 2)5 dx.

3 Finding the constant of integration

Use a known point.

WORKED EXAMPLE 3 Curve through a point (standard)

A curve has gradient dy/dx = 3x2 − 4x and passes through (2, 5). Find its equation.

Solution

y = x3 − 2x2 + cA1

5 = 8 − 8 + c, c = 5M1

y = x3 − 2x2 + 5A1

Practice exercise 2

1. f′(x) = ex + 1 and f(0) = 4. Find f(x).

2. f″(x) = 6x, f′(1) = 0 and f(1) = 2. Find f(x).

4 Reverse chain rule and substitution

Spot the derivative of the inside.

If an integrand has the form g′(x) f(g(x)), the integral is F(g(x)) + c, where F is an antiderivative of f. In practice: look for a function and (a multiple of) its derivative. Examples: ∫2x(x2 + 1)4 dx = (x2 + 1)5/5 + c; ∫g′(x)/g(x) dx = ln|g(x)| + c.

Substitution makes this systematic: let u = g(x), so du = g′(x) dx; rewrite the integral in u, integrate, then substitute back.

WORKED EXAMPLE 4 By inspection (standard)

Find ∫x2ex³ dx.

Solution

Inside x3 has derivative 3x2M1

= ex³/3 + cA1

WORKED EXAMPLE 5 By substitution (standard)

Find ∫x√(x2 + 9) dx using u = x2 + 9.

Solution

du = 2x dx, so x dx = du/2M1

∫½u1/2 du = u3/2/3A1

= (x2 + 9)3/2/3 + cA1

WORKED EXAMPLE 6 A log integral (exam level)

Find ∫(6x + 3)/(x2 + x + 1) dx.

Solution

Numerator = 3 × derivative of the denominatorM1

= 3 ln(x2 + x + 1) + cA1

Practice exercise 3

1. Find ∫cos x sin3x dx.

2. Find ∫x/(x2 + 4) dx.

3. Find ∫(2x + 1)ex² + x dx.

5 Definite integrals

Evaluate between limits.

This is the fundamental theorem of calculus. No + c is needed, because it cancels. Properties: ∫ from a to a = 0; reversing the limits changes the sign; ∫ from a to b + ∫ from b to c = ∫ from a to c; constants can be taken outside.

WORKED EXAMPLE 7 Exact definite integral (standard)

Find the exact value of ∫ from 0 to π/2 of (sin x + 2x) dx.

Solution

[−cos x + x2] from 0 to π/2M1

= (0 + π2/4) − (−1 + 0)A1

= π2/4 + 1A1

WORKED EXAMPLE 8 Exact log answer (exam level)

Find ∫ from 1 to 3 of 2/(2x + 1) dx exactly.

Solution

[ln|2x + 1|] from 1 to 3M1

= ln 7 − ln 3 = ln(7/3)A1

GDC tip

Use the numerical integration function (∫ dx) for definite integrals on Paper 2. For areas, integrate the absolute value, or split at the roots.

Practice exercise 4

1. Evaluate ∫ from 1 to 4 of √x dx.

2. Evaluate ∫ from 0 to 1 of e2x dx exactly.

3. Given ∫ from 1 to 3 of f(x) dx = 5, find ∫ from 1 to 3 of (2f(x) + 1) dx.

6 Area under a curve

Split at the x-axis.

For f(x) ≥ 0 on [a, b], the area between the curve and the x-axis is ∫ from a to b of f(x) dx. If part of the curve is below the axis, the integral of that part is negative. To find total area, split at the roots and add the absolute values (or integrate |f(x)| with technology).

WORKED EXAMPLE 9 Area crossing the axis (exam level)

Find the total area enclosed between y = x2 − 4 and the x-axis for 0 ≤ x ≤ 3.

Solution

Root at x = 2; below axis on [0, 2]M1

∫ from 0 to 2 = [x3/3 − 4x] = 8/3 − 8 = −16/3; ∫ from 2 to 3 = (9 − 12) − (8/3 − 8) = 7/3A1 A1

Area = 16/3 + 7/3 = 23/3A1

Figure 2. The region between y = x2 − 4 and the x-axis, split at x = 2.

Common mistake

Integrating straight from 0 to 3 gives −3, not the area 23/3, because the negative part cancels some of the positive part.

Practice exercise 5

1. Find the area under y = 6x − x2 between x = 0 and x = 6.

2. Find the area enclosed by y = sin x and the x-axis for 0 ≤ x ≤ 2π.

7 Area between two curves

Top minus bottom.

Find the intersection points (the limits), decide which curve is on top, and integrate top minus bottom. If the curves cross within the interval, split at the crossing points.

WORKED EXAMPLE 10 Area between curves (standard)

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

Solution

x2 = x + 2: x = −1, 2A1

∫ from −1 to 2 of (x + 2 − x2) dx = [x2/2 + 2x − x3/3]M1

= (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 9/2A1

Figure 3. The region enclosed by y = x2 and y = x + 2.

Practice exercise 6

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

2. Use technology to find the area enclosed by y = ex and y = x + 3.

8 Kinematics with integration

From acceleration to velocity to displacement.

Integrating velocity gives displacement; the constant is found from the initial position. The definite integral ∫ from t1 to t2 of v dt gives the displacement over that interval, while ∫|v| dt gives the total distance travelled.

WORKED EXAMPLE 11 Distance and displacement (exam level)

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

Solution

Displacement = ∫ from 0 to 3 of v dt = [t3/3 − 2t2 + 3t] = 9 − 18 + 9 = 0 mM1 A1

v changes sign at t = 1 and t = 3: ∫ from 0 to 1 = 4/3, ∫ from 1 to 3 = −4/3M1

Distance = 4/3 + 4/3 = 8/3 mA1

Practice exercise 7

1. a(t) = 6t − 4, v(0) = 2, s(0) = 0. Find s(t).

2. Use technology to find the distance travelled in the first 5 seconds by a particle with v(t) = 3 cos t.

9 Indefinite integrals: more practice

Rewrite, then integrate.

Expand products, split fractions with a single-term denominator, and write roots as powers before integrating. Always include + c.

WORKED EXAMPLE 12 Fraction (standard)

Find ∫(x3 − 2x + 5)/x2 dx.

Solution

∫(x − 2/x + 5x−2) dxM1

= x2/2 − 2 ln|x| − 5/x + cA1 A1

WORKED EXAMPLE 13 Roots (standard)

Find ∫(√x + 1/√x) dx.

Solution

∫(x1/2 + x−1/2) dxM1

= (2/3)x3/2 + 2x1/2 + cA1

Practice exercise 8

1. Find ∫(2x + 1)2 dx.

2. Find ∫(4 − x)/√x dx.

3. Find ∫(ex + 1)2 dx.

10 Integrals of linear compositions

Divide by the coefficient of x.

∫f(ax + b) dx = F(ax + b)/a + c. This covers (ax + b)n, eax+b, sin(ax + b), cos(ax + b) and 1/(ax + b).

WORKED EXAMPLE 14 Linear compositions (standard)

Find (a) ∫(5 − 2x)4 dx; (b) ∫cos(3x − 1) dx; (c) ∫4/(2x + 3) dx.

Solution

(a) −(5 − 2x)5/10 + cA1

(b) sin(3x − 1)/3 + cA1

Practice exercise 9

1. Find ∫e1−3x dx.

2. Find ∫sin(πx/2) dx.

11 Integration by inspection (reverse chain rule)

Spot f′(x) multiplying f(x).

If an integrand is (a multiple of) g′(x) times a function of g(x), integrate the outer function and substitute back. Check by differentiating.

WORKED EXAMPLE 15 Inspection (exam level)

Find ∫x(x2 + 3)5 dx and ∫sin x cos4x dx.

Solution

(x2 + 3)6/12 + cA1

−cos5x/5 + cA1

Practice exercise 10

1. Find ∫3x2ex³ dx.

2. Find ∫(2x + 1)/(x2 + x + 5) dx.

12 Definite integrals with exact answers

Evaluate carefully.

Evaluate F(b) − F(a), keeping exact forms (fractions, π, ln, e). Bracket negative terms when subtracting.

WORKED EXAMPLE 16 Exact (standard)

Evaluate ∫ from 0 to π/3 of (2 sin x + cos 2x) dx.

Solution

[−2 cos x + ½ sin 2x] from 0 to π/3M1

= (−1 + √3/4) − (−2)A1

= 1 + √3/4A1

Practice exercise 11

1. Evaluate ∫ from 1 to e of (1 + 1/x) dx.

2. Evaluate ∫ from 0 to ln 2 of e2x dx.

13 Properties of definite integrals

Splitting and reversing.

∫ from a to b of kf = k∫ from a to b of f; ∫ from a to b + ∫ from b to c = ∫ from a to c; ∫ from b to a = −∫ from a to b. These let you use given integral values to find others.

WORKED EXAMPLE 17 Using properties (exam level)

∫ from 1 to 4 of f(x) dx = 7 and ∫ from 1 to 6 of f(x) dx = 10. Find ∫ from 4 to 6 of (3f(x) + 2) dx.

Solution

∫ from 4 to 6 of f = 3A1

3(3) + 2(2)M1

= 13A1

Practice exercise 12

1. Given ∫ from 0 to 5 of g(x) dx = 8, find ∫ from 5 to 0 of 2g(x) dx.

14 Areas under curves: more practice

Positive and negative regions.

Total area requires splitting at the roots and taking absolute values; a GDC can integrate |f(x)| directly.

WORKED EXAMPLE 18 Cubic (exam level)

Find the total area between y = x3 − 4x and the x-axis.

Solution

Roots −2, 0, 2; odd functionM1

∫ from −2 to 0 = 4A1

Total area 8A1

Practice exercise 13

1. Find the area enclosed by y = sin x and the x-axis for 0 ≤ x ≤ 3π/2.

2. Find the total area between y = x2 − x − 2 and the x-axis for 0 ≤ x ≤ 3.

15 Areas between curves: more practice

Upper minus lower.

Find intersections, identify which function is above, and integrate the difference. When curves cross more than once inside the interval, split at each crossing.

Figure 4. Two regions between y = x3 and y = x.

WORKED EXAMPLE 19 Two regions (exam level)

Find the total area enclosed by y = x3 and y = x.

Solution

Intersections x = −1, 0, 1A1

∫ from 0 to 1 of (x − x3) dx = 1/4M1 A1

Total 1/2 (by symmetry)A1

Practice exercise 14

1. Find the area between y = ex and y = e−x for 0 ≤ x ≤ 1.

2. Find the area between y = 4 − x2 and y = x + 2.

16 Kinematics with integration

From acceleration to displacement.

v = ∫a dt and s = ∫v dt, using initial conditions to find constants. Displacement is ∫v dt; distance is ∫|v| dt.

WORKED EXAMPLE 20 Acceleration (exam level)

a(t) = 6 − 2t, v(0) = 4, s(0) = 0. Find v(t), s(t) and the time the particle is at rest.

Solution

v = 6t − t2 + 4A1

s = 3t2 − t3/3 + 4tA1

v = 0: t2 − 6t − 4 = 0: t = 6.61M1 A1

Practice exercise 15

1. Find the distance travelled in the first 8 seconds.

17 Accumulation problems

Totals from rates.

If r(t) is a rate (litres per minute, people per hour), then ∫ from a to b of r(t) dt is the total change over [a, b]. Add the initial amount to find the final amount.

WORKED EXAMPLE 21 Water tank (standard)

Water enters a tank at r(t) = 30 − 0.5t L/min for 0 ≤ t ≤ 40, starting with 200 L. Find the volume at t = 40.

Solution

∫ from 0 to 40 = 1200 − 400 = 800M1 A1

Volume 1000 LA1

Practice exercise 16

1. A town's population grows at 120e0.02t people per year. Find the increase over the first 10 years.

18 Trapezoidal-style estimates and technology

When no antiderivative is known.

Some integrals, such as ∫e−x² dx, have no elementary antiderivative. Evaluate definite integrals numerically with the GDC; for data, a sum of trapezia estimates the area.

WORKED EXAMPLE 22 Numerical (standard)

Evaluate ∫ from 0 to 1 of e−x² dx.

Solution

GDC: 0.747A1

Practice exercise 17

1. Evaluate ∫ from 0 to π of sin(x2) dx with technology.

19 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 23 Mixed problem (exam level)

f′(x) = 6x2 − 4x and f(1) = 5. (a) Find f(x). (b) Find the area between y = f′(x) and the x-axis for 0 ≤ x ≤ 1.

Solution

(a) f = 2x3 − 2x2 + 5A1 A1

(b) f′ < 0 on (0, 2/3): splitM1

∫ from 0 to 2/3 = −8/27 and ∫ from 2/3 to 1 = 8/27: total area 16/27A1

Practice exercise 18

1. Find k > 0 such that ∫ from 0 to k of (2x + 1) dx = 12.

20 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 24 Paper 1 question (exam level)

(a) Find ∫x√(x2 + 1) dx. (b) Hence find the exact value of ∫ from 0 to √3 of x√(x2 + 1) dx.

Solution

(a) (x2 + 1)3/2/3 + cM1 A1

(b) (8 − 1)/3 = 7/3A1

Practice exercise 19

1. Find the area enclosed by y = 2x and y = x2 − 3.

21 Integration and differentiation together

The fundamental theorem.

d/dx ∫ from a to x of f(t) dt = f(x), and ∫ from a to b of f′(x) dx = f(b) − f(a). These link the two halves of calculus and allow quick evaluation when f is known.

WORKED EXAMPLE 25 FTC (standard)

Find ∫ from 0 to 2 of d/dx(x2ex) dx.

Solution

= [x2ex] from 0 to 2M1

= 4e2A1

Practice exercise 20

1. Find d/dx of ∫ from 1 to x of ln t dt.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

Let f(x) = 3x2 − 6x and g(x) = 6 − 3x.

(a) Find the x-coordinates of the points of intersection of the graphs. [3]

(b) Find the area of the region enclosed by the two graphs. [4]

Markscheme

Part	Answer	Marks
(a)	3x2 − 6x = 6 − 3x	M1
	3x2 − 3x − 6 = 0, x2 − x − 2 = 0	A1
	x = −1, 2	A1
(b)	∫ from −1 to 2 of (6 − 3x − 3x2 + 6x) dx	M1
	[6x + 3x2/2 − x3] from −1 to 2	A1
	(12 + 6 − 8) − (−6 + 3/2 + 1)	A1
	13.5	A1

Examiner's comments. Identifying g as the upper function between the roots is essential.

Question 2 (Paper 2 style) (about 10 minutes)

A car accelerates from rest with v(t) = 10(1 − e−0.2t) m/s.

(a) Find the acceleration when t = 5. [2]

(b) Find the distance travelled in the first 10 seconds. [2]

Markscheme

Part	Answer	Marks
(a)	a = 2e−0.2t	M1
	0.736 m/s2	A1
(b)	∫ from 0 to 10 of v dt	M1
	56.8 m	A1
(c)	approaches 10 m/s	A1

Examiner's comments. Technology is expected for (b).

23 Common misconceptions and links

Students often think…	Actually…
forgetting + c	include it for indefinite integrals
∫1/x dx = x0/0	ln\|x\|
area = integral even below the axis	split at roots
displacement = distance	distance uses \|v\|

How 5.5 connects to the rest of the course

Topic	Connection
5.1 Differentiation	reverse process
5.8 Kinematics	velocity and acceleration
4.14 Continuous RVs (HL)	probability as area
5.16 Further integration (HL)	parts, volumes

24 Check your understanding

Section A: multiple choice

1. ∫x4 dx =

A. x5/5 + c B. 4x3 C. x5 + c D. 5x5

2. ∫e2x dx =

A. 2e2x + c B. e2x + c C. e2x/2 + c D. ex²

3. ∫cos x dx =

A. −sin x + c B. sin x + c C. cos x + c D. −cos x + c

4. ∫ from 0 to 2 of 3x2 dx =

A. 12 B. 6 C. 4 D. 8

5. Distance travelled is

A. ∫v dt B. ∫|v| dt C. ∫a dt D. |∫v dt|

Section B: structured

6. Find the area between y = 4 − x2 and the x-axis. [3]

7. Find f(x) if f′(x) = 1/x and f(1) = 3. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. 32/3.

7. ln x + 3.

25 Summary and topic map

Idea	Remember
Antiderivative	∫f dx = F + c
Power rule	xn+1/(n + 1), n ≠ −1
Standard	1/x → ln\|x\|; ex; sin → −cos; cos → sin
Linear inside	divide by a
Reverse chain	g′(x)f(g(x))
Definite	F(b) − F(a)
Areas	split at roots; top minus bottom
Kinematics	s = ∫v dt; distance ∫\|v\| dt

Before the exam, make sure you…

•include + c
•rewrite as powers
•split areas at the axis
•use |v| for distance

Integration on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 5. Topic map for Integration.

26 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. x4/4 − x2 + 7x + c

2. x + 2 ln|x| − 1/x + c

3. 2e2x − (1/3)sin 3x + c

4. (3x − 2)6/18 + c

Practice exercise 2

1. ex + x + 3

2. f′ = 3x2 − 3, f = x3 − 3x + 4

Practice exercise 3

1. sin4x/4 + c

2. ½ ln(x2 + 4) + c

3. ex² + x + c

Practice exercise 4

1. 14/3

2. (e2 − 1)/2

3. 12

Practice exercise 5

1. 36

2. 4

Practice exercise 6

1. 4/3

2. intersections −2.95 and 1.51: area 5.69

Practice exercise 7

1. v = 3t2 − 4t + 2; s = t3 − 2t2 + 2t

2. ∫ from 0 to 5 of |3 cos t| dt = 9.12 m

Practice exercise 8

1. (4/3)x3 + 2x2 + x + c

2. 8√x − (2/3)x3/2 + c

3. e2x/2 + 2ex + x + c

Practice exercise 9

1. −e1−3x/3 + c

2. −(2/π)cos(πx/2) + c

Practice exercise 10

1. ex³ + c

2. ln(x2 + x + 5) + c

Practice exercise 11

1. e

2. 3/2

Practice exercise 12

1. −16

Practice exercise 13

1. 2 + 1 = 3

2. ∫ from 0 to 2 = −10/3; from 2 to 3 = 11/6: 31/6

Practice exercise 14

1. e + 1/e − 2 = 1.09

2. 9/2

Practice exercise 15

1. ∫ from 0 to 8 of |v| dt = 69.2 (technology)

Practice exercise 16

1. 6000(e0.2 − 1) = 1328

Practice exercise 17

1. 0.773

Practice exercise 18

1. k2 + k = 12: k = 3

Practice exercise 19

1. intersections −1, 3: 32/3

Practice exercise 20

1. ln x

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