Additional Mathematics notes, chapter 3
Integration
Textbook Additional Mathematics Form 5 (DLP), pages 80–117
What this chapter covers
- Explain integration as the reverse process of differentiation and find an integral from a known derivative (3.1)
- Find indefinite integrals of algebraic functions, including (ax + b)ⁿ, and find the equation of a curve from its gradient function (3.2)
- Evaluate definite integrals and use their properties (3.3)
- Find areas of regions under a curve, beside the y-axis, and between a curve and a line or another curve (3.3)
- Find the volume of revolution of a region about the x-axis or the y-axis (3.3)
- Solve problems involving integration in areas, volumes and rates (3.4)
3.1Integration as the inverse of differentiation
Integration is the reverse of differentiation. If d/dx [f(x)] = f′(x), then ∫ f′(x) dx = f(x) (plus a constant). The function f(x) is called the antiderivative of f′(x).
| Process | Starting from | Result |
|---|---|---|
| Differentiation | y = 4x² | dy/dx = 8x |
| Integration | dy/dx = 8x | y = 4x² (+ c) |
Idea from textbook p. 82-83. The symbol ∫ is a stretched letter "S" for "sum".
Example (like Self-Exercise 3.1, p. 83): use a known derivative
Given d/dx (x⁵ + 3x) = 5x⁴ + 3, find ∫ (5x⁴ + 3) dx.Integration undoes differentiation, so ∫ (5x⁴ + 3) dx = x⁵ + 3x + c.
Example: a multiple of a known derivative
Given d/dx [(2x + 1)³] = 6(2x + 1)², find ∫ 18(2x + 1)² dx.18(2x + 1)² = 3 × 6(2x + 1)², so ∫ 18(2x + 1)² dx = 3(2x + 1)³ + c.
To check any integral, differentiate your answer. You must get back the expression inside the integral.
Given d/dx (3x⁴ − x²) = 12x³ − 2x, find ∫ (6x³ − x) dx.
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6x³ − x = ½(12x³ − 2x), so the integral is ½(3x⁴ − x²) + c = 32x⁴ − 12x² + c.
3.2Indefinite integral
The basic rules
Indefinite integral of axn: ∫ axn dx = a xn+1n + 1 + c, where n ≠ −1. The constant c is the constant of integration. For a constant, ∫ a dx = ax + c.
- Add 1 to the index.
- Divide by the new index.
- Add the constant c (only once for the whole expression).
Why c? Differentiating 5x, 5x + 2 and 5x − 3 all give 5, so integrating 5 can only tell us the answer is 5x + some constant.
| Integral | Working | Answer |
|---|---|---|
| ∫ 12 dx | constant | 12x + c |
| ∫ x³ dx | x3+1 ÷ (3 + 1) | x⁴4 + c |
| ∫ 2x−2 dx | 2x−1 ÷ (−1) | −2x−1 + c |
| ∫ √x dx = ∫ x1/2 dx | x3/2 ÷ 32 | 23x3/2 + c |
n = −1 is the one case the rule cannot handle (it would divide by 0). Also do not forget "+ c" in an indefinite integral.
Sums and differences
Integrate term by term: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx. Only one constant of integration is written for the whole answer.
Example (textbook p. 87): expand first
∫ (x − 2)(x + 6) dx = ∫ (x² + 4x − 12) dx = x³3 + 2x² − 12x + c.∫ x(3 + 1x) dx = ∫ (3x + 1) dx = 3x²2 + x + c.
You cannot integrate a product by integrating each factor. ∫ (x − 2)(x + 6) dx is not (∫ (x − 2) dx)(∫ (x + 6) dx). Expand the brackets first.
Find ∫ (4x³ − 6x + 5) dx.
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x⁴ − 3x² + 5x + c
Integrals of the form (ax + b)n
A bracket raised to a high power is hard to expand, so we do not expand it. Use the substitution u = ax + b, so du/dx = a and dx = du/a.
∫ (ax + b)n dx = (ax + b)n+1a(n + 1) + c, where n ≠ −1.
Example (like textbook Example 6–7, p. 89)
∫ (3x + 5)⁵ dx = (3x + 5)⁶3 × 6 + c = (3x + 5)⁶18 + c.∫ 1(5x − 3)⁵ dx = ∫ (5x − 3)−5 dx = (5x − 3)−45 × (−4) + c = −120(5x − 3)⁴ + c.
Forgetting to divide by a, the coefficient of x inside the bracket. ∫ (2x − 5)⁴ dx = (2x − 5)⁵10 + c, not (2x − 5)⁵5 + c.
Find ∫ (4x + 1)³ dx.
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(4x + 1)⁴4 × 4 + c = (4x + 1)⁴16 + c
Equation of a curve from its gradient function
Integrate the gradient function dy/dx to get y. Then use a point on the curve to find c.
- y = ∫ (dy/dx) dx, with + c.
- Substitute the known x and y.
- Solve for c and write the equation.
Example (textbook p. 90)
dy/dx = 15x² + 4x − 3 and the curve passes through (−1, 2).y = 5x³ + 2x² − 3x + c. At x = −1, y = 2: 2 = −5 + 2 + 3 + c, so c = 2.
The curve is y = 5x³ + 2x² − 3x + 2. When x = 1, y = 5 + 2 − 3 + 2 = 6.
The same method works for rates: if ds/dt is given and s is known at one time, integrate and find c from that time.
dy/dx = 6x − 4 and y = 3 when x = 1. Find y when x = 2.
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y = 3x² − 4x + c. 3 = 3 − 4 + c, so c = 4. When x = 2: y = 12 − 8 + 4 = 8.
3.3Definite integral
Definite integral: if ∫ f(x) dx = g(x) + c, then ∫ab f(x) dx = [g(x)]ab = g(b) − g(a). The constant c cancels, so it is not written.
Example (like textbook Example 10, p. 92)
Evaluate ∫13 (2x + 1) dx.= [x² + x]13 = (9 + 3) − (1 + 1) = 10.
Properties
| Property | In words |
|---|---|
| ∫aa f(x) dx = 0 | Same limits give 0 |
| ∫ab f(x) dx = −∫ba f(x) dx | Swapping the limits changes the sign |
| ∫ab k f(x) dx = k ∫ab f(x) dx | A constant multiple comes outside |
| ∫ac f(x) dx + ∫cb f(x) dx = ∫ab f(x) dx | Joining two intervals |
| ∫ab [f(x) ± g(x)] dx = ∫ab f(x) dx ± ∫ab g(x) dx | Add or subtract integrals |
Example (like textbook Self-Exercise 3.5, p. 94): find an unknown
Given ∫25 f(x) dx = 12 and ∫25 [h f(x) − 3] dx = 51, find h.12h − [3x]25 = 51, so 12h − 9 = 51 and h = 5.
A constant inside an integral still integrates to a term in x: ∫25 3 dx = 3(5 − 2) = 9, not 3.
Given ∫15 f(x) dx = 7 and ∫13 f(x) dx = 3, find ∫35 f(x) dx.
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7 − 3 = 4
Area under a curve
The area under y = f(x) from x = a to x = b is cut into n thin strips of width δx. The strip areas add up to Σ yi δx. As the strips become infinitely thin (n → ∞, δx → 0) the sum becomes the exact area:
Area = lim Σ yi δx = ∫ab y dx
Regions above and below the x-axis
- A region above the x-axis gives a positive integral; below gives a negative integral.
- An area is always positive. Take the size (ignore the minus sign) of each part.
- If a region lies on both sides, find each part separately, then add the sizes.
Example (textbook p. 98, Example 15 method): total area
For y = x² − 3x: A = |∫03 (x² − 3x) dx| = |x³3 − 3x²2|03 = |9 − 13.5| = 92.B = ∫34 (x² − 3x) dx = 116.
Total area = 92 + 116 = 193 units². The single integral from 0 to 4 would give −83, which is wrong for the area.
Integrating straight across the x-axis. The positive and negative parts cancel and the answer is not the area.
Area between a curve and the y-axis
For a region between a curve x = g(y) and the y-axis from y = a to y = b, the area is ∫ab x dy. Write x in terms of y first. A region to the right of the y-axis is positive; to the left it is negative (use the size).
Area between a curve and a line, or two curves
- Solve the two equations together to find where they meet. These give the limits.
- Area = ∫ (upper curve − lower curve) dx between the limits.
A quick sketch tells you which curve is on top. Check by testing one x value between the limits.
Find the area under y = 3x² from x = 1 to x = 2.
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[x³] from 1 to 2 = 8 − 1 = 7 units²
Volume of revolution
Turn a region through 360° about an axis. Cut the solid into thin discs (cylinders) of thickness δx. Each has volume π y² δx, so the total volume is the limit of the sum.
About the x-axis: V = ∫ab π y² dx
About the y-axis: V = ∫ab π x² dy
About the y-axis: V = ∫ab π x² dy
- Square y (or x) before integrating. (x + 1)² must be expanded or integrated as (ax + b)n.
- Volume is always positive, so give the answer in terms of π.
- For a region between a line and a curve revolved, find the volume for each, then subtract. Do not square the difference of the heights.
Using ∫ π (y₁ − y₂)² dx for the region between two curves. The correct form is π ∫ (y₁² − y₂²) dx.
Find the volume when the region under y = √x from x = 0 to x = 4 is revolved about the x-axis.
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V = π ∫04 x dx = πx²2 from 0 to 4 = 8π units³
3.4Applications of integration
Integration solves problems on areas, volumes, and quantities built up from a rate (growth, depreciation, flow, production). Use the 4-step approach from the textbook (p. 111): understand, plan, carry out, check.
Example: filling a tank
Water flows into a tank at dV/dt = 6t + 4 litres per minute. The tank is empty at t = 0.V = ∫ (6t + 4) dt = 3t² + 4t + c, and c = 0 because V = 0 at t = 0.
Time for 64 litres: 3t² + 4t − 64 = 0, (3t + 16)(t − 4) = 0, so t = 4 minutes (the negative root is rejected).
Check the answer is sensible: a time must be positive, a volume or area must be positive, and units must match (cm² per second times seconds gives cm²).
The cross-section of a bowl is the curve y = ax², with width 12 cm and depth 2 cm. Find a.
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The rim point is (6, 2): 2 = a(36), so a = 118.