Definite Integration

This lesson covers definite integration — evaluating integrals between given limits to find exact numerical values — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to understand how to evaluate definite integrals, find areas under curves, and handle areas below the x-axis.

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What Is a Definite Integral?

A definite integral has upper and lower limits of integration. It produces a numerical value rather than a function.

The notation is:

∫ from a to b of f(x) dx

where a is the lower limit and b is the upper limit.

Evaluating a Definite Integral

To evaluate a definite integral:

1. Find the indefinite integral (antiderivative) of f(x) — call it F(x).
2. Evaluate F(b) - F(a).

This is written using square bracket notation:

∫ from a to b of f(x) dx = [F(x)] from a to b = F(b) - F(a)

Note: There is no constant of integration c when evaluating definite integrals. The constants would cancel: [F(x) + c] from a to b = (F(b) + c) - (F(a) + c) = F(b) - F(a).

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Worked Examples

Example 1: Evaluate ∫ from 1 to 3 of 2x dx

• Integrate: [x²] from 1 to 3
• Evaluate: (3)² - (1)² = 9 - 1 = 8

Example 2: Evaluate ∫ from 0 to 2 of (3x² + 4x - 1) dx

• Integrate: [x³ + 2x² - x] from 0 to 2
• Evaluate at x = 2: (8) + (8) - (2) = 14
• Evaluate at x = 0: (0) + (0) - (0) = 0
• Answer: 14 - 0 = 14

Example 3: Evaluate ∫ from 1 to 4 of (2√x + 3/x²) dx

• Rewrite: 2x^(1/2) + 3x⁻²
• Integrate: [(2)(2/3)x^(3/2) + 3(-1)x⁻¹] = [(4/3)x^(3/2) - 3/x] from 1 to 4
• At x = 4: (4/3)(8) - 3/4 = 32/3 - 3/4 = 128/12 - 9/12 = 119/12
• At x = 1: (4/3)(1) - 3/1 = 4/3 - 3 = 4/3 - 9/3 = -5/3
• Answer: 119/12 - (-5/3) = 119/12 + 20/12 = 139/12

Example 4: Evaluate ∫ from 0 to π of sin x dx

• Integrate: [-cos x] from 0 to π
• Evaluate: (-cos π) - (-cos 0) = (-(-1)) - (-(1)) = 1 + 1 = 2

Exam Tip: Be very careful with negative signs when substituting limits. Write out each step clearly — do not try to do it all in your head.

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The Fundamental Theorem of Calculus

The result we are using is the Fundamental Theorem of Calculus, which states:

If F'(x) = f(x), then ∫ from a to b of f(x) dx = F(b) - F(a)

This theorem provides the crucial link between differentiation and integration. It tells us that the definite integral of a function can be evaluated using any antiderivative of that function.

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Area Under a Curve

The definite integral ∫ from a to b of f(x) dx gives the signed area between the curve y = f(x), the x-axis, and the vertical lines x = a and x = b.

Area Above the x-axis

When f(x) ≥ 0 for all x in [a, b], the definite integral gives the actual area:

Area = ∫ from a to b of f(x) dx

Example: Find the area under y = x² between x = 0 and x = 3.

• Area = ∫ from 0 to 3 of x² dx = [x³/3] from 0 to 3 = 27/3 - 0 = 9

Area Below the x-axis

When f(x) ≤ 0 for all x in [a, b], the integral gives a negative value. The actual area is the absolute value:

Area = |∫ from a to b of f(x) dx| = -∫ from a to b of f(x) dx (when f(x) ≤ 0)

Example: Find the area enclosed between y = x² - 4 and the x-axis between x = -2 and x = 2.

Note that x² - 4 ≤ 0 for -2 ≤ x ≤ 2 (the curve is below the x-axis in this region).

• ∫ from -2 to 2 of (x² - 4) dx = [x³/3 - 4x] from -2 to 2
• At x = 2: 8/3 - 8 = 8/3 - 24/3 = -16/3
• At x = -2: -8/3 + 8 = -8/3 + 24/3 = 16/3
• Integral = -16/3 - 16/3 = -32/3
• Area = |-32/3| = 32/3

Exam Tip: When a question asks for the "area" (not the integral), always give a positive answer. If the curve is below the x-axis, the integral will be negative — take the absolute value.

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Areas Partly Above and Partly Below the x-axis

When a curve crosses the x-axis within the interval [a, b], you must split the integral at the x-intercepts and calculate each part separately.

Example: Find the total area enclosed between y = x(x - 2) and the x-axis for 0 ≤ x ≤ 3.

Step 1: Find x-intercepts — x(x - 2) = 0 gives x = 0 and x = 2.

Step 2: Determine where the curve is above/below the x-axis:

• For 0 ≤ x ≤ 2: y = x² - 2x ≤ 0 (below x-axis)
• For 2 ≤ x ≤ 3: y = x² - 2x ≥ 0 (above x-axis)

Step 3: Calculate each area separately:

Region 1 (below): ∫ from 0 to 2 of (x² - 2x) dx = [x³/3 - x²] from 0 to 2 = (8/3 - 4) - 0 = -4/3 Area₁ = 4/3

Region 2 (above): ∫ from 2 to 3 of (x² - 2x) dx = [x³/3 - x²] from 2 to 3 = (9 - 9) - (8/3 - 4) = 0 + 4/3 = 4/3 Area₂ = 4/3

Total area = 4/3 + 4/3 = 8/3

Critical Warning: If you simply calculate ∫ from 0 to 3 of (x² - 2x) dx = [x³/3 - x²] from 0 to 3 = 0, you get zero! The positive and negative parts cancel. This is why you MUST split at x-intercepts when finding areas.

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Properties of Definite Integrals

Several useful properties follow from the definition:

1. Reversing limits: ∫ from a to b of f(x) dx = -∫ from b to a of f(x) dx

2. Splitting the interval: ∫ from a to b of f(x) dx = ∫ from a to c of f(x) dx + ∫ from c to b of f(x) dx

3. Zero width: ∫ from a to a of f(x) dx = 0

4. Constant multiple: ∫ from a to b of kf(x) dx = k ∫ from a to b of f(x) dx

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Definite Integrals with Trigonometric and Exponential Functions

Example 1: Evaluate ∫ from 0 to π/2 of cos x dx

• Integrate: [sin x] from 0 to π/2
• Evaluate: sin(π/2) - sin(0) = 1 - 0 = 1

Example 2: Evaluate ∫ from 0 to 1 of eˣ dx

• Integrate: [eˣ] from 0 to 1
• Evaluate: e¹ - e⁰ = e - 1

Example 3: Evaluate ∫ from 1 to e of (1/x) dx

• Integrate: [ln|x|] from 1 to e
• Evaluate: ln(e) - ln(1) = 1 - 0 = 1

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Summary

• A definite integral ∫ from a to b of f(x) dx is evaluated as F(b) - F(a), where F is the antiderivative.
• No constant of integration is needed for definite integrals.
• The definite integral gives the signed area between the curve and the x-axis.
• If the curve is below the x-axis, the integral is negative — take the absolute value for the area.
• When a curve crosses the x-axis, split the integral at x-intercepts and add the absolute values of each part.
• Be careful with arithmetic when substituting limits, especially with fractions and negative values.

A-Level Deep Dive: Definite Integration

Spec mapping

Edexcel 9MA0 specification section 10 — Integration, sub-strands 10.1–10.5 covers the Fundamental Theorem of Calculus; evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves; understand and use integration as the limit of a sum (refer to the official specification document for exact wording). Definite integration is examined in 9MA0-01 (Pure Mathematics 1) and 9MA0-02 (Pure Mathematics 2) and underpins applied content in 9MA0-03 Section 7 (Mechanics — kinematics with variable acceleration) and 9MA0-03 Section 5 (Statistics — continuous probability density functions). The Fundamental Theorem of Calculus (FTC), $\int_a^b f(x)\,dx = F(b) - F(a)$∫ab​f(x)dx=F(b)−F(a) where $F'(x) = f(x)$F′(x)=f(x), is not listed in the Edexcel formula booklet — it must be applied from memory.

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Worked example with full mark scheme

Question (8 marks): Evaluate $\displaystyle \int_1^4 \left(3x^2 - \dfrac{2}{\sqrt{x}}\right)\,dx$∫14​(3x2−x​2​)dx, giving your answer in exact form.

Solution with mark scheme:

Step 1 — rewrite the integrand using fractional indices.

$\int_1^4 \left(3x^2 - 2x^{-1/2}\right)\,dx$∫14​(3x2−2x−1/2)dx

M1 — converting $\dfrac{2}{\sqrt{x}}$x​2​ to $2x^{-1/2}$2x−1/2 before integrating. Trying to integrate $\dfrac{2}{\sqrt{x}}$x​2​ directly without converting to fractional-index form is the most common reason candidates lose this opening mark.

Step 2 — integrate term by term using the power rule.

For $3x^2$3x2: $\int 3x^2\,dx = \dfrac{3x^3}{3} = x^3$∫3x2dx=33x3​=x3.

For $-2x^{-1/2}$−2x−1/2: $\int -2x^{-1/2}\,dx = \dfrac{-2x^{1/2}}{1/2} = -4x^{1/2} = -4\sqrt{x}$∫−2x−1/2dx=1/2−2x1/2​=−4x1/2=−4x​.

M1 — applying $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$∫xndx=n+1xn+1​ to at least one term correctly.

A1 — both antiderivative terms correct: $F(x) = x^3 - 4\sqrt{x}$F(x)=x3−4x​. (No $+ C$+C is required for a definite integral, but writing it does not lose marks.)

Step 3 — apply the FTC: substitute the upper and lower limits.

$\left[x^3 - 4\sqrt{x}\right]_1^4 = F(4) - F(1)$[x3−4x​]14​=F(4)−F(1)

M1 — correct setup of the bracket evaluation, with upper limit minus lower limit.

Step 4 — substitute the upper limit $x = 4$x=4.

$F(4) = 4^3 - 4\sqrt{4} = 64 - 4 \cdot 2 = 64 - 8 = 56$F(4)=43−44​=64−4⋅2=64−8=56.

A1 — correct evaluation at $x = 4$x=4.

Step 5 — substitute the lower limit $x = 1$x=1.

$F(1) = 1^3 - 4\sqrt{1} = 1 - 4 = -3$F(1)=13−41​=1−4=−3.

Step 6 — combine.

$F(4) - F(1) = 56 - (-3) = 59$F(4)−F(1)=56−(−3)=59

A1 — final exact answer $59$59, with substitution back at the end clearly displayed.

B1 — communication mark, awarded for the answer being given in exact form (an integer here, not a decimal approximation).

Total: 8 marks (M3 A3 B1, distributed as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The curve $C$C has equation $y = 6\sqrt{x} - x$y=6x​−x for $x \geq 0$x≥0.

(a) Find the $x$x-coordinates of the points where $C$C meets the $x$x-axis. (2)

(b) Find the exact area of the finite region bounded by $C$C and the $x$x-axis. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — setting $6\sqrt{x} - x = 0$6x​−x=0 and factorising as $\sqrt{x}(6 - \sqrt{x}) = 0$x​(6−x​)=0.
• A1 (AO1.1b) — solutions $x = 0$x=0 and $x = 36$x=36.

(b)

• M1 (AO3.1a) — recognising the area is $\displaystyle\int_0^{36}(6\sqrt{x} - x)\,dx$∫036​(6x​−x)dx with limits taken from part (a).
• M1 (AO1.1b) — integrating to $F(x) = 4x^{3/2} - \dfrac{x^2}{2}$F(x)=4x3/2−2x2​ (using $\int 6x^{1/2}\,dx = 4x^{3/2}$∫6x1/2dx=4x3/2).
• A1 (AO1.1b) — substituting limits: $F(36) = 4(216) - 648 = 864 - 648 = 216$F(36)=4(216)−648=864−648=216, $F(0) = 0$F(0)=0.
• A1 (AO2.5) — final exact area $216$216 square units, with units stated.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is a textbook synoptic 9MA0 question — solving an equation (section 2), then integrating (section 10), then interpreting geometrically as area.

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Synoptic links

Connects to:

1. Fundamental Theorem of Calculus (within section 10): the FTC $\int_a^b f(x)\,dx = F(b) - F(a)$∫ab​f(x)dx=F(b)−F(a) is the conceptual bridge between differentiation and integration. It says definite integration can be computed by first finding any antiderivative $F$F and evaluating at the endpoints — turning a "limit of a sum" into a two-point arithmetic problem.

2. Section 10.4 — area between two curves: $\int_a^b (f(x) - g(x))\,dx$∫ab​(f(x)−g(x))dx where $f(x) \geq g(x)$f(x)≥g(x) on $[a,b]$[a,b] generalises the basic "area under a curve" result. The trick: always integrate (upper curve − lower curve), then use the FTC as before.

3. Section 10.5 — trapezium rule: when $f$f has no elementary antiderivative (e.g. $\int_0^1 e^{-x^2}\,dx$∫01​e−x2dx), the trapezium rule $\int_a^b f(x)\,dx \approx \dfrac{h}{2}\left[f(x_0) + 2\sum f(x_i) + f(x_n)\right]$∫ab​f(x)dx≈2h​[f(x0​)+2∑f(xi​)+f(xn​)] provides a numerical approximation. Definite integration provides the exact value when the FTC is usable; the trapezium rule fills in when it is not.

4. 9MA0-03 Mechanics, section 7 — kinematics: displacement is the definite integral of velocity over time: $s = \int_{t_1}^{t_2} v(t)\,dt$s=∫t1​t2​​v(t)dt. A particle's net displacement from $t = 0$t=0 to $t = 5$t=5 when $v(t) = 3t^2 - 12t$v(t)=3t2−12t is computed using the FTC exactly as above.