10 min readPathway sheet1-mark MCQ, 2-mark and 3-mark questions; every area question in the next chapter uses it

Definite integrals and the fundamental theorem of calculus

If F is any antiderivative of f, then ∫ f(x) dx from a to b = F(b) − F(a). The answer is a number, and no C is needed.

Second fundamental theoremIf f is continuous on [a, b] and F′(x) = f(x), then ∫ f(x) dx from a to b = [F(x)] from a to b = F(b) − F(a)

First fundamental theoremIf f is continuous on [a, b], the area function A(x) = ∫ f(t) dt from a to x has A′(x) = f(x) for every x in [a, b]

MeaningFor f(x) ≥ 0, ∫ f(x) dx from a to b is the area between y = f(x), the x-axis and the lines x = a and x = b

SubstitutionWhen you put t = g(x), change the limits to g(a) and g(b), and do not return to x

What a definite integral is

In ∫ f(x) dx from a to b, a is the lower limit and b is the upper limit. For a function with f(x) ≥ 0, the integral is the area under the curve y = f(x) from x = a to x = b. The area function A(x) = ∫ f(t) dt from a to x gives this area up to a moving line at x. The first fundamental theorem says A′(x) = f(x), so the area function is an antiderivative of f. The second fundamental theorem turns this into a method: find any antiderivative F and work out F(b) − F(a). Your textbook states both theorems without proof.

How to evaluate

1
Find an antiderivative F(x) by any method. Leave out C.
2
Write [F(x)] from a to b.
3
Put x = b, then x = a.
4
Subtract: F(b) − F(a). Simplify to a number.
5
If you substituted t = g(x), change the limits too: x = a gives t = g(a), and x = b gives t = g(b).

Indefinite vs definite integral

Indefinite: ∫ f(x) dxA family of functions F(x) + C. It needs + C, and the answer contains x.

Definite: ∫ f(x) dx from a to bA number, F(b) − F(a). There is no C: it cancels, since [F(b) + C] − [F(a) + C] = F(b) − F(a).

The letter used does not matter: ∫ f(x) dx from a to b = ∫ f(t) dt from a to b.

Example 1: direct use of the theorem

Problem

Evaluate ∫ (4x³ + cos x) dx from 0 to π/2.

Working

Antiderivative: F(x) = x⁴ + sin x.
F(π/2) = (π/2)⁴ + sin(π/2) = π⁴/16 + 1.
F(0) = 0 + sin 0 = 0.
Integral = F(π/2) − F(0) = π⁴/16 + 1.

Answer

π⁴/16 + 1

Example 2: substitution with new limits

Problem

Evaluate ∫ (tan⁻¹ x)/(1 + x²) dx from 0 to √3.

Working

Put t = tan⁻¹ x. Then dt = dx/(1 + x²).
Change the limits: x = 0 gives t = 0; x = √3 gives t = π/3.
The integral becomes ∫ t dt from 0 to π/3.
= [t²/2] from 0 to π/3 = (π/3)²/2 − 0 = π²/18.

Answer

π²/18

Example 3: partial fractions inside

Problem

Evaluate ∫ dx/((x + 1)(x + 2)) from 0 to 1.

Working

Partial fractions: 1/((x + 1)(x + 2)) = 1/(x + 1) − 1/(x + 2).
Antiderivative: log|x + 1| − log|x + 2| = log|(x + 1)/(x + 2)|.
At x = 1: log(2/3). At x = 0: log(1/2).
Subtract: log(2/3) − log(1/2) = log((2/3) × 2) = log(4/3).

Answer

log(4/3)

In real life

A home electricity meter keeps adding up the power used at each moment. The units on the bill (kWh) are the definite integral of power over the billing period.

Exam angle

1 marks: “If ∫ 3x² dx from 0 to a = 8, find a.” [x³] from 0 to a = a³ = 8, so a = 2.

2 marks: “Evaluate ∫ sin³ x cos x dx from 0 to π/2.” Put t = sin x, dt = cos x dx; the limits become 0 to 1. ∫ t³ dt from 0 to 1 = 1/4.

3 marks: “Evaluate ∫ x log x dx from 1 to e.” By parts with u = log x: [(x²/2) log x − x²/4] from 1 to e = (e²/2 − e²/4) − (0 − 1/4) = (e² + 1)/4.

Remember it

Upper minus lower, no + C. Change the limits when you change the variable.