Definite integrals and areas

Maths: Analysis and Approaches · Calculus. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A definite integral has limits and gives a number, not a function. Integrate, substitute the upper limit, substitute the lower limit, and subtract. The constant C cancels, so it is left out.

The definite integral measures the area between a curve and the x-axis, but with a sign: regions above the axis count as positive and regions below count as negative. For a real area, find where the curve crosses the axis, integrate each part separately and add the sizes.

For the area between two curves, first find where they meet. These x-values are the limits. Then integrate the upper curve minus the lower curve.

Rules to remember

∫_a^b f(x) dx = F(b) − F(a), where F′(x) = f(x)
∫_a^b k f(x) dx = k∫_a^b f(x) dx, and the integral of a sum is the sum of the integrals
∫_a^b k dx = k(b − a) for a constant k
Area between a curve and the x-axis = ∫_a^b |f(x)| dx
Below the axis the integral is negative: take its size as the area
Area between two curves = ∫_a^b (upper − lower) dx, with a and b where the curves meet
An area is never negative

Common mistake

Integrating straight across a point where the curve crosses the x-axis. The positive and negative parts cancel and the answer is too small, so split the integral at the crossing point.

Worked example

Find the area enclosed between y = 3x − x² and y = x.

They meet where 3x − x² = x, so 2x − x² = 0, x(2 − x) = 0, x = 0 or x = 2
At x = 1 the curve gives 2 and the line gives 1, so the curve is the upper one
Area = ∫₀² (3x − x² − x) dx = ∫₀² (2x − x²) dx = [x² − x³/3] from 0 to 2
= (4 − 8/3) − 0 = 4/3

Answer: 4/3 square units

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Topic 5: Calculus

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