Definite integrals and areas

NUST NET Maths · Maths · Integration. 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: integrate, substitute the upper limit, then subtract the value at the lower limit. No constant c is needed. Its properties often give the answer with almost no working, so look at the limits and the symmetry of the function before integrating.

The integral gives area with a sign: parts of the curve below the x-axis count as negative. For a real area, split the interval where the curve crosses the axis and add the sizes. Between two curves, integrate upper minus lower between their intersection points.

A separable differential equation is solved by moving all y terms to one side and all x terms to the other, then integrating both sides.

Rules to remember

∫_a^b f(x) dx = F(b) − F(a), where F′(x) = f(x)
∫_b^a f dx = −∫_a^b f dx, and ∫_a^b f dx = ∫_a^c f dx + ∫_c^b f dx
∫_−a^a f dx = 0 if f is odd, and = 2 ∫₀^a f dx if f is even
∫₀^a f(x) dx = ∫₀^a f(a − x) dx; hence ∫₀^π/2 sin² x dx = ∫₀^π/2 cos² x dx = π/4
Area between two curves = ∫ (upper − lower) dx between the points where they meet
Shortcut: the area between a parabola y = ax² + bx + c and a straight line (or the x-axis) that meet at x = α and x = β is |a|(β − α)³/6
Separable equation: write g(y) dy = f(x) dx and integrate both sides with one constant; dy/dx = ky gives y = Ae^kx

Common mistake

Integrating straight across a point where the curve crosses the x-axis when the question asks for area. The positive and negative parts cancel, so split the integral at the crossing and add the sizes of the parts.

Worked example

Find the area enclosed between the curve y = x² and the line y = x + 2.

They meet where x² = x + 2: x² − x − 2 = (x − 2)(x + 1) = 0, so x = −1 and x = 2.
Area = ∫₋₁² (x + 2 − x²) dx = [x²/2 + 2x − x³/3] from −1 to 2.
= (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 10/3 − (−7/6) = 27/6 = 9/2.
Shortcut check: |a|(β − α)³/6 = 1 × 3³/6 = 27/6 = 9/2.

Answer: 9/2 square units

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