Example 1: Using odd and even functions
Evaluate ∫−22 (x³ cos x + x²) dx.
x³ cos xis odd (odd × even), so its integral over[−2, 2]is 0x²is even, so its integral is2∫02 x² dx2 × [x³/3]02 = 2 × 8/3 = 16/3
Answer: 16/3
A definite integral is the limit of a sum, evaluated by the Fundamental Theorem of Calculus: ∫ab f(x) dx = F(b) − F(a). When you substitute u = g(x), change the limits to g(a) and g(b). Over [−a, a] an odd function integrates to 0 and an even function to twice its integral over [0, a].
∫ab f(x) dx = F(b) − F(a)∫ab f(x) dx = −∫ba f(x) dx∫ab = ∫ac + ∫cb∫−aa f(x) dx = 0∫−aa f(x) dx = 2∫0a f(x) dx∫ab u dv = [uv]ab − ∫ab v duEvaluate ∫−22 (x³ cos x + x²) dx.
x³ cos x is odd (odd × even), so its integral over [−2, 2] is 0x² is even, so its integral is 2∫02 x² dx2 × [x³/3]02 = 2 × 8/3 = 16/3Answer: 16/3
Evaluate ∫01 x / (x² + 1) dx.
u = x² + 1, du = 2x dxx = 0 → u = 1, x = 1 → u = 2∫12 ½ · (1/u) du = ½ [ln u]12= ½ ln 2Answer: ½ ln 2
u = g(x), the limits must be values of u. Keeping the x-limits while integrating in u gives a wrong number even when the antiderivative is right.−a and a, test each term for odd or even first. Examiners often include a term that looks impossible to integrate but is odd, so it contributes 0.How to study: Click any question below to instantly load it on the coordinate grid and check your steps.