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Evaluate a definite integral using an antiderivative

Consider the definite integral ∫−22(x3−4x) dx.\int_{-2}^{2} (x^3 - 4x)\, dx.

Use the Fundamental Theorem of Calculus (Part 2) to evaluate ∫−22(x3−4x) dx.\int_{-2}^{2} (x^3 - 4x)\, dx. Find an antiderivative F(x)F(x) of f(x)=x3−4xf(x) = x^3 - 4x, and compute F(2)−F(−2)F(2) - F(-2). What is the exact value of the integral?

Hint

Find an antiderivative term by term using the power rule for integration, then evaluate it at the upper limit and subtract its value at the lower limit: F(2)−F(−2)F(2) - F(-2).

Step-by-step solution

An antiderivative of f(x)=x3−4xf(x) = x^3 - 4x is F(x)=x44−2x2,F(x) = \frac{x^4}{4} - 2x^2, since F′(x)=x3−4xF'(x) = x^3 - 4x. By FTC Part 2, ∫−22(x3−4x) dx=F(2)−F(−2)\int_{-2}^{2}(x^3-4x)\,dx = F(2) - F(-2).

Compute F(2)=244−2(22)=4−8=−4F(2) = \frac{2^4}{4} - 2(2^2) = 4 - 8 = -4, and F(−2)=(−2)44−2(−2)2=4−8=−4F(-2) = \frac{(-2)^4}{4} - 2(-2)^2 = 4 - 8 = -4. So the integral equals F(2)−F(−2)=−4−(−4)=0F(2) - F(-2) = -4 - (-4) = 0.

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