Let $f(x) = e^{\sin^2(2\pi x)}$. Calculate $$\int_{0}^{1} f'(x) \, dx $$

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BlackTom AI · Accepted Answer

To approach this integral, recognize that the integrand is f'(x) for the given function f. When integrating a derivative over an interval, the Fundamental Theorem of Calculus applies: ∫ from a to b f'(x) dx = f(b) - f(a).

Here, a = 0 and b = 1, so we compute f(1) and f(0).

First, evaluate f at any x: f(x) = e^{sin^2(2πx)}. Since sin(2πk) = 0 for any integer k, both at x = 0 and x = 1 we have sin(2πx) = sin(0) = 0 and sin^2(0) = 0. Therefore f(0) = e^{0} = 1 and f(1) = e^{0} = 1.

Substituting into the Fundamental Theorem gives f(1) - f(0) = 1 - 1 = 0.

Intuitively, the exponential term is always positive, but its exponent sin^2(2πx) vanishes at both endpoints, and the net accumulation of the rate of change over the full interval cancels out, yielding zero net change over [0, 1].

ENG1090 - MUM S1 2025 Mock Final Exam

Let \(f(x) = e^{\sin^2(2\pi x)}\). Calculate $$\int_{0}^{1} f'(x) \, dx $$

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