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Evaluate a definite integral of sine using the Fundamental Theorem of Calculus

Use the Fundamental Theorem of Calculus, Part 2, to evaluate the definite integral ∫0πsin⁡(x) dx.\int_0^{\pi} \sin(x)\, dx. Give the exact value.

Hint

Recall that an antiderivative of sin⁡(x)\sin(x) is −cos⁡(x)-\cos(x). Apply ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) using F(x)=−cos⁡(x)F(x) = -\cos(x), being careful with signs when substituting the limits.

Step-by-step solution

Since ddx[−cos⁡(x)]=sin⁡(x)\frac{d}{dx}[-\cos(x)] = \sin(x), an antiderivative of sin⁡(x)\sin(x) is F(x)=−cos⁡(x)F(x) = -\cos(x). By FTC Part 2, ∫0πsin⁡(x) dx=F(π)−F(0)=−cos⁡(x)∣0π.\int_0^{\pi} \sin(x)\,dx = F(\pi) - F(0) = -\cos(x)\Big|_0^{\pi}.

Substituting the limits gives −cos⁡(π)−(−cos⁡(0))=−(−1)−(−1)=1+1=2-\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2. Therefore ∫0πsin⁡(x) dx=2\int_0^{\pi} \sin(x)\,dx = 2.

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