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Chapter 7: Problem 31

Evaluate the following definite integrals. $$\int_{0}^{\pi} x \sin x d x$$

Short Answer

Expert verified
Answer: The value of the definite integral is \(\pi\).

Step by step solution

01

Find the antiderivative of the integrand using integration by parts

Integration by parts is a technique used to integrate products of functions. It's based on the product rule of differentiation. It is given by the rule: $$\int u \, dv = uv - \int v\,du$$ Choose \(u\) and \(dv\): - \(u = x\), which means \(du = dx\) - \(dv = \sin{x}\,dx\), which means \(v = -\cos{x}\) Apply the integration by parts rule: $$\int x \sin{x} \,dx = \left(-x\cos{x} \right) - \left( \int -\cos{x} \,dx \right)$$ Now, integrate \(-\cos{x}\) with respect to \(x\): $$\int -\cos{x} \,dx = -\sin{x} + C$$ So, the combined antiderivative is: $$\int x \sin{x} \,dx = -x\cos{x} + \sin{x} + C$$
02

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that: $$\int_{a}^{b} f(x) \,dx = F(b) - F(a)$$ Where \(F(x)\) is the antiderivative of \(f(x)\). In our case: - \(f(x) = x\sin{x}\) - \(F(x) = -x\cos{x}+\sin{x}\) - The interval \([a, b] = [0, \pi]\) Now apply the theorem: $$\int_{0}^{\pi} x\sin{x} \,dx = F(\pi) - F(0)$$
03

Substitute the bounds and evaluate

Plug in the bounds \(\pi\) and \(0\) into \(F(x)\): $$F(\pi) = -\pi\cos{\pi} + \sin{\pi} = -\pi(-1) + 0 = \pi$$ $$F(0) = -0\cos{0} + \sin{0} = 0$$ Now, substitute these values into the equation: $$\int_{0}^{\pi} x\sin{x} \,dx = F(\pi) - F(0) = \pi - 0 = \pi$$ So, the value of the definite integral is \(\pi\).

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