Problem 32 - Essential Calculus

Question

Evaluate the definite integral. Use a graphing utility to confirm your result. $$ \int_{0}^{\pi} x \sin 2 x d x $$

Step-by-Step Solution

Answer

The value of the integral is \(-\frac{\pi}{2}\).

Step 1: Identify the functions for Integration by Parts

According to the formula of Integration by parts, \(\int u dv = uv - \int v du\). Identify the two functions in the integral where, typically, \(u\) is chosen to be a function that simplifies when differentiated, and \(dv\) is a function that simplifies when integrated. Here, let \(u = x\) and \(dv = \sin2x dx\).

Step 2: Find du and v

Differentiate \(u\) and integrate \(dv\). Differentiating \(x\) with respect to \(x\) gives \(du = dx\). Integrating \(\sin2x\) with respect to \(x\) gives \(v = -\frac{1}{2}\cos2x\).

Step 3: Apply the formula of Integration by Parts

Substitute \(u\), \(v\), and \(du\) into the integration by parts formula. We get \(\int x \sin2x dx = x(-\frac{1}{2}\cos2x) - \int -\frac{1}{2}\cos2x dx = -\frac{1}{2}x\cos2x + \frac{1}{4}\sin2x + C\).

Step 4: Evaluate the definite integral

Substitute the limits of integration, that are \(0\) and \(\pi\), into the antiderivative. Upon substituting these values, we get \(-\frac{1}{2}\pi\cos2\pi + \frac{1}{4}\sin2\pi - (-\frac{1}{2}*0*\cos2*0 + \frac{1}{4}\sin2*0) = -\frac{\pi}{2}\).