/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Evaluate the definite integral. ... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

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

Step by step solution

01

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 \(dv = \sin2x dx\).
02

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 \(v = -\frac{1}{2}\cos2x\).
03

Apply the formula of Integration by Parts

Substitute \(u\), \(v\), and \(du\) into the integration by parts formula. We get \(\int x \sin2x dx = x(-\frac{1}{2}\cos2x) - \int -\frac{1}{2}\cos2x dx = -\frac{1}{2}x\cos2x + \frac{1}{4}\sin2x + C\).
04

Evaluate the definite integral

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

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