<p>The value of <span class="math-tex">\(\sqrt{\pi\overset{2008}{\underset0{\int x\vert\sin\pi x\vert dx}}}\)</span> is equal to:</p>
<p style="display:inline">2008</p>
<p style="display:inline"><span class="math-tex">\(\sqrt{2008}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\pi \sqrt{2008}\)</span></p>
<p style="display:inline">1004</p>
Step-by-Step Solution
Key Concept: Use substitution πx = t to transform the integral, then exploit the periodicity of |sin t| to express the integral as a sum of integrals over periods [0, π], [π, 2π], etc., where |sin t| = sin t. The integral over each period [nπ, (n+1)π] yields the formula ∫[nπ to (n+1)π] t sin t dt = 2 + 2nπ.
<p>Put <span class="math-tex">$\pi x=t$</span><br />
<span class="math-tex">$\Rightarrow \quad d x=\frac{d t}{\pi}$</span><br />
<span class="math-tex">$I=\frac{1}{\pi} \frac{\pi}{\pi} \int_{0}^{2008 \pi} t|\sin t| d t=\frac{1}{\pi} \int_{0}^{2008 \pi} t|\sin t| d t$</span> ..... (i)<br />
<span class="math-tex">$I=\frac{1}{\pi} \int_{0}^{2008 \pi}(2008 \pi-t)|\sin t| d t$</span> .... (ii)<br />
(i) + (ii) <span class="math-tex">$\Rightarrow \quad 2 I=\frac{2008 \pi}{\pi} \int_{0}^{2008 \pi}|\sin t| d t$</span> <span class="math-tex">$=(2008)^{2} \cdot \int_{0}^{\pi}|\sin t| d t$</span><br />
I =(2008)<sup>2</sup>;<br />
Hence, here <span class="math-tex">$\sqrt{I}=2008$</span></p>
Correct Answer: A