Question

The value of $\sqrt{\pi \int_0^{2008} x|\sin \pi x|dx}$ is equal to

• A. $\sqrt{2008}$
• B. $\pi\sqrt{2008}$
• C. 1004
• D. 2008

Answer 1

Answer: (D)

2008

Answer 2

To find the value of $\sqrt{\pi \int_0^{2008} x|\sin \pi x|dx}$, we first need to evaluate the integral $I = \int_0^{2008} x|\sin \pi x|dx$.

The function $|\sin \pi x|$ is periodic with period 1. We can split the integral over the intervals $[n, n+1]$ for $n = 0, 1, \dots, 2007$. $I = \sum_{n=0}^{2007} \int_n^{n+1} x|\sin \pi x|dx$.

Let's consider the integral over a single interval $[n, n+1]$. Let $x = n+u$, so $dx = du$. When $x=n$, $u=0$. When $x=n+1$, $u=1$. $\int_n^{n+1} x|\sin \pi x|dx = \int_0^1 (n+u)|\sin \pi (n+u)|du$. We have $\sin \pi (n+u) = \sin(n\pi + \pi u) = \sin(n\pi)\cos(\pi u) + \cos(n\pi)\sin(\pi u) = (-1)^n \sin(\pi u)$. So, $|\sin \pi (n+u)| = |(-1)^n \sin(\pi u)| = |\sin(\pi u)|$. For $u \in [0, 1]$, $\pi u \in [0, \pi]$, so $\sin(\pi u) \ge 0$. Thus, $|\sin(\pi u)| = \sin(\pi u)$. The integral becomes $\int_0^1 (n+u)\sin(\pi u)du$. We can split this into two integrals: $\int_0^1 n\sin(\pi u)du + \int_0^1 u\sin(\pi u)du$.

First integral: $n\int_0^1 \sin(\pi u)du = n \left[-\frac{\cos(\pi u)}{\pi}\right]_0^1 = n \left(-\frac{\cos(\pi)}{\pi} - (-\frac{\cos(0)}{\pi})\right) = n \left(-\frac{-1}{\pi} + \frac{1}{\pi}\right) = n \left(\frac{1}{\pi} + \frac{1}{\pi}\right) = \frac{2n}{\pi}$.

Second integral: $\int_0^1 u\sin(\pi u)du$. We use integration by parts: $\int v dw = vw - \int w dv$. Let $v=u$ and $dw=\sin(\pi u)du$. Then $dv=du$ and $w=\int \sin(\pi u)du = -\frac{\cos(\pi u)}{\pi}$. $\int_0^1 u\sin(\pi u)du = \left[-\frac{u\cos(\pi u)}{\pi}\right]_0^1 - \int_0^1 \left(-\frac{\cos(\pi u)}{\pi}\right)du$ $= \left(-\frac{1\cos(\pi)}{\pi} - (-\frac{0\cos(0)}{\pi})\right) + \frac{1}{\pi}\int_0^1 \cos(\pi u)du$ $= \frac{1}{\pi} + \frac{1}{\pi}\left[\frac{\sin(\pi u)}{\pi}\right]_0^1$ $= \frac{1}{\pi} + \frac{1}{\pi^2}(\sin(\pi) - \sin(0)) = \frac{1}{\pi} + \frac{1}{\pi^2}(0 - 0) = \frac{1}{\pi}$.

So, $\int_n^{n+1} x|\sin \pi x|dx = \frac{2n}{\pi} + \frac{1}{\pi} = \frac{2n+1}{\pi}$.

Now, we sum this expression from $n=0$ to $n=2007$: $I = \sum_{n=0}^{2007} \frac{2n+1}{\pi} = \frac{1}{\pi} \sum_{n=0}^{2007} (2n+1)$. The sum $\sum_{n=0}^{2007} (2n+1) = 1 + 3 + 5 + \dots + (2 \times 2007 + 1) = 1 + 3 + 5 + \dots + 4015$. This is the sum of the first 2008 odd numbers. The sum of the first $N$ odd numbers is $N^2$. Here, $N=2008$. So the sum is $2008^2$.

So, $I = \frac{1}{\pi} \times 2008^2$.

The question asks for the value of $\sqrt{\pi I}$. $\sqrt{\pi I} = \sqrt{\pi \times \left(\frac{1}{\pi} \times 2008^2\right)} = \sqrt{2008^2}$. Since 2008 is a positive number, $\sqrt{2008^2} = 2008$.