Question

The value of \[ \sum_{r=1}^{20}\sqrt{\left|\pi\left(\int_0^r x|\sin \pi x|\,dx\right)\right|} \] is:

Hint:

Absolute trigonometric functions often simplify drastically when periodicity is exploited.

Answer 1

Correct Answer: 210

Step 1: Understanding the Concept:
Evaluate the integral $I_r = \int_{0}^{r} x |\sin \pi x| dx$ first, then simplify the summation.
Step 2: Detailed Explanation:
Let $I_r = \int_{0}^{r} x |\sin \pi x| dx$. Put $\pi x = t \implies dx = \frac{dt}{\pi}$.
$I_r = \frac{1}{\pi^2} \int_{0}^{r\pi} t |\sin t| dt = \frac{1}{\pi^2} \sum_{k=1}^{r} \int_{(k-1)\pi}^{k\pi} t |\sin t| dt$.
In the interval $[(k-1)\pi, k\pi]$, $|\sin t| = (-1)^{k-1} \sin t$.
$\int_{(k-1)\pi}^{k\pi} t (-1)^{k-1} \sin t dt = (-1)^{k-1} [-t\cos t + \sin t]_{(k-1)\pi}^{k\pi}$
$= (-1)^{k-1} [(-k\pi \cos k\pi) - (-(k-1)\pi \cos(k-1)\pi)]$
$= (-1)^{k-1} [k\pi (-1)^{k-1} + (k-1)\pi (-1)^{k-1}] = (2k-1)\pi$.
Summing over $k$: $\int_{0}^{r\pi} t |\sin t| dt = \pi \sum_{k=1}^{r} (2k-1) = \pi r^2$.
So, $I_r = \frac{1}{\pi^2} \cdot \pi r^2 = \frac{r^2}{\pi}$.
The term inside the summation is $\left\lfloor \sqrt{\pi \cdot \frac{r^2}{\pi}} \right\rfloor = \lfloor r \rfloor = r$.
Sum $= \sum_{r=1}^{20} r = \frac{20 \times 21}{2} = 210$.
Step 3: Final Answer:
The value of the sum is 210.