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

Concept: The function \( |\sin \pi x| \) is periodic with period \(1\). On every interval \([n,n+1]\), \[ |\sin \pi x|=\sin(\pi(x-n)) \]
Step 1: Evaluate the integral over one unit interval \[ \int_0^1 x|\sin\pi x|\,dx =\int_0^1 x\sin(\pi x)\,dx =\frac{1}{\pi} \]
Step 2: Extend to \( r \in \mathbb{N} \) \[ \int_0^r x|\sin\pi x|\,dx=\frac{r^2}{2\pi} \]
Step 3: Substitute in the summation \[ \sqrt{\left|\pi\cdot\frac{r^2}{2\pi}\right|} =\frac{r}{\sqrt2} \] \[ \sum_{r=1}^{20}\frac{r}{\sqrt2} =\frac{1}{\sqrt2}\cdot\frac{20\cdot21}{2} =105\sqrt2 \] Final Answer: \[ \boxed{105\sqrt2} \]