Question

The value of \(S = \sum_{r=1}^{20} \sqrt{\pi \int_{0}^{r} x |\sin \pi x| dx}\) is :

• A. 210
• B. 201
• C. 120
• D. 102

Hint:

For integrals like \(\int_0^n x f(x) dx\) where \(f(x)\) is periodic with period 1, the value is often proportional to \(n^2\). Breaking the integral into unit intervals usually reveals a simple arithmetic progression.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves the integration of a periodic function multiplied by \(x\) over an interval \([0, r]\), followed by a summation. The function \(|\sin \pi x|\) has a period of 1.
Step 2: Key Formula or Approach:
We use the property of integrals of periodic functions and integration by parts:
\[ \int_0^r x f(x) dx = \sum_{k=0}^{r-1} \int_k^{k+1} x f(x) dx \] where \(f(x) = |\sin \pi x|\).
Step 3: Detailed Explanation: Let \(I = \int_k^{k+1} x |\sin \pi x| dx\). Let \(x = u + k\), then \(dx = du\):
\[ I = \int_0^1 (u+k) |\sin \pi(u+k)| du = \int_0^1 (u+k) \sin \pi u du \] since \(\sin \pi(u+k)\) is either \(\sin \pi u\) or \(-\sin \pi u\), but the absolute value ensures positivity.
\[ I = \int_0^1 u \sin \pi u du + k \int_0^1 \sin \pi u du \] Using integration by parts on the first term:
\[ \int_0^1 u \sin \pi u du = \left[ -u \frac{\cos \pi u}{\pi} \right]_0^1 + \int_0^1 \frac{\cos \pi u}{\pi} du = \left( \frac{1}{\pi} - 0 \right) + 0 = \frac{1}{\pi} \] For the second term:
\[ \int_0^1 \sin \pi u du = \left[ -\frac{\cos \pi u}{\pi} \right]_0^1 = \frac{1 - (-1)}{\pi} = \frac{2}{\pi} \] So, \(I = \frac{1}{\pi} + \frac{2k}{\pi} = \frac{2k+1}{\pi}\).
Now, the full integral for \(r\):
\[ \int_0^r x |\sin \pi x| dx = \sum_{k=0}^{r-1} \frac{2k+1}{\pi} = \frac{1}{\pi} [1 + 3 + 5 + \dots + (2r-1)] = \frac{r^2}{\pi} \] Substitute into the summation \(S\):
\[ S = \sum_{r=1}^{20} \sqrt{\pi \cdot \frac{r^2}{\pi}} = \sum_{r=1}^{20} r = \frac{20 \times 21}{2} = 10 \times 21 = 210 \] Step 4: Final Answer:
The value of \(S\) is 210.