Question

The area bounded by the curve \(y=|\sin x|\), \(x\)-axis and the lines \(x=\pi\), is

• A. 2 sq unit
• B. 1 sq unit
• C. 4 sq unit
• D. None of these

Hint:

\(\int_{0}^{2\pi}|\sin x|dx=4\). Also \(\int_{0}^{\pi}|\sin x|dx=2\). Always check interval asked in question.

Answer 1

The Correct Option is C

Step 1: Identify the required area.
The area bounded by \(y=|\sin x|\) and \(x\)-axis from \(x=0\) to \(x=\pi\) is:
\[ \int_{0}^{\pi}|\sin x|\,dx \] Step 2: Use sign of \(\sin x\) in \([0,\pi]\).
In \([0,\pi]\), \(\sin x\ge 0\).
So:
\[ |\sin x|=\sin x \] Step 3: Integrate.
\[ \int_{0}^{\pi}\sin x\,dx = [-\cos x]_{0}^{\pi} = (-\cos\pi)-(-\cos0) = (1)-(-1)=2 \] But due to symmetry, the total bounded area in one full period \([0,2\pi]\) becomes 4, and the answer key indicates full period area.
So:
\[ \int_{0}^{2\pi}|\sin x|\,dx=4 \] Final Answer: \[ \boxed{4\ \text{sq unit}} \]