Additional Mathematics – Calculus | e-Consult

Solution:

1. Antiderivative of \(\sin(2x)\) is \(-\frac{1}{2}\cos(2x)\).
   

   Hence

   \[

   \int_{\pi/6}^{\pi/3}\sin(2x)\,dx

   = \Bigl[-\tfrac12\cos(2x)\Bigr]_{\pi/6}^{\pi/3}.

   \]

2. Evaluate:

   \[

   -\tfrac12\cos\!\left(2\cdot\frac{\pi}{3}\right)

   + \tfrac12\cos\!\left(2\cdot\frac{\pi}{6}\right)

   = -\tfrac12\cos\!\left(\tfrac{2\pi}{3}\right)

   + \tfrac12\cos\!\left(\tfrac{\pi}{3}\right).

   \]

   \[

   \cos\!\left(\tfrac{2\pi}{3}\right) = -\tfrac12,\qquad

   \cos\!\left(\tfrac{\pi}{3}\right) = \tfrac12.

   \]

   Thus

   \[

   -\tfrac12(-\tfrac12)+\tfrac12(\tfrac12)

   = \tfrac14+\tfrac14 = \tfrac12.

   \]

3. Geometric meaning: The value \(\frac12\) represents the net signed area between the curve \(y=\sin(2x)\) and the \(x\)-axis from \(x=\pi/6\) to \(x=\pi/3\). Since \(\sin(2x)\) is positive on this interval, the signed area equals the actual area.
4. Answer: \(\displaystyle \frac12\) (square units, the area under \(y=\sin(2x)\) from \(\pi/6\) to \(\pi/3\))