The value of \(\int_{0}^{\frac{\pi}{6}}\displaystyle\cos 3x\,dx\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{3}\)
An antiderivative of \(\cos 3x\) is \(\displaystyle\frac{\sin 3x}{3}\). So
\( \displaystyle \int_{0}^{\pi/6}\cos 3x\,dx = \left[\frac{\sin 3x}{3}\right]_{0}^{\pi/6} \)
\( \displaystyle = \frac{1}{3}\sin\frac{\pi}{2} - \frac{1}{3}\sin 0 \)
\( \displaystyle = \frac{1}{3}(1) - 0 = \frac{1}{3} \)
Why the other options are incorrect
Option (1) 1 — Incorrect. It leaves out the factor \(\frac{1}{3}\) that comes from the 3 inside \(\cos 3x\).
Option (3) \(-\displaystyle\frac{1}{3}\) — Incorrect. The antiderivative of cos is +sin, not −sin; also \(\cos 3x \gt 0\) on \((0, \frac{\pi}{6})\), so the value is positive.
Option (4) 3 — Incorrect. It multiplies by 3 instead of dividing by 3.