The area bounded by the curve \( y=\tan x,\ x=-\displaystyle\frac{\pi}{6},\ x=\frac{\pi}{6} \) and \( y=0 \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \log_e\displaystyle\frac{4}{3} \) Sq. units
tan x is negative on \( \left(-\frac{\pi}{6},0\right) \) and positive on \( \left(0,\frac{\pi}{6}\right) \). Since tan x is an odd function, the two parts have equal areas, so
\( \displaystyle \text{Area}=2\int_{0}^{\pi/6}\tan x\,dx \)
\( \displaystyle =2\Big[\log_e|\sec x|\Big]_{0}^{\pi/6} \)
\( \displaystyle =2\left(\log_e\frac{2}{\sqrt{3}}-\log_e 1\right) \)
\( \displaystyle =2\log_e\frac{2}{\sqrt{3}} \)
\( \displaystyle =\log_e\left(\frac{2}{\sqrt{3}}\right)^2 \)
\( \displaystyle =\log_e\frac{4}{3} \)
Why the other options are incorrect
Option (2) \( \log_e\displaystyle\frac{2}{3} \) Sq. units — Incorrect. \( \log_e\frac{2}{3} \) is negative (since \( \frac{2}{3}\lt 1 \)), so it cannot be an area.
Option (3) \( \log_e\displaystyle\frac{2}{\sqrt{3}} \) Sq. units — Incorrect. \( \log_e\frac{2}{\sqrt{3}} \) is the area of only one of the two parts, from 0 to π/6.
Option (4) \( \log_e 2 \) Sq. units — Incorrect. The area is \( \log_e\frac{4}{3} \), not \( \log_e 2 \).