The area bounded by curves \( y=\tan x \), \( x \)-axis and \( x=-\displaystyle\frac{\pi}{3} \) and \( x=\displaystyle\frac{\pi}{3} \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \log_{e}4 \) sq.units
For \( -\frac{\pi}{3}\lt x\lt 0 \), \( \tan x \) is negative, so that part of the region lies below the x-axis and its area must be counted as positive.
Since \( \tan x \) is an odd function, the two halves are equal:
\( \displaystyle \text{Area}=2\int_{0}^{\pi/3}\tan x\,dx \)
\( \displaystyle =2\Big[\log_{e}\sec x\Big]_{0}^{\pi/3} \)
\( \displaystyle =2\log_{e}2 \)
\( \displaystyle =\log_{e}4 \)
Why the other options are incorrect
Option (1) 0 sq.units — Incorrect. This is the signed integral \( \int_{-\pi/3}^{\pi/3}\tan x\,dx \), in which the two halves cancel; it is not an area.
Option (3) \( \log_{e}\sqrt{2} \) sq.units — Incorrect. \( \log_{e}\sqrt{2} \) is only one quarter of the area.
Option (4) \( \log_{e}2 \) sq.units — Incorrect. \( \log_{e}2 \) is the area on one side of the origin only.