KCET · Maths · Area Under Curves
In the interval \((0, \pi / 2)\) area lying between the curves \(y=\tan x\) and \(y=\cot x\) and the \(X\)-axis is
- A \(2 \log 2\) sq units
- B \(4 \log 2\) sq units
- C \(\log 2\) sq units
- D \(3 \log 2\) sq units
Answer & Solution
Correct Answer
(C) \(\log 2\) sq units
Step-by-step Solution
Detailed explanation
Given, \(c_1 \rightarrow y=\tan x, x \in\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]\)
\(c_2 \rightarrow y=\cot x \in\left[\frac{\pi}{6}, \frac{\pi}{6}\right]\)
The region is symmetric about lines \(x=\frac{\pi}{4}\) and its first portion is bounded between the lines \(x=0\) and \(x=\frac{\pi}{4}\)
\(\therefore\) Required Area \(=2 \int_0^{\pi / 4} \tan x d y\)
\(\begin{aligned} & =2[\log (\sec x)]_0^{\frac{\pi}{4}} \\ & =2\left[\log \left(\sec \frac{\pi}{4}\right)-\log (\sec 0)\right] \\ & =2 \log \sqrt{2}=\log 2 \text { sq units. }\end{aligned}\)
\(c_2 \rightarrow y=\cot x \in\left[\frac{\pi}{6}, \frac{\pi}{6}\right]\)
The region is symmetric about lines \(x=\frac{\pi}{4}\) and its first portion is bounded between the lines \(x=0\) and \(x=\frac{\pi}{4}\)
\(\therefore\) Required Area \(=2 \int_0^{\pi / 4} \tan x d y\)
\(\begin{aligned} & =2[\log (\sec x)]_0^{\frac{\pi}{4}} \\ & =2\left[\log \left(\sec \frac{\pi}{4}\right)-\log (\sec 0)\right] \\ & =2 \log \sqrt{2}=\log 2 \text { sq units. }\end{aligned}\)
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