The area of the closed region bounded by y = ^ -1 x, y = cosec^ - 1 x and the line x - 1 = 0 is

Correct answer: A. \ _ e (3 + 2 2 ) - (π)/(2)\ sq . units Integrating along y -axis, we get A = 2 ∫ _ 0 ^ (π)/(4) (sec y - 1)dy = 2 [ _ e | y + tan y| - y] _ 0 ^ (π)/(4) = 2[ _ e | 2 + 1| - (π)/(4)] = \ _ e (3 + 2 2 ) - (π)/(2)\ sq. units

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Area Under Curves

NTA Abhyas 2020 · Mathematics · Area Under Curves

The area of the closed region bounded by y = ^ -1 x, y = cosec^ - 1...

The area of the closed region bounded by

y = \sec^{-1} x, y = cosec^{- 1}x

and the line

x - 1 = 0

Options

A $\left\{\log_{e} \left(3 + 2\sqrt{2}\right) - \dfrac{\pi}{2}\right\}sq . units$
B $\left\{\dfrac{\pi}{2} - \log_{e} \left(3 + 2\sqrt{2}\right)\right\}sq . units$
C $\pi - 3\log_{\mathrm{e}} 3sq. units$
DNone of these

Correct answer

A. $\left\{\log_{e} \left(3 + 2\sqrt{2}\right) - \dfrac{\pi}{2}\right\}sq . units$

Step-by-step solution

Integrating along

y

-axis, we get

A = 2\mathop{∫}_{0}^{\dfrac{\pi}{4}}\left(sec y - 1\right)dy

= 2{\left[\log_{\mathrm{e}}\left|\sec y + tan y\right| - y\right]}_{0}^{\dfrac{\pi}{4}}

= 2\left[\log_{\mathrm{e}} \left|\sqrt{2} + 1\right| - \dfrac{\pi}{4}\right] = \left\{\log_{\mathrm{e}} \left(3 + 2\sqrt{2}\right) - \dfrac{\pi}{2}\right\}sq. units

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