The area of the region (in square units) above the x -axis bounded by the curve y = ,0 ≤ x ≤ (π)/(2) and the tangent to the curve at x = (π)/(4) & is&

Correct answer: A. (1)/(2) ( 2 - (1)/(2)) Required Area = ∫ _ 0 ^ (π)/(4) ( x)dx - (Areaoftriangleformedbytangent, x = (π)/(4)andx - axis) = [ (sec(π)/(4)) - ( 0)] - (1)/(2) × 1((π)/(4) - x _ 0 ) = log 2 - (1)/(4) (where x _ 0 = (π)/(4) - (1)/(2) is x -interept of tangent) = (1)/(2) 2 - (1)/(4)

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 region (in square units) above the x -axis bounded...

The area of the region (in square units) above the

x

-axis bounded by the curve

y = \text\tan\text{x}\text{,}0 \le x \le \dfrac{π}{2}

and the tangent to the curve at

x = \dfrac{\pi}{4}

& is&

The area of the region (in square units) above the x -axis bounded by the curve y = ,0 ≤ x ≤ (π)/(2) and the tangent to

Options

A $\dfrac{1}{2} \left(\text{\log 2} - \dfrac{1}{2}\right)$
B $\dfrac{1}{2} \left(1 + \text{\log 2}\right)$
C $\dfrac{1}{2} \left(1 - \text{\log 2}\right)$
D $\dfrac{1}{2} \left(\text{\log 2} + \dfrac{1}{2}\right)$

Correct answer

A. $\dfrac{1}{2} \left(\text{\log 2} - \dfrac{1}{2}\right)$

Step-by-step solution

Required Area

= \mathop{∫}_{0}^{\dfrac{\pi}{4}}\left(\tan x\right)dx - \left(Areaoftriangleformedbytangent, x = \dfrac{\pi}{4}andx - axis\right)

= \left[\log \left(sec\dfrac{\pi}{4}\right) - \log \left(\sec 0\right)\right] - \dfrac{1}{2} \times 1\left(\dfrac{\pi}{4} - {x}_{0}\right)

= log \sqrt{2} - \dfrac{1}{4}

(where

{x}_{0} = \dfrac{\pi}{4} - \dfrac{1}{2}

x

-interept of tangent)

= \dfrac{1}{2}\log 2 - \dfrac{1}{4}

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