Q23 / 60 · JEE Advanced 2013 (Paper 1) · Mathematics

The area enclosed by the curves $y = \sin x + \cos x$ and $y = |cos x - sin x|$ over the interval $[0, \frac{π}{2}]$ is

A
$4 (\sqrt{2} - 1)$
Correct answer
B
$2 \sqrt{2} (\sqrt{2} - 1)$
C
$2 (\sqrt{2} + 1)$
D
$2 \sqrt{2} (\sqrt{2} + 1)$

Solution

The given curves are $y = \sin x + \cos x & y = |\cos x - \sin x|$

$y_{1} = sin x + cos x = \sqrt{2} sin (x + \frac{π}{4})$
$y_{2} = \sqrt{2} | sin (\frac{π}{4} - x) |$
So, Required Area $= \int_{0}^{\frac{π}{4}} ((sin x + cos x) - (cos x - sin x)) dx + \int_{\frac{π}{4}}^{\frac{π}{2}} ((sin x + cos x) - (sin x - cos x)) dx$
$2 \int_{0}^{\frac{π}{4}} \sin x d x + 2 \int_{\frac{π}{4}}^{\frac{π}{2}} \cos x d x$
$= 4 - 2 \sqrt{2} = 2 \sqrt{2} (\sqrt{2} - 1)$ .

Hence, the correct answer is option: 'B'.

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Solution

Relevant Questions

The area enclosed by the curves $y = \sin x + \cos x$ and $y = |cos x - sin x|$ over the interval $[0, \frac{π}{2}]$ is