Question

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

• A. $4(\sqrt2 - 1) $
• B. $2\sqrt2(\sqrt2 - 1) $
• C. $2(\sqrt2 + 1) $
• D. $2\sqrt2(\sqrt2 + 1) $

Answer 1

The Correct Option is B

To find the bounded area between y = f(x) and y = g(x)
between x= a to x = b.
$ \therefore Area bounded = \int \limits_a^c [g(x) - f(x)] dx + \int \limits_c^b [f(x) - g(x)] dx $
$\ \ \ \ \ \ = \int \limits_a^b |f(x) - g(x)| dx $
and $\ \ \ \ \ \ \ \ g(x) = y = |cos x - sin x| $
Here, $\ \ \ \ \ \ f(x) = y = sin x + cos x , when\\0 \le x \le \frac{\pi}{2} $
$ \bigg \{ \begin{array}
\ cos x - sin x, 0 \le x \le \frac{\pi}{4} \\
sin x - cos x, \frac{\pi}{4} \le x \le \frac{\pi}{2} \\
\end{array}$
could be shown as
$ \therefore Area \ bounded = \int \limits_0^{\pi/4} {(sin x + cos x) - (cos x - sinn x)} dx$
$\ \ \ \ \ + \int \limits_{\pi/4}^{\pi/2} {(sin x + cos x) - (sin x - cos x)} dx $
$ \ \ \ \ \ \ \ = \int \limits_0^{\pi/4} 2 sin x dx + \int \limits_{\pi/4}^{\pi/2} 2 cos x dx $
$ = -2 [cos x]_0^{\pi/4} + 2[sin x. n]_{\pi/4}^{\pi/2} $
$ 4 - 2 \sqrt2 = 2 \sqrt2 (\sqrt2 - 1)$ sq units