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

Answer: (B)

$2\sqrt2(\sqrt2 - 1)$

Answer 2

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