Question

Area bounded by the curve $y=\sin x-\cos x$ in $[0,\frac{\pi }{2}]$ in sq. units is:

• A. 2
• B. $2\sqrt{2}$
• C. $2(\sqrt{2}-1)$
• D. $(\sqrt{2}-1)$

Answer 1

Answer: (C)

$2(\sqrt{2}-1)$

$y=\sin x-\cos x$ in $[0,\frac{\pi }{2}]$
$y=\sqrt{2}\sin (x-\pi /4)$
$\stackrel{\pi /4}{\underset{0}{\int }}|\sqrt{2}\sin (x-\pi /4)|dx+\stackrel{\pi /2}{\underset{\pi /4}{\int }}\sqrt{2}\sin (x-\pi /4)dx$
$=|\sqrt{2}[-\cos (x-\pi /4\left)\right]{|}_0^{\pi /4}+\sqrt{2}(-\cos (x-\pi /4{|}_{\pi /4}^{\pi /2})+c$
$\sqrt{2}|-(1-1/\sqrt{2})|+\sqrt{2}(-\frac{1}{\sqrt{2}}+1)$
$2\sqrt{2}(1-1/\sqrt{2})=2(\sqrt{2}-1)squnits.$