Question

If $A_1$ is the area bounded by $y=\cos x,y=\sin x$ & $x=0$ and $A_2$ the area bounded by $y=\cos x,y=\sin x,y=0$ in $(0,\frac{\pi }{2})$ then $\frac{A_1}{A_2}$ equals to :

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $1$
• D. None of these

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}$

$A_1={\int }_0^{\frac{\pi }{4}}(\cos x-\sin x)dx=(\sin x+\cos x{)}_0^{\frac{\pi }{4}}$

$=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1)=(\sqrt{2}-1)$

$A_2={\int }_{\frac{\pi }{4}}^0\sin xdx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cos xdx={[-\cos x]}_0^{\frac{\pi }{4}}+(\sin x{)}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$

$=(1-\frac{1}{\sqrt{2}})+(1-\frac{1}{\sqrt{2}})=(2-\sqrt{2})=\sqrt{2}(\sqrt{2}-1)$

Then, $A_1:A_2=1:\sqrt{2}=\frac{1}{\sqrt{2}}$