The line $x = \frac{\pi}{4}$ divides the area of the region bounded by $y = \sin{x,\ }\ y = \cos x\ and\$ $x$-axis $\left( 0 \leq x \leq \frac{x}{2} \right)$ into two regions of areas $A_{1}\ and\ A_{2}$. Then $A_{1}:A_{2}$ equals

Question

Dr. Abubacker · Accepted Answer

(d)
$Area,\ \ A_{1} = \int_{0}^{\frac{\pi}{4}}{\sin{x\ dx}}$
[image:8_image60.jpeg]
$= - {\lbrack\cos{xbrack}}_{0}^{\pi/4}$
$= 1 - \frac{1}{\sqrt{2}}$
$= \frac{\sqrt{2} - 1}{\sqrt{2}}$
$and\ area,\ A_{2} = \int_{\pi/4}^{\pi/2}{\cos{x\ dx}}$
$= {\lbrack\sin{xbrack}}_{\pi/4}^{\pi/2} = \left\lbrack 1 - \frac{1}{\sqrt{2}} ightbrack = \frac{\sqrt{2} - 1}{\sqrt{2}}$
$	herefore\ A_{1}:A_{2} = \frac{\sqrt{2} - 1}{\sqrt{2}}:\frac{\sqrt{2} - 1}{\sqrt{2}} = 1:1$

Q. The line $x = \frac{\pi}{4}$ divides the area of the region bounded by $y = \sin{x,\ }\ y = \cos x\ and\$ $x$-axis $\left( 0 \leq x \leq \frac{x}{2} \right)$ into two regions of areas $A_{1}\ and\ A_{2}$. Then $A_{1}:A_{2}$ equals

Explanation

Reviewed by Dr. Abubacker

Want more practice?