Question

The line $x=\frac{\mathrm{\pi }}{4}$ divides the area of the region bounded by $y=\sin \left(x\right)$, $y=\cos \left(x\right)$ and $x$-axis $\left(0\le x\le \frac{\mathrm{\pi }}{2}\right)$ into two regions of areas $A_1$ and $A_2$. Then, $A_1:A_2$ equals to

• A. $4:1$
• B. $3:1$
• C. $2:1$
• D. $1:1$

Answer 1

The correct option is D

$1:1$

Explanation for the correct option:

Step-1 Area bounded by $\sin x$

The graph of the situation is given in the figure

Area $A_1={\int }_0^{\frac{\mathrm{\pi }}{4}}\sin \left(x\right)dx$

$$\begin{gathered} A_1={\int }_0^{\frac{\mathrm{\pi }}{4}}\sin \left(x\right)dx \\ ={\left|-\cos \left(x\right)\right|}_0^{\frac{\mathrm{\pi }}{4}} \\ =\left|\cos \left(\frac{\mathrm{\pi }}{4}\right)-\cos \left(0\right)\right| \\ =\left|\frac{1}{\sqrt{2}}-1\right| \\ {A}_1=1-\frac{1}{\sqrt{2}} \end{gathered}$$

Step-2 Area bounded by $\cos x$

Area $A_2={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\cos \left(x\right)dx$

$$\begin{gathered} A_2={\left|\sin \left(x\right)\right|}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}} \\ =\sin \left(\frac{\mathrm{\pi }}{2}\right)-\sin \left(\frac{\mathrm{\pi }}{4}\right) \\ =1-\frac{1}{\sqrt{2}} \end{gathered}$$

Step-3 Required ratio:

The ratio will be,

$$\begin{gathered} A_1:A_2=\frac{1-{\displaystyle \frac{1}{\sqrt{2}}}}{1-{\displaystyle \frac{1}{\sqrt{2}}}} \\ {A}_1:A_2=1:1 \end{gathered}$$

Thus, $A_1:A_2=1:1$.

Hence, option D is correct.