Question

The area of the region bounded by the y-axis, $y = \cos x, y = \sin x$, when $0 \leq x \leq \frac{\pi}{4}$, is

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

Answer 1

Answer: (C)

$(\sqrt{2}-1)$

Answer 2

The area of the region bounded by the y-axis, $y = \cos x$, and $y = \sin x$, when $0 \leq x \leq \frac{\pi}{4}$, is given by the integral of the difference between the upper curve and the lower curve over the given interval.

In the interval $[0, \frac{\pi}{4}]$, we have $\cos x \geq \sin x$. The y-axis corresponds to $x=0$. The upper boundary is $y = \cos x$ and the lower boundary is $y = \sin x$. The interval for $x$ is $[0, \frac{\pi}{4}]$.

The area $A$ is given by the definite integral: $A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx$

Now, we evaluate the integral: $\int (\cos x - \sin x) dx = \int \cos x dx - \int \sin x dx = \sin x - (-\cos x) + C = \sin x + \cos x + C$

Applying the limits of integration: $A = [\sin x + \cos x]_{0}^{\frac{\pi}{4}}$ $A = (\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) - (\sin(0) + \cos(0))$ $A = (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) - (0 + 1)$ $A = (\frac{2\sqrt{2}}{2}) - 1$ $A = \sqrt{2} - 1$

The area of the region is $(\sqrt{2}-1)$ square units.