Question

The area of the region bounded by the y-axis, y = cos x, y = sin x, when 0 ≤ x ≤$\frac {π}{4}$, 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) sq. units

Answer 2

The region is bounded by the y-axis on one side, so we integrate with respect to x.
The curve y = cos(x) is above the curve y = sin(x) in the given interval.
Let's denote the area as A. To calculate it, we can set up the following integral:
A = ∫[0, $\frac {π}{4}$] (cos(x) - sin(x)) dx
To evaluate this integral, we take the antiderivative of cos(x) and sin(x):
A = [sin(x) + cos(x)]|[0, $\frac {π}{4}$]
Now we substitute the limits of integration:
A = [sin$\frac {π}{4}$ + cos$\frac {π}{4}$] - [sin(0) + cos(0)] = [$\frac {√2}{2}$ + $\frac {√2}{2}$] - [0 + 1] = $\sqrt {2}$ - 1
Therefore, the area of the region bounded by the y-axis, y = cos(x), and y = sin(x) for 0 ≤ x ≤ π/4 is ($\sqrt 2$ - 1) square units.
Hence, the correct answer is option (C) ($\sqrt 2$- 1) sq. units.