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

The Correct Option is C

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.