Calculating the Area Enclosed Between Curves y=sin x and y=cos x

This problem asks us to find the area enclosed between the curves \(y = \sin x\) and \(y = \cos x\) over the interval \(0 \le x \le \frac{\pi}{2}\). This is a classic application of geometric application of integrals in calculus.

Finding Intersection Points for Area Enclosed Between Curves

First, we need to find where the curves \(y = \sin x\) and \(y = \cos x\) intersect within the given interval \(0 \le x \le \frac{\pi}{2}\). They intersect when \(\sin x = \cos x\). Dividing by \(\cos x\) (assuming \(\cos x \ne 0\)), we get \(\tan x = 1\). In the interval \(0 \le x \le \frac{\pi}{2}\), the only solution is \(x = \frac{\pi}{4}\). This point divides the interval into two parts, where the 'upper' curve might change, affecting the calculation of the area enclosed between the curves.

Determining Which Curve is Above in the Subintervals

We need to see which function has a larger value in each subinterval to correctly set up the integration for the area calculation.

• For \(0 \le x \le \frac{\pi}{4}\): Consider a point like \(x = \frac{\pi}{6}\). \(\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\) and \(\sin(\frac{\pi}{6}) = \frac{1}{2}\). Since \(\frac{\sqrt{3}}{2} > \frac{1}{2}\), we have \(\cos x \ge \sin x\) in this interval.
• For \(\frac{\pi}{4} \le x \le \frac{\pi}{2}\): Consider a point like \(x = \frac{\pi}{3}\). \(\cos(\frac{\pi}{3}) = \frac{1}{2}\) and \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\). Since \(\frac{\sqrt{3}}{2} > \frac{1}{2}\), we have \(\sin x \ge \cos x\) in this interval.

Setting up the Definite Integrals using Integration

The total area enclosed between the curves is the sum of the areas of the regions in each subinterval. The formula for the area between two curves \(f(x)\) and \(g(x)\) from \(a\) to \(b\) where \(f(x) \ge g(x)\) is \(\int_a^b (f(x) - g(x)) dx\). Using this concept from calculus and geometric application of integrals:

Total Area = Area from \(0\) to \(\frac{\pi}{4}\) + Area from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\)

Total Area \( = \int_0^{\pi/4} (\cos x - \sin x) dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) dx \)

Evaluating the Integrals and Finding the Total Area

Now we perform the integration for each part:

First integral:

\( \int_0^{\pi/4} (\cos x - \sin x) dx \)

The antiderivative of \(\cos x\) is \(\sin x\) and the antiderivative of \(-\sin x\) is \(\cos x\). So, the definite integral is:

\( [\sin x + \cos x]_0^{\pi/4} \)

Evaluating at the limits:

\( = (\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) - (\sin(0) + \cos(0)) \)

\( = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) \)

\( = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \)

Second integral:

\( \int_{\pi/4}^{\pi/2} (\sin x - \cos x) dx \)

The antiderivative of \(\sin x\) is \(-\cos x\) and the antiderivative of \(-\cos x\) is \(-\sin x\). So, the definite integral is:

\( [-\cos x - \sin x]_{\pi/4}^{\pi/2} \)

Evaluating at the limits:

\( = (-\cos(\frac{\pi}{2}) - \sin(\frac{\pi}{2})) - (-\cos(\frac{\pi}{4}) - \sin(\frac{\pi}{4})) \)

\( = (-0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) \)

\( = -1 - (-\frac{2}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1 \)

Total Area = (Result of first integral) + (Result of second integral)

Total Area \( = (\sqrt{2} - 1) + (\sqrt{2} - 1) \)

Total Area \( = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1) \)

Thus, the area enclosed between the curves \(y=\sin x\) and \(y=\cos x\) from \(0\) to \(\frac{\pi}{2}\) is \(2(\sqrt{2} - 1)\).