Finding Area Bounded by sin x and cos x Curves

The problem asks us to calculate the area of the region enclosed by two trigonometric curves, \(y = \sin x\) and \(y = \cos x\), and the vertical lines \(x = \frac{\pi}{4}\) and \(x = \frac{\pi}{2}\).

To find the area between two curves \(y = f(x)\) and \(y = g(x)\) from \(x=a\) to \(x=b\), we need to evaluate the definite integral of the absolute difference between the functions over the interval \([a, b]\). The formula is given by:

\( \text{Area} = \int_{a}^{b} |f(x) - g(x)| \, dx \)

In this specific problem, \(f(x) = \sin x\), \(g(x) = \cos x\), \(a = \frac{\pi}{4}\), and \(b = \frac{\pi}{2}\). We need to determine which function has a greater value over the interval \([\frac{\pi}{4}, \frac{\pi}{2}]\).

Let's analyze the values of \(\sin x\) and \(\cos x\) at the endpoints of the interval:

• At \(x = \frac{\pi}{4}\), \(\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\) and \(\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\). The curves intersect at this point.
• At \(x = \frac{\pi}{2}\), \(\sin(\frac{\pi}{2}) = 1\) and \(\cos(\frac{\pi}{2}) = 0\).

For values of \(x\) between \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\), the sine function increases from \(\frac{\sqrt{2}}{2}\) to 1, while the cosine function decreases from \(\frac{\sqrt{2}}{2}\) to 0. This means that over the interval \([\frac{\pi}{4}, \frac{\pi}{2}]\), \(\sin x \ge \cos x\).

Therefore, the area is given by the integral:

\( \text{Area} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx \)

Now, we evaluate the definite integral:

The antiderivative of \(\sin x\) is \(-\cos x\).

The antiderivative of \(\cos x\) is \(\sin x\).

So, the antiderivative of \((\sin x - \cos x)\) is \(-\cos x - \sin x\).

Now, apply the limits of integration:

\( \text{Area} = [-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \)

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

Substitute the values of \(\sin\) and \(\cos\) at these angles:

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

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

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

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

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

The area of the region bounded by the curves \(y = \sin x\), \(y = \cos x\) and the lines \(x = \frac{\pi}{4}\), \(x = \frac{\pi}{2}\) is \( \sqrt{2} - 1 \) square units.

Additional Information: Area Between Curves

The concept of finding the area between curves is a fundamental application of definite integrals in calculus. Here are some key points to remember:

• To find the area between \(y = f(x)\) and \(y = g(x)\) over \([a, b]\), first determine which function is the upper curve and which is the lower curve over the interval. If they cross, you may need to split the integral into multiple parts.
• If \(f(x) \ge g(x)\) on \([a, b]\), the area is \( \int_{a}^{b} (f(x) - g(x)) \, dx \).
• If \(g(x) \ge f(x)\) on \([a, b]\), the area is \( \int_{a}^{b} (g(x) - f(x)) \, dx \).
• Using the absolute value \( \int_{a}^{b} |f(x) - g(x)| \, dx \) handles cases where the curves cross within the interval, as it automatically calculates the sum of the absolute differences over subintervals. However, it's often easier to find the intersection points and set up separate integrals.
• For areas bounded by curves defined as \(x = f(y)\) and \(x = g(y)\) over a vertical interval \([c, d]\) on the y-axis, the formula is \( \int_{c}^{d} |f(y) - g(y)| \, dy \), determining which function is the rightmost curve and which is the leftmost.

Understanding the graphs of common functions like \(\sin x\) and \(\cos x\) is helpful for quickly identifying which curve is above the other in a given interval.