How would the area change if the interval was [0,π][0,*π*]?

If the interval is extended to \([0, \pi]\), we need to find the area of the region enclosed by the curves \( y = \sin(x) \) and \( y = \cos(x) \) over this interval.

### Step 1: Determine the Points of Intersection
As before, \( \sin(x) = \cos(x) \) at \( x = \frac{\pi}{4} \). Over the interval \([0, \pi]\), the curves intersect again at \( x = \frac{5\pi}{4} \), but we focus only on the region within \([0, \pi]\).

### Step 2: Set Up the Integral
We now split the interval into two parts:
1. From \( x = 0 \) to \( x = \frac{\pi}{4} \), where \( \cos(x) \) is above \( \sin(x) \).
2. From \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \), where \( \sin(x) \) is above \( \cos(x) \).
3. From \( x = \frac{\pi}{2} \) to \( x = \pi \), where \( \sin(x) \) is above \( \cos(x) \).

### Step 3: Compute the Integrals

#### For \( x \in [0, \frac{\pi}{4}] \):
\[
A_1 = \int_{0}^{\frac{\pi}{4}} (\cos(x) - \sin(x)) \, dx = \sqrt{2} - 1
\]

#### For \( x \in [\frac{\pi}{4}, \frac{\pi}{2}] \):
\[
A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin(x) - \cos(x)) \, dx = -1 + \sqrt{2}
\]

#### For \( x \in [\frac{\pi}{2}, \pi] \):
Here, \( \sin(x) \) is always greater than \( \cos(x) \).
\[
A_3 = \int_{\frac{\pi}{2}}^{\pi} (\sin(x) - \cos(x)) \, dx
\]
Let's compute \( A_3 \):
\[
A_3 = \left[-\cos(x) - \sin(x)\right]_{\frac{\pi}{2}}^{\pi} = \left[-\cos(\pi) - \sin(\pi)\right] - \left[-\cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)\right]
\]
\[
= \left[1 - 0\right] - \left[0 - 1\right] = 1 + 1 = 2
\]

### Step 4: Add the Three Areas Together

\[
A = A_1 + A_2 + A_3 = (\sqrt{2} - 1) + (-1 + \sqrt{2}) + 2 = 2\sqrt{2} - 2 + 2 = 2\sqrt{2}
\]

Thus, the area of the region enclosed by the curves \( y = \sin(x) \) and \( y = \cos(x) \) on the interval \([0, \pi]\) is:

\[
\boxed{2\sqrt{2}}
\]

This result shows that the area is larger when the interval is extended to \([0, \pi]\) compared to \([0, \frac{\pi}{2}]\).

Do you have any questions, or would you like further details?

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Here are 5 related questions to explore:

1. How would the area change if the interval was \( [0, 2\pi] \)?
2. Can you find the area between \( y = \sin(2x) \) and \( y = \cos(2x) \) on \([0, \pi]\)?
3. What happens if we only consider the interval \([\frac{\pi}{2}, \pi]\)?
4. How do the areas compare if we consider \( y = \sin(x) \) and \( y = -\cos(x) \) on \([0, \pi]\)?
5. What if the curves were \( y = \sin(x) \) and \( y = \sin(x - \frac{\pi}{4}) \) on \([0, \pi]\)?

**Tip:** Extending the interval often requires recalculating areas based on the changing positions of the curves relative to each other. Always check which curve is above the other within each subinterval.

Explore how to calculate the area between the curves y = sin(x) and y = cos(x) over the interval [0, π]. This detailed solution includes step-by-step integration and analysis of the regions where each curve is above the other. Suitable for advanced high school students studying calculus and trigonometry.

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