find the area of the region enclosed by the curves y=sinx amd y=cos x on [0,pi/2]

To find the area of the region enclosed by the curves \( y = \sin(x) \) and \( y = \cos(x) \) on the interval \([0, \frac{\pi}{2}]\), follow these steps:

### Step 1: Determine the Points of Intersection
Set \( \sin(x) = \cos(x) \) to find the points where the curves intersect.

\[
\sin(x) = \cos(x)
\]

Divide both sides by \( \cos(x) \) (assuming \( \cos(x) \neq 0 \)):

\[
\tan(x) = 1
\]

The solution for \( x \) within the interval \([0, \frac{\pi}{2}]\) is:

\[
x = \frac{\pi}{4}
\]

### Step 2: Set Up the Integral
The area \( A \) between two curves \( y = f(x) \) and \( y = g(x) \) from \( a \) to \( b \) is given by:

\[
A = \int_{a}^{b} \left| f(x) - g(x) \right| \, dx
\]

For the interval \([0, \frac{\pi}{4}]\):

\[
A_1 = \int_{0}^{\frac{\pi}{4}} (\cos(x) - \sin(x)) \, dx
\]

For the interval \([\frac{\pi}{4}, \frac{\pi}{2}]\):

\[
A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin(x) - \cos(x)) \, dx
\]

### Step 3: Compute the Integrals

\[
A_1 = \int_{0}^{\frac{\pi}{4}} (\cos(x) - \sin(x)) \, dx = \left[\sin(x) + \cos(x)\right]_{0}^{\frac{\pi}{4}}
\]
\[
= \left(\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)\right) - \left(\sin(0) + \cos(0)\right)
\]
\[
= \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1) = \sqrt{2} - 1
\]

\[
A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin(x) - \cos(x)) \, dx = \left[-\cos(x) - \sin(x)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}
\]
\[
= \left[-\cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right)\right] - \left[-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right]
\]
\[
= \left[-0 - 1\right] - \left[-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right] = -1 + \sqrt{2}
\]

### Step 4: Add the Two Areas Together

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

Thus, the area of the region enclosed by the curves \( y = \sin(x) \) and \( y = \cos(x) \) on the interval \([0, \frac{\pi}{2}]\) is:

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

Would you like further details or have any questions?

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Here are 5 related questions to explore:

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

**Tip:** When comparing areas under curves, always identify the points of intersection to correctly set up your integral limits.

Discover how to find the area enclosed by y = sin(x) and y = cos(x) on the interval [0, π/2]. This calculus problem involves setting up integrals to calculate the area between curves. Suitable for advanced high school students and those studying calculus and trigonometry.

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