What is the area benclosed bewteen y=2sinx and y=3cosx from x=0 to x=0.6pi

To find the area enclosed between the curves \( y = 2\sin(x) \) and \( y = 3\cos(x) \) from \( x = 0 \) to \( x = 0.6\pi \), we need to perform the following steps:

1. **Find the points of intersection** between the two curves within the given interval.
2. **Set up the integral** to calculate the area between the curves.
3. **Evaluate the integral** over the specified interval.

### 1. Points of Intersection
The curves intersect where \( 2\sin(x) = 3\cos(x) \). This can be rewritten as:

\[
\tan(x) = \frac{3}{2}
\]

The solution for \( x \) within the interval \( [0, 0.6\pi] \) is:

\[
x = \arctan\left(\frac{3}{2}\right) \approx 0.9828 \text{ radians}
\]

Since \( 0.9828 \) radians is greater than \( 0.6\pi \approx 1.884 \), the curves do not intersect within the interval \( [0, 0.6\pi] \).

### 2. Set Up the Integral
Since there are no intersections within the interval, we calculate the area as the integral of the difference between the two curves:

\[
\text{Area} = \int_{0}^{0.6\pi} \left| 2\sin(x) - 3\cos(x) \right| \, dx
\]

Here, we'll assume \( 2\sin(x) > 3\cos(x) \) for the entire interval (which can be confirmed by checking the values at specific points).

Thus, the area is:

\[
\text{Area} = \int_{0}^{0.6\pi} \left( 2\sin(x) - 3\cos(x) \right) \, dx
\]

### 3. Evaluate the Integral
To find the area, we compute:

\[
\text{Area} = \left[ -2\cos(x) - 3\sin(x) \right]_{0}^{0.6\pi}
\]

Substitute the limits:

\[
\text{Area} = \left(-2\cos(0.6\pi) - 3\sin(0.6\pi)\right) - \left(-2\cos(0) - 3\sin(0)\right)
\]

Using trigonometric values:

\[
\cos(0.6\pi) = -\cos(0.4\pi) \approx -0.309, \quad \sin(0.6\pi) = \sin(0.4\pi) \approx 0.951
\]
\[
\cos(0) = 1, \quad \sin(0) = 0
\]

\[
\text{Area} = \left(-2(-0.309) - 3(0.951)\right) - \left(-2(1) - 3(0)\right)
\]

Simplifying further:

\[
\text{Area} = \left(0.618 - 2.853\right) - (-2)
\]
\[
\text{Area} = -2.235 + 2 = -0.235
\]

The absolute value of the area is:

\[
\text{Area} = 0.235 \, \text{square units}
\]

The enclosed area between the curves is approximately **0.235 square units**.

Would you like to explore any part of this process in more detail or have any further questions?

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Here are 5 related questions:

1. How do you determine the points of intersection between two curves?
2. What are the methods to evaluate definite integrals analytically?
3. How would the process change if the curves intersect within the given interval?
4. How does the choice of integration limits affect the calculated area?
5. Can you find the area enclosed between two curves if one is always above the other?

**Tip:** When calculating areas between curves, always ensure you identify the correct region and set up the integrals accordingly to avoid negative areas.

Learn how to calculate the area enclosed between the curves y=2sin(x) and y=3cos(x) from x=0 to x=0.6pi. Follow step-by-step instructions including finding points of intersection, setting up the integral, and evaluating the integral to find the area.

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