area bounded between sin2x and sin(-2x) in interval -pi to pi

To find the area bounded between the curves \( y = \sin(2x) \) and \( y = \sin(-2x) \) over the interval \( [-\pi, \pi] \), we can follow these steps:

### 1. **Understand the Curves:**
   - \( y = \sin(2x) \) and \( y = \sin(-2x) \) are trigonometric functions.
   - Since \( \sin(-2x) = -\sin(2x) \), the curves are symmetric about the x-axis.

### 2. **Identify Points of Intersection:**
   - The curves intersect where \( \sin(2x) = \sin(-2x) \).
   - This simplifies to \( \sin(2x) = -\sin(2x) \), or \( \sin(2x) = 0 \).
   - The solutions for \( \sin(2x) = 0 \) within \( [-\pi, \pi] \) are \( x = -\frac{\pi}{2}, 0, \frac{\pi}{2} \).

### 3. **Set Up the Integral:**
   - The area between the curves can be found by integrating the absolute difference between the functions over the given interval.

\[
\text{Area} = 2 \times \int_{0}^{\frac{\pi}{2}} (\sin(2x) - (-\sin(2x))) \, dx
\]

Since \( \sin(2x) = -\sin(-2x) \), the integrand becomes:

\[
\text{Area} = 2 \times \int_{0}^{\frac{\pi}{2}} 2\sin(2x) \, dx
\]

### 4. **Compute the Integral:**

\[
\text{Area} = 4 \times \int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx
\]

The integral of \( \sin(2x) \) is:

\[
\int \sin(2x) \, dx = -\frac{1}{2}\cos(2x)
\]

Applying the limits:

\[
\text{Area} = 4 \times \left[-\frac{1}{2}\cos(2x)\right]_{0}^{\frac{\pi}{2}}
\]

Substitute the limits:

\[
\text{Area} = 4 \times \left[-\frac{1}{2}\cos\left(\pi\right) + \frac{1}{2}\cos\left(0\right)\right]
\]

Simplifying further:

\[
\text{Area} = 4 \times \left[-\frac{1}{2}(-1) + \frac{1}{2}(1)\right] = 4 \times \left[\frac{1}{2} + \frac{1}{2}\right] = 4 \times 1 = 4
\]

### Final Answer:
The area bounded between the curves \( y = \sin(2x) \) and \( y = \sin(-2x) \) over the interval \( [-\pi, \pi] \) is \( \boxed{4} \) square units.

Would you like any further details or have any questions?

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Here are five related questions:

1. How would the area change if the interval was different, say from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \)?
2. What would the area be if we considered the curves \( y = \cos(2x) \) and \( y = \cos(-2x) \)?
3. How can we find the area if the curves were shifted vertically?
4. What is the significance of symmetry in finding the area between curves?
5. How does the result change if one curve was \( y = 2\sin(2x) \) and the other \( y = -\sin(2x) \)?

**Tip:** Always check the symmetry of functions before solving integrals; it can significantly simplify the calculations.

Learn how to calculate the area bounded between the curves sin(2x) and sin(-2x) over the interval [-π, π]. Detailed step-by-step solution using trigonometric properties and integration. Suitable for advanced high school students or calculus enthusiasts.

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