Property 1: f(x)>or= 0 for all xE[a,b]. Property 2: integral (a,b) f(x) dx=1. Find the value of K that function f(x)= K[Sin[x/2]+10] is a probability density function in the interval xE[0,π]

To make \( f(x) = K[\sin(x/2) + 10] \) a probability density function (PDF) on the interval \( x \in [0, \pi] \), it must satisfy the two properties of a PDF:

1. \( f(x) \geq 0 \) for all \( x \in [0, \pi] \).
2. \( \int_{0}^{\pi} f(x) \, dx = 1 \).

### Step 1: Check Non-negativity
The function \( f(x) = K[\sin(x/2) + 10] \) is non-negative over the interval \( [0, \pi] \) because \( \sin(x/2) \) oscillates between 0 and 1 in this interval, and adding 10 ensures that the expression \( \sin(x/2) + 10 \) is always positive.

### Step 2: Determine the Normalization Constant \( K \)
To satisfy the second property, we need to find \( K \) such that:

\[
\int_{0}^{\pi} K[\sin(x/2) + 10] \, dx = 1
\]

Let's evaluate this integral:

\[
K \int_{0}^{\pi} [\sin(x/2) + 10] \, dx = 1
\]

This integral can be split into two parts:

\[
K \left(\int_{0}^{\pi} \sin(x/2) \, dx + \int_{0}^{\pi} 10 \, dx \right) = 1
\]

#### Part 1: Evaluate \( \int_{0}^{\pi} \sin(x/2) \, dx \)

To solve this, we use substitution. Let:

\[
u = \frac{x}{2}, \quad du = \frac{1}{2} dx, \quad dx = 2 du
\]

Changing the limits accordingly:
- When \( x = 0 \), \( u = 0 \).
- When \( x = \pi \), \( u = \frac{\pi}{2} \).

Now the integral becomes:

\[
\int_{0}^{\pi} \sin(x/2) \, dx = 2 \int_{0}^{\frac{\pi}{2}} \sin(u) \, du
\]

The integral of \( \sin(u) \) is:

\[
\int \sin(u) \, du = -\cos(u)
\]

So:

\[
2 \int_{0}^{\frac{\pi}{2}} \sin(u) \, du = 2[-\cos(u)]_{0}^{\frac{\pi}{2}} = 2[-\cos(\frac{\pi}{2}) + \cos(0)] = 2[0 + 1] = 2
\]

#### Part 2: Evaluate \( \int_{0}^{\pi} 10 \, dx \)

This is a straightforward integral:

\[
\int_{0}^{\pi} 10 \, dx = 10x \Big|_{0}^{\pi} = 10\pi
\]

### Combine the Results

Adding both parts together:

\[
K \left(2 + 10\pi \right) = 1
\]

\[
K (10\pi + 2) = 1
\]

Solve for \( K \):

\[
K = \frac{1}{10\pi + 2}
\]

### Conclusion

The value of \( K \) that makes \( f(x) = K[\sin(x/2) + 10] \) a valid probability density function on the interval \( [0, \pi] \) is:

\[
K = \frac{1}{10\pi + 2}
\]

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How does changing the interval affect the value of \( K \)?
2. What happens if the function \( f(x) \) was \( K\sin(x) \) instead?
3. How does the constant \( 10 \) in \( f(x) \) affect the distribution of the probability density function?
4. What are the properties of a probability density function for non-uniform distributions?
5. How can we generalize this process for any function \( f(x) \)?

**Tip**: Always check both the non-negativity and normalization conditions when defining or verifying a probability density function.

Learn how to determine the value of K that makes the function f(x) = K[Sin(x/2) + 10] a valid probability density function on the interval [0, π]. This involves checking non-negativity and normalization conditions using integral calculus and properties of probability density functions.

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