To solve this problem involving the probability density function (PDF) $P(x) = kx^2$ for $-3 \leq x \leq 3$ and $0$ elsewhere, let's walk through the steps systematically.

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Step 1: Find the value of $k$

The total probability for a PDF must integrate to 1 over the entire range. That is,
$\int_{-3}^{3} P(x) \, dx = 1$
Since $P(x) = kx^2$ within the interval $-3 \leq x \leq 3$, we solve:
$\int_{-3}^{3} kx^2 \, dx = 1$

First, compute the integral: $\int_{-3}^{3} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-3}^{3} = \left( \frac{3^3}{3} \right) - \left( \frac{(-3)^3}{3} \right)$
$= \left( \frac{27}{3} \right) - \left( \frac{-27}{3} \right) = 9 + 9 = 18$

Now, multiply by $k$ and equate to 1: $k \cdot 18 = 1 \quad \Rightarrow \quad k = \frac{1}{18}$
So, the value of $k$ is $\frac{1}{18}$.

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Step 2: Evaluate $P(1 \leq x \leq 2)$

We need to compute the probability within the interval $1 \leq x \leq 2$: $P(1 \leq x \leq 2) = \int_1^2 kx^2 \, dx$ Substitute $k = \frac{1}{18}$: $P(1 \leq x \leq 2) = \frac{1}{18} \int_1^2 x^2 \, dx$ Now, compute the integral: $\int_1^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^2 = \left( \frac{2^3}{3} \right) - \left( \frac{1^3}{3} \right)$ $= \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ So: $P(1 \leq x \leq 2) = \frac{1}{18} \times \frac{7}{3} = \frac{7}{54}$

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Step 3: Evaluate $P(x \leq 2)$

We need to compute the probability from the lower bound $-3$ to $2$: $P(x \leq 2) = \int_{-3}^2 kx^2 \, dx$ Substitute $k = \frac{1}{18}$: $P(x \leq 2) = \frac{1}{18} \int_{-3}^2 x^2 \, dx$ We have already computed $\int_{-3}^{3} x^2 \, dx = 18$ in Step 1, and now we compute the integral from $-3$ to $2$: $\int_{-3}^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-3}^{2} = \left( \frac{2^3}{3} \right) - \left( \frac{(-3)^3}{3} \right)$ $= \frac{8}{3} - \frac{-27}{3} = \frac{8}{3} + \frac{27}{3} = \frac{35}{3}$ So: $P(x \leq 2) = \frac{1}{18} \times \frac{35}{3} = \frac{35}{54}$

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Final Answers

1. Value of $k$:
   $k = \frac{1}{18}$

2. Probability $P(1 \leq x \leq 2)$:
   $P(1 \leq x \leq 2) = \frac{7}{54}$

3. Probability $P(x \leq 2)$:
   $P(x \leq 2) = \frac{35}{54}$

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Do you have any further questions? Would you like a breakdown of any step?

Here are 5 follow-up questions you might consider:

1. What is the meaning of a probability density function (PDF)?
2. How does the integral of a PDF relate to probability?
3. Can a PDF have values greater than 1?
4. What happens if the PDF is not normalized to 1?
5. How do we interpret cumulative probability $P(X \leq a)$ for continuous distributions?

Tip:
For continuous distributions, probabilities of individual points like $P(X = 2)$ are always zero because the probability is calculated over intervals!