The probability density function of a random variable is given by $f_X(x) = \frac{3}{208}x^2$ Find $E(X)$

Question

The probability density function of a random variable is given by $$f_X(x) = \frac{3}{208}x^2$$ whenever $2 < x < 6$. $0$ otherwise

$(a)$ Compute $E(X)$

$(b)$ Compute $E[X(X+1)]$

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My attempt

$(a)$

$$E(X) = \int_{2}^{6} x\frac{3}{208}x^2dx = \frac{3}{208}\cdot\frac{1}{4}(x^4)\bigg|_{2}^{6} = 60/13$$

$(b)$

$$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$

$$E(X^2) = \int_{2}^{6}x^2 \frac{3}{208}x^2 dx = \frac{3}{208}\frac{1}{5}(x^5)\bigg|_{2}^{6} = 1452/65$$

Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 \approx 26.9538 $$

Would this be right?

Answer 1

The probability density function of a random variable is given by $$f_X(x) = \frac{3}{208}x^2$$ whenever $2 < x < 6$. $0$ otherwise

$(a)$ Compute $E(X)$

$(b)$ Compute $E[X(X+1)]$

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$(a)$

$$E(X) = \int_{2}^{6} x\frac{3}{208}x^2dx = \frac{3}{208}\cdot\frac{1}{4}(x^4)\bigg|_{2}^{6} = 60/13$$

$(b)$

$$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$

$$E(X^2) = \int_{2}^{6}x^2 \frac{3}{208}x^2 dx = \frac{3}{208}\frac{1}{5}(x^5)\bigg|_{2}^{6} = 1452/65$$

Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 \approx 26.9538 $$