1. **Problem statement:** We have a random variable $X$ with values between 0 and 1 and probability density function (pdf) $f(x) = 2x$. We want to compute the variance of $X$, denoted $\mathrm{Var}(X)$, and the variance of $X^2$, denoted $\mathrm{Var}(X^2)$. 2. **Recall the variance formula:** For any random variable $Y$, $\mathrm{Var}(Y) = E(Y^2) - [E(Y)]^2$, where $E(Y)$ is the expected value of $Y$. 3. **Step 1: Compute $E(X)$:** $$E(X) = \int_0^1 x \cdot f(x) \, dx = \int_0^1 x \cdot 2x \, dx = \int_0^1 2x^2 \, dx = 2 \cdot \frac{x^3}{3} \Big|_0^1 = \frac{2}{3}.$$ 4. **Step 2: Compute $E(X^2)$:** $$E(X^2) = \int_0^1 x^2 \cdot f(x) \, dx = \int_0^1 x^2 \cdot 2x \, dx = \int_0^1 2x^3 \, dx = 2 \cdot \frac{x^4}{4} \Big|_0^1 = \frac{1}{2}.$$ 5. **Step 3: Compute $\mathrm{Var}(X)$:** $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = \frac{1}{2} - \left(\frac{2}{3}\right)^2 = \frac{1}{2} - \frac{4}{9} = \frac{9}{18} - \frac{8}{18} = \frac{1}{18}.$$ 6. **Step 4: Compute $E(X^2)$ again for $X^2$ variable:** Here, the variable is $Y = X^2$. We need $E(Y)$ and $E(Y^2) = E(X^4)$. 7. **Compute $E(X^4)$:** $$E(X^4) = \int_0^1 x^4 \cdot f(x) \, dx = \int_0^1 x^4 \cdot 2x \, dx = \int_0^1 2x^5 \, dx = 2 \cdot \frac{x^6}{6} \Big|_0^1 = \frac{1}{3}.$$ 8. **Compute $E(X^2)$ again (already found):** $E(X^2) = \frac{1}{2}$. 9. **Step 5: Compute $\mathrm{Var}(X^2)$:** $$\mathrm{Var}(X^2) = E(X^4) - [E(X^2)]^2 = \frac{1}{3} - \left(\frac{1}{2}\right)^2 = \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}.$$ **Final answers:** $$\mathrm{Var}(X) = \frac{1}{18}, \quad \mathrm{Var}(X^2) = \frac{1}{12}.$$