If the function f defined by f(x) = begin{cas | vedclass

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(A) Since $f(x)$ is the p.d.f. of a random variable $X$,the total area under the curve must be $1$.$\int_{0}^{1} f(x) dx = 1 \Rightarrow \int_{0}^{1}

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$\frac{1}{2}$

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(A) Since $f(x)$ is the p.d.f. of a random variable $X$,the total area under the curve must be $1$.$\int_{0}^{1} f(x) dx = 1 \Rightarrow \int_{0}^{1} K(x-x^2) dx = 1$$K \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 1 \Rightarrow K \left( \frac{1}{2} - \frac{1}{3} \right) = 1$$K \left( \frac{1}{6} \right) = 1 \Rightarrow K = 6$Now,we calculate $P(X $= 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{\frac{1}{2}} = \left[ 3x^2 - 2x^3 \right]_{0}^{\frac{1}{2}}$$= 3 \left( \frac{1}{2} \right)^2 - 2 \left( \frac{1}{2} \right)^3 = 3 \left( \frac{1}{4} \right) - 2 \left( \frac{1}{8} \right)$$= \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

If the function $f$ defined by $f(x) = \begin{cases} K(x-x^2) & \text{if } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}$ is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(X < \frac{1}{2})$ is

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