Choose c so that f is a density function

Question

Let the function f : R→R with c > 0 be defined by $$ f(x) := \begin{cases} cx, & \text{for $0\le x\le1$} \\ cx^2, & \text{for $1\lt x\lt2$} \\\ 0, & \text{otherwise} \end{cases} $$ i. Choose c such that f is a density function. Ok so here I made the integral of the functions equal to 1. I get 2 values of c. $C= 2$ or $\frac7 3$

Is it normal to have 2 values of C? How do I know which one I should use?

ii. Find $P[X > 1]$ and $P[−1/2 < X < 1/2]$.

again which c or row should I use? and when I integrate, is it$ P[X\gt1]=\int_1^2 (cx)dx? $ And for $P[-\frac{1}{2}\lt x \lt \frac{1}{2}]=\int_{-\frac1 2}^{\frac 12} (cx)dx$?

Answer 1

Guide:

• You should not have two values of $c$.

• Solve for $c$ in $$c\left( \int_0^1 x\, dx + \int_1^2 x^2\, dx\right)=1$$

• When $x>1$, use the rule in $1<x<2$.

• $$P(-\frac12 < x< \frac12) = P(0\le x < \frac12)$$