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# Decide whether the functions defined as follows are probability density functions on the indicated intervals. If not, tell why. $f(x)=2 x^{2}$; [-1, 1]

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From the definition:

The function $f$ is a probability density function of a random variable $x$ in the interval $[a,b]$ if :

\begin{align*} &1.f(x)\geq0 \text{ for all x in interval} [a,b]\\ &2.\int_{a}^{b} f(x)\,dx=1. \end{align*}

We need to check these to conditions.

First condition:

$f(x)=2\cdot x^{2}$

On the interval $[-1,1]$, $x^{2}\geq0$, any number squared is positive number, so this condition is fulfilled.

Second condition:

\begin{align*} \int_{-1}^{1} 2x^{2}\,dx&=2\int_{-1}^{1} x^2\,dx\\ &=2\frac{x^{2+1}}{2+1}\bigg|_{-1}^{1}\\ &=2\frac{x^{3}}{3}\bigg|_{-1}^{1}\\ &=\frac{2\,x^{3}}{3}\bigg|_{-1}^{}\\ &=\frac{2\cdot 1^{3}}{3}-\frac{2\cdot -1^{3}}{3}\\ &=\frac{2}{3}-\frac{-2}{3}\\ &=\frac{2}{3}+\frac{2}{3}\\ &=\frac{4}{3}\\ \end{align*}

Second condition is not fulfilled.

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