Determine whether the function is a probability density function over the given interval.
-$$f ( x ) = 5 \sin 5 x \text { over } \left[ \frac { \pi } { 10 } , \frac { \pi } { 5 } \right]$$

Question

Quizplus · Accepted Answer

For a function to be a probability density function over an interval, it must satisfy the following conditions:

1. $f(x) \geq 0$, for all $x$ in the interval
2. $\int_{-\infty}^{\infty} f(x)\ dx = 1$, where the integral is taken over the interval

Notice that $f(x) = 5\sin(5x)$ is non-negative over the interval $\left[\frac{\pi}{10}, \frac{\pi}{5}ight]$ since the sine function is bounded between $-1$ and $1$.

Next, we need to check whether the integral of $f(x)$ over the given interval is equal to 1.

\begin{align*}
\int_{\pi/10}^{\pi/5} 5\sin(5x)dx &= -\frac{1}{5}\cos(5x)\Bigg|_{\pi/10}^{\pi/5}\
&= -\frac{1}{5}\left[\cos\left(\frac{\pi}{2}ight) - \cos\left(\frac{\pi}{2}ight)ight]\
&= \frac{1}{5}
\end{align*}

Since the integral of $f(x)$ over the interval equals 1, we can conclude that $f(x)$ is a probability density function over the interval $\left[\frac{\pi}{10}, \frac{\pi}{5}ight]$. Therefore, the answer is A).

Determine Whether the Function Is a Probability Density Function Over f(x)=5sin⁡5x over [π10,π5]f ( x ) = 5 \sin 5 x \text { over } \left[ \frac { \pi } { 10 } , \frac { \pi } { 5 } \right]f(x)=5sin5x over [10π​,5π​]

Determine Whether the Function Is a Probability Density Function Over $f ( x ) = 5 \sin 5 x \text { over } \left[ \frac { \pi } { 10 } , \frac { \pi } { 5 } \right]$