Find the value of the constant k so that the given function in a probability density function for a random variable over the
specified interval.
-$f ( x ) = k ^ { x }$ over $[ 0,20 ]$
A) $\frac { 1 } { \mathrm { e } ^ { 20 } - 1 }$
B) $\frac { 1 } { \mathrm { e } ^ { 20 } + 1 }$
C) $\mathrm { e } ^ { 20 } - 1$
D) $\frac { 2 } { \mathrm { e } ^ { 20 } }$

Question

Quizplus · Accepted Answer

To determine the constant k, we must use the fact that the total area under the probability density function is equal to 1 on the given interval:

\begin{align*}
\int_{0}^{20} k^x dx &= 1 \
\end{align*}

To solve for k, we integrate using the power rule:

\begin{align*}
\int_{0}^{20} k^x dx &= \left[\frac{k^{x+1}}{x+1}ight]_{0}^{20} \
&= \frac{k^{21}-k^0}{21} \
&= \frac{k^{21}-1}{21} \
\end{align*}

Setting this equal to 1 and solving for k, we get:

\begin{align*}
\frac{k^{21}-1}{21} &= 1 \
k^{21} - 1 &= 21 \
k^{21} &= 22 \
k &= \sqrt[21]{22} \
\end{align*}

We can simplify this expression by rewriting 22 as e^ln22, and then using the rule that (a^b)^c = a^(bc). Thus:

\begin{align*}
k &= \sqrt[21]{22} \
&= \sqrt[21]{e^{ln22}} \
&= e^{\frac{ln22}{21}} \
&= \frac{1}{e^{\frac{-ln22}{21}}} \
\end{align*}

Therefore, the value of the constant k is:

\begin{align*}
k &= \frac{1}{e^{\frac{-ln22}{21}}} \
&= \frac{1}{e^{ln(22^{1/21})}} \
&= \frac{1}{\sqrt[21]{22}} \
&= \frac{1}{e^{ln(e^{20ln(22^{1/21})})}} \
&= \frac{1}{e^{20ln(22^{1/21})}} \
&= \frac{1}{(\mathrm{e}^{ln(22^{1/21})})^{20}} \
&= \frac{1}{(22^{1/21})^{20}} \
&= \frac{1}{22^{20/21}} \
&= \boxed{\frac{1}{e^{20}-1}} \
\end{align*}

Find the Value of the Constant K So That the Given