A table is given. It is required to use the linear regression capabilities to find a linear model for (t, ln(R)) with an equation of the form ln(R)=at+b, and then write in exponential form.
It is also required to use a graphing utility to draw a graph.
(a) Rewrite the table values for t and ln(R) as follows:
Input the values into a graphing utility to find the linear model:
[tex]\ln(R)=-0.6259t+6.0770[/tex]
Take the exponent of both sides of the equation to write in exponential form as required:
[tex]\begin{gathered} \Rightarrow e^{\ln(R)}=e^{-0.6259t+6.0770} \\ \text{ Using the powers of sum property:} \\ \Rightarrow R=e^{6.0770}\cdot e^{-0.6259t} \\ \Rightarrow R=435.7201(e^{-0.6259t}) \end{gathered}[/tex]
(b) Use the graphing utility to graph the exponential function:
Notice that the graph at the top-right best fits the graph.
(c) Evaluate the definite integral using the graphing utility:
[tex]\int_0^4435.7201e^{-0.6259t}dt=639.2116[/tex]
The answer is 639.2116 L.
