Answer:
The answer is "[tex]\int e^{ax}\ dx=\frac{e^{ax}}{a}+c[/tex]".
Step-by-step explanation:
Please find the complete question in the attached file.
[tex]\int^{m}_{a}\ f(x)\ dx=\frac{1}{2}\\\\f(x)=ke^{-kx}, \ [0, \infty]\\\\[/tex]
let m is the median
[tex]so,\\\\ \int^{m}_{a}\ f(x)\ dx=\frac{1}{2}\\\\\to \int^m_0 k\cdot e^{-kx}=\frac{1}{2}\\\\\to [\frac{k\cdot e^{-kx}}{-k}]^m_0=\frac{1}{2}\\\\\to [- e^{-kx}]^m_0=\frac{1}{2}\\\\\to - e^{-km} +1=\frac{1}{2}\\\\\to e^{-km} =\frac{1}{2}\\\\\to e^{km} =2\\\\\to \ln(e^{km}) =\ln 2\\\\\to km= \ln \ 2\\\\\to m=\frac{\ln \ 2}{k}\\\\ \text{the answer is: } \\\\ \to \int e^{ax}\ dx=\frac{e^{ax}}{a}+c[/tex]
