At this stage, can I simple plug in 1 and 0 into the exponential to get the answer?
By doing so, I would get 4/3. But the model answer is 4/[3xln(5)].
Thanks!
You can use a small substitution if you want, let $\displaystyle y\cdot \ln(5)=t \Rightarrow dy=\frac{dt}{\ln(5)}$ therefore:
$\displaystyle \frac{1}{3\cdot \ln(5)}\int_{0}^{1} e^{t}dt= ...$
You can use a small substitution if you want, let $\displaystyle y\cdot \ln(5)=t \Rightarrow dy=\frac{dt}{\ln(5)}$ therefore:
$\displaystyle \frac{1}{3\cdot \ln(5)}\int_{0}^{1} e^{t}dt= ...$
This agrees with the result that you would get by first rewriting the integrand as $\displaystyle e^{\ln(5^y)} = 5^y$, then integrating. Of course, the "rule" for integrating $\displaystyle a^x$ for a > 0 comes from writing it in base e!