Calculus Area Indefinite Integral 
I calculated the anti-derivative but plugging in for x is confusing.
can someone show me the evaluation step-by-step please
 A: $$
\int_{[0, \log(8)/2]} e^{2x} - 8 \mathrm{d}x = \left[\frac{1}{2} e^{2x} - 8 x \right]_{x=0}^{x = \log(8)/2} \\ \hspace{67mm}= \underbrace{\left(\frac{1}{2} e^{\log(8)} - 4 \log 8 \right)}_{\text{Plugging in $x = \log(8)/2$}} - \underbrace{\left(\frac{1}{2} e^0 - 8\cdot 0 \right)}_{\text{Plugging in $x = 0$}} \\\hspace{20mm} = \frac{7}{2} - 4 \log(8)
$$
EDIT: This edit is to address your questions in the comments. It seems like perhaps you are simply have trouble with basic algebra, rather than calculus. 
Here are some questions for you that if you solve, will finish all the algebra. 
What is $2(\frac{\log(8)}{2})$, $8(\frac{\log(8)}{2})$? What about $e^0$? What about $e^{\log(8)}$? It is important that you can figure these things out!
A: The antiderivative is $\frac{1}{2}e^{2x} - 8x$.  Evaluating at the two endpoints gives $\frac{1}{2}e^{\ln8} - \frac{8}{2}\ln8 - \frac{1}{2}e^{0} = \frac{7}{2} - 4 \ln8$
A: So, for the lower bound, we have $e^{0}/2 = 1/2$, and then for the upper bound $x=ln(8)/2$ we have $8 \; ln(8)/2 + e^{ln(8)}/2$ , which is equal to $4+12ln(2)$. You just subtract $1/2$ from this value to calculate the integral.
