Evaluating $\int 4 e^x - x^3 + 6x - 8 \ dx$ The problem asks to integrate the following: $$\int 4 e^x - x^3 + 6x - 8 \ dx$$ 
So far I got
$$4e^x-(x^4/4)+(6x^2/2)+8x+c$$
Is this correct?

Hints:
Use the additivity of the integral: Given functions $\{ f_i \}_{i=1}^n $, we have
$$ \int \sum f_i = \sum \int f_i $$
In your case, $f_1 = 4e^x, f_2 = -x^3, f_3 = 6x, f_4 = -8$. These are fairly easy to integrate, but here are some formulae:
$$ \int e^x = e^x + K $$
$$ \int x^n = \frac{ x^{n+1} }{n+1} + K , \; \;\;\; (n \neq1)$$
Obviously, if $K$ is a constant, 
$$ \int K f = K \int f $$
$$ \text{Answer:}  \;\;\;\; 4e^x - \frac{ x^4}{4} + 3x^2 - 8x + K $$
