Question

Primitive function for: 1. f(x)=4e2x+e−x, sådan att F(0)=2

Answer 1

The primitive function for $f(x) = 4e^{2x} + e^{-x}$ is given by it's indefinite integral.

To calculate it, recall:

$\frac{d}{dx}e^{ax}=ae^{ax}$

$\int c f(x)dx = c\int f(x)dx$

Let's begin

$\int 4e^{2x}+e^{-x}dx \\4\int e^{2x}dx+ \int e^{-x}dx\\4\frac{e^{2x}}{2}+\frac{e^{-x}}{-1} + c$

$F(x) = 2e^{2x}-e^{-x} +c$

To find the costant, we just need to use the fact that $F(0)=2$

$F(0) = 2 \\4e^{0}+e^{0} + c =2\\5+c =2\\c=-3$

Therefore,

$\boxed{F(x) = 2e^{2x} - e^{-x} -3 }$