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:

`(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(e^(2x))/(2)+(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 }`