Answer
$F(x)=2e^x-3x^2-1$
Work Step by Step
Given $f(x)=2e^x-6x$ and $F(0)=1$
The antiderivative of $2e^x$ is $2e^x$ and the antiderivative of $6x$ is $\frac{6}{1+1}x^{1+1}=\frac{6}{2}x^2=3x^2$.
Then,
$f(x)$ has the antiderivative $F(x)=2e^x-3x^2+C$.
Substituting $F(0)=1$,
$2e^0-3\cdot 0^2+C=1$
$2\cdot 1-3\cdot 0+C=1$
$2-0+C=1$
$C=-1$
Thus, $F(x)=2e^x-3x^2-1$.
Graph $f$ and $F$:
