Using knowledge of derivatives, what is $f(x)$ if $f'(x) = 4x + e^x$ with initial condition $f(0)=2$? How to solve this with knowledge of derivatives? Is it asking to solve this without using anti-derivatives?
 A: Integrate both sides to get $f(x)$ and the equation becomes 
$$f(x) = 2x^2 + e^x + c.$$
Put $x=0$ and $f(x=0)=2$ to get $c=1$. Substitute the value of $c$ and get 
$$f(x)=2x^2+e^x+1.$$
A: Probably this is the intention of the exercise: because $f'(x)=g(x)+h(x)$ then $f(x)=G(x)+H(x)+C$ for some constant $C$ where $G$ and $H$ are antiderivatives of $g$ and $h$ respectively.
I mean: "knowledge of derivatives" seems to me to search for such $G$ and $H$ with derivatives $g$ and $h$, knowing that if $f(x)=G(x)+H(x)+C$ then $f'(x)=g(x)+h(x)$.
Then for any valid pair $G$ and $H$ the constant $C$ is defined by the value $f(0)=2$.
A: You absolutely need antiderivatives here. You have to find the antiderivative of the function $f'(x)$ which will give you a family of functions that differ only by a constant. One of those functions is going to be the function you're looking for. Then, use the initial condition that you're given ($f(0)=2$) to determine the consonant of integration you get when you're done integrating $f'(x)$.
$$
f(x)=\int\left(4x+e^x\right)\,dx=2x^2+e^x+C,\\
f(0)=2\cdot0^2+e^0+C=2\implies
C=1.\\
\therefore\ f(x)=2x^2+e^x+1.
$$
A: The First  Fundametal Theorem of Integral Calculus asserts that the antiderivative of $4x+\mathrm e^x$ which takes the value $2$ at $x=0$ is the function
$$f(x)=2+\int_0^x(4tx+\mathrm e^t)\,\mathrm dt.$$
(The integral from $0$ is the antiderivative which vanishes at $x=0$).
