If $\frac{d}{dx}[f(x)] = ax + b$ and $f(0) = 0$, then $f(x)$ is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{ax^2}{2} + bx$
$\frac{ax^2 + bx}{2}$
$\frac{d}{dx}[f(x)] = ax + b$
Integral both side:
$\int \frac{d}{dx}[f(x)] dx = \int (ax + b) dx$
$f(x) = \frac{ax^2}{2} + bx + c$
at $x=0$,
$f(x)=0$, $⇒0 = \frac{a(0)^2 }{2}+ b(0) + c$
$0 = c$
$∴f(x) = \frac{ax^2}{2} + bx$