# Thread: last question i promise- another antiderivative

1. ## last question i promise- another antiderivative

Find f. f '(x) = √x(9 + 5x)
f(1) = 11
f =

thats what I got, and obviously its wrong.. what am I doing wrong here?

2. Hello, rsanford!

$f'(x) \:= \:\sqrt{x}\,(9 + 5x),\quad f(1) \,=\,11$

Find $f(x)$

We have: . $f'(x) \;=\;9x^{\frac{1}{2}} + 5x^{\frac{3}{2}}$

Integrate: . $f(x)\;=\;\int\left(9x^{\frac{1}{2}} + 5x^{\frac{3}{2}}\right)\,dx \;=\;6x^{\frac{3}{2}} + 2x^{\frac{5}{2}} + C \;=\;2x^{\frac{3}{2}}(3 + x) + C$

Since $f(1) = 11$, we have: . $2\!\cdot\!1^{\frac{3}{2}}(3 + 1) + C \:=\:11 \quad\Rightarrow\quad C \,=\,3$

Therefore: . $f(x)\;=\;2x^{\frac{3}{2}}(3 + x) + 3$