Question

Find \(f\left( u \right)\).

\(f'\left( u \right) = \frac{{{u^2} + \sqrt u }}{u}\), \(f\left( 1 \right) = 3\)

Short answer

The required function is, \(f\left( u \right) = \frac{{{u^2} - \sqrt u }}{2} + 2\).

Step-by-step solution

Find anti-derivative

Given that \(f'\left( u \right) = \frac{{{u^2} + \sqrt u }}{u}\).

\(f'\left( u \right) = u + \frac{1}{{\sqrt u }}\)

The general anti-derivative of the function is

\(\begin{aligned}{l}f\left( u \right) &= \frac{{{u^2}}}{2} - \frac{{\sqrt u }}{2} + C\\f\left( u \right) &= \frac{{{u^2} - \sqrt u }}{2} + C\end{aligned}\)

Find the constant

Now, \(f\left( u \right) = \frac{{{u^2} - \sqrt u }}{2} + C\)

And \(f\left( 1 \right) = 3\), then

\(\begin{aligned}{c}\frac{{{1^2} + \sqrt 1 }}{2} + C &= 3\\\frac{{1 + 1}}{2} + C &= 3\\1 + C &= 3\\C &= 2\end{aligned}\)

Hence \(f\left( u \right) = \frac{{{u^2} - \sqrt u }}{2} + 2\).