Question

Find \(f\)

\(f'\left( t \right) = t + \frac{1}{{{t^3}}},\;t > 0,\;f\left( 1 \right) = 6\)

Short answer

Required function is \(f\left( t \right) = \frac{{{t^2}}}{2} - \frac{1}{{2{t^2}}} + 6\).

Step-by-step solution

 Step 1: Find a general antiderivative of \(f'\)

Here we have,

\(f'\left( t \right) = t + \frac{1}{{{t^3}}}\)

Now, the general antiderivative of \(f'\left( t \right) = t + \frac{1}{{{t^3}}}\) is,

\(f\left( t \right) = \frac{{{t^2}}}{2} - \frac{1}{{2{t^2}}} + C\)

Find the value of constant \(C\)

Now, we have, \(f\left( t \right) = \frac{{{t^2}}}{2} - \frac{1}{{2{t^2}}} + C\)

Also, we have \(f\left( 1 \right) = 6\)

Now, by putting \(f\left( 1 \right) = 6\) in \(f\left( t \right) = \frac{{{t^2}}}{2} - \frac{1}{{2{t^2}}} + C\) we get,

\(\begin{aligned}{l} \Rightarrow 6 &= \frac{{{{\left( 1 \right)}^2}}}{2} - \frac{1}{{2{{\left( 1 \right)}^2}}} + C\\ \Rightarrow 6 &= \frac{1}{2} - \frac{1}{2} + C\\ \Rightarrow C &= 6\end{aligned}\)

So, required function is \(f\left( t \right) = \frac{{{t^2}}}{2} - \frac{1}{{2{t^2}}} + 6\).