Question

Find \(f\)

\(f'\left( t \right) = \frac{4}{{\left( {1 + {t^2}} \right)}},\;f\left( 1 \right) = 0\)

Short answer

Required function is \(f\left( t \right) = 4{\tan ^{ - 1}}t - \pi \).

Step-by-step solution

Find a general antiderivative of \(f'\)

Here we have,

\(f'\left( t \right) = \frac{4}{{\left( {1 + {t^2}} \right)}}\)

Now, the general antiderivative of \(f'\left( t \right) = \frac{4}{{\left( {1 + {t^2}} \right)}}\) is,

\(f\left( t \right) = 4{\tan ^{ - 1}}t + C\)

Find the value of constant \(C\)

Now, we have, \(f\left( t \right) = 4{\tan ^{ - 1}}t + C\)

Also, we have \(f\left( 1 \right) = 0\)

Now, by putting \(f\left( 1 \right) = 0\) in \(f\left( t \right) = 4{\tan ^{ - 1}}t + C\) we get,

\(\begin{aligned}{c} \Rightarrow 0 &= 4{\tan ^{ - 1}}\left( 1 \right) + C\\ \Rightarrow 0 &= 4\left( {\frac{\pi }{4}} \right) + C\\ \Rightarrow C &= - \pi \end{aligned}\)

So, required function is \(f\left( t \right) = 4{\tan ^{ - 1}}t - \pi \).