Question

Find \(f\)

\(f'''\left( t \right) = {e^t} + {t^{ - 4}}\)

Short answer

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

Step-by-step solution

Find a general antiderivative of \(f'''\)

Here we have,

\(f'''\left( t \right) = {e^t} + {t^{ - 4}}\)

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

\(\begin{aligned}{c}f''\left( t \right) &= {e^t} + \frac{{{t^{ - 3}}}}{{ - 3}} + C\\ &= {e^t} - \frac{{{t^{ - 3}}}}{3} + C\end{aligned}\)

Find a general antiderivative of \(f''\)

Now, we have, \(f''\left( t \right) = {e^t} - \frac{{{t^{ - 3}}}}{3} + C\)

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

\(\begin{aligned}{c}f'\left( t \right) &= {e^t} - \frac{{{t^{ - 2}}}}{{3( - 2)}} + Ct + D\\ &= {e^t} + \frac{{{t^{ - 2}}}}{6} + Ct + D\end{aligned}\)

Find a general antiderivative of \(f'\)

Now, we have,\(f'\left( t \right) = {e^t} + \frac{{{t^{ - 2}}}}{6}Ct + D\)

Now, the general antiderivative of \(f'\left( t \right) = {e^t} + \frac{{{t^{ - 2}}}}{6} + Ct + D\) is,

\(\begin{aligned}{c}f\left( t \right) &= {e^t} + \frac{{{t^{ - 1}}}}{{6( - 1)}} + C\frac{{{t^2}}}{2} + Dt + E\\ &= {e^t} - \frac{{{t^{ - 1}}}}{6} + C\frac{{{t^2}}}{2} + Dt + E\end{aligned}\)

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