Question

Find \(f\)

\(f'''\left( t \right) = {\mathop{\rm cost}\nolimits} \)

Short answer

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

Step-by-step solution

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

Here we have,

\(f'''\left( t \right) = \cos t\)

Now, the general antiderivative of \(f'''\left( t \right) = \cos t\) is,

\(f''\left( t \right) = \sin t + C\)

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

Now, we have, \(f''\left( t \right) = \sin t + C\)

Now, the general antiderivative of \(f''\left( t \right) = \sin t + C\) is,

\(f'\left( t \right) = - \cos t + Ct + D\)

Find a general antiderivative of \(f'\)

Now, we have,\(f'\left( t \right) = - \cos t + Ct + D\)

Now, the general antiderivative of \(f'\left( t \right) = - \cos t + Ct + D\) is,

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

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