Question

Find f. (Use C for the constant of the first antiderivative, D for the constant of the second antiderivative and E for the constant of the third antiderivative.) f '''(t) = t − 5 cos(t)

Answer 1

Given that,

f'''(t) = t — 5cos(t)

We want to find f(t) so we need to integrate this function three times to get f(t)

First anti derivative

∫ f'''(t) dt = ∫ (t —5cos(t)) dt

Note, the integral of cos(t) is sin(t), and the integral of sin(t) is —Cos(t)

Integrating third derivatives decreases it to second derivatives i.e. f'''(t) to f''(t)

f''(t) = t²/2 — 5sin(t) + C

Where C is the first anti derivative constant

Second anti derivative

f''(t) = t²/2 — 5sin(t) + C

∫ f''(t) = ∫ (t²/2 — 5sin(t) + C) dt

f'(t) = t³/6 + 5Cos(t) + Ct + D

Where D is the second anti derivative constant

Third anti derivative

f'(t) = t³/6 + 5Cos(t) + Ct + D

∫ f'(t) = ∫ t³/6 + 5Cos(t) + Ct + D) dt

f(t) = t⁴/24 + 5Sin(t) + Ct²/2 + Dt + E

Where E is the third anti derivative constant.

So the required f(t) function is

f(t) = t⁴/24 + 5Sin(t) + ½Ct² + Dt + E

Answer 2

`f(t) = (1)/(24)\cdot t^(4) + 5\cdot \sin t +(1)/(2)\cdot C \cdot t^(2) + D\cdot t + E`

Explanation:

The second derivative is found by integrating it:

`f''(t) = (1)/(2)\cdot t^(2) -5\cdot \sin t + C`

The first derivative is:

`f' (t) = (1)/(6)\cdot t^(3)+5\cdot \cos t + C\cdot t + D`

Lastly, the function is:

`f(t) = (1)/(24)\cdot t^(4) + 5\cdot \sin t +(1)/(2)\cdot C \cdot t^(2) + D\cdot t + E`