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

Answer:

$f(t) = \frac{1}{24}\cdot t^{4} + 5\cdot \sin t +\frac{1}{2}\cdot C \cdot t^{2} + D\cdot t + E$

Step-by-step explanation:

The second derivative is found by integrating it:

$f''(t) = \frac{1}{2}\cdot t^{2} -5\cdot \sin t + C$

The first derivative is:

$f' (t) = \frac{1}{6}\cdot t^{3}+5\cdot \cos t + C\cdot t + D$

Lastly, the function is:

$f(t) = \frac{1}{24}\cdot t^{4} + 5\cdot \sin t +\frac{1}{2}\cdot C \cdot t^{2} + D\cdot t + E$