Question

Evaluate the indefinite integral. ft-53 at 1250 + c 5 - + 2 - 125t + c 312 - 30t + 75 + 6 O r4 - 15+ 7512 - 125t + c

Answer 1

Given:

$∫(t-5)^3\text{ dt}$

Let's evaluate the integral:

$\begin{gathered} ∫(t-5)^3\text{ dt} \\ \\ =\frac{1}{4}(t-5)^4+c \\ \\ ^{} \end{gathered}$

Thus, we have:

$\frac{1}{4}(t^4-20t^3+150t^2-500t+625)+C$

Expand using distributive property:

$\begin{gathered} (\frac{t^4}{4}-\frac{20t^3}{4}+\frac{150t^2}{4}-\frac{500t}{4}+\frac{625}{4})+C \\ \\ \\ =\frac{t^4}{4}-5t^3+\frac{75t^2}{2}-125t+C \end{gathered}$

ANSWER:

$=\frac{t^4}{4}-5t^3+\frac{75}{2}t^2-125t+C$