Answer :
The initial expression is:
[tex]\int (t-5)^3\text{ dt}[/tex]First we can expant cube so:
[tex]\int t^3-15t^2+75t-125[/tex]and now we can integrate so:
[tex]\frac{t^4}{4}-5t^3+\frac{75t^2}{2}-125t+C[/tex]So is option B
The initial expression is:
[tex]\int (t-5)^3\text{ dt}[/tex]First we can expant cube so:
[tex]\int t^3-15t^2+75t-125[/tex]and now we can integrate so:
[tex]\frac{t^4}{4}-5t^3+\frac{75t^2}{2}-125t+C[/tex]So is option B