evaluate the definite intergral top ia 4 and bottom is 1 (t+1)dt

Question

all methods to be shown

Answer 1

integral((1+t)dt)=integral(dt)+integral(tdt) for 1<t<4 is [t+t^2/2](1<t<4)=(4+4^2/2)-(1+1/2)=12-3/2=21/2=10.5 or 10 1/2.

General formula for integrating t^ndt (where n is not -1) is t^(n+1)/(n+1).