Ok, I know what to do (take antiderivative and plug in points from a certain interval) but the final answer i get is 10.5 (which is wrong to the system). Can someone please check my work and let me know the correct answer? it's driving me nuts

My work (the boxes show the antiderivative and the interval from 1 to 0. I know that you have to plug in 1 to the antiderivative and 0 and subtract the two. The same goes for the interval 3 to 1. You then would add the two together):

3 | t^{2}+ 3t− 4|dt0 = +

1 − t^{2}− 3t+ 4dt0

3 t^{2}+ 3t− 4dt1 Antiderivatve= -1/3 t^3 -3/2 t^2 +4t |1 and 0 intervals

Antiderivative= 1/3t^3 +3/2 t^2 -4t | 3 and 1 intervals

Interval 1 subtracted by interval of 0:

[-1/3 (1)^3 -3/2 (1)^2 +4(1)] - [0]

Added with

The interval of 3 subtracted by interval of 1:

[1/3(3)^3 +3/2(3)^2-4(3) ] - [1/3(1)^3 +3/2(1)^2-4(1)