F(-4) = 6, f(0) = -6, f(1) = -4
Since
, we can rewrite the integral as

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

Both integrals are quite immediate: you only need to use the power rule

to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.
The answer is 82 because 5+7=12 then 3+4=7 plus the 1 = 82
0.95 rounded to the nearest hundredth (I assume that's what you mean) would still be 0.95. Since I saw your other question, I think you're looking for 0.095 rounded to the nearest hundredth, which is 0.10.