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.
Answer:
<u><em>I want brainliest</em></u>
Step-by-step explanation:
7.6 x 8 = 60.8
so the answer is either b or c
7.6 per hour
or
7.6h
makes the answer b
Answer:
D
Step-by-step explanation:
group them first :
( x3+5x2) and ( -6x-30)
then simply by gcf ( greatest common factor) :
x2(x+5) and -6(x+5)
and just add them together:
x2(x+5)-6(x+5)
bonus :
it can be written as (x2-6)(x+5)
/ means divide, so U( an unknown value ) / 6 = 9. to get the answer, simply multiply 6 x 9