Question

Evaluate the following integrals a. ∫_0^2▒(3x^3-x^2+2)dx b.
Resolved Question:
Evaluate the following integrals
a. ∫_0^2▒(3x^3-x^2+2)dx
b. ∫_0^1▒(e^2x-x^2 )dx
c. ∫▒〖1/(x+3) dx〗
2. In a test run, a new train travels along a straight-line track. Data obtained from the speedometer indicate that the velocity of the train at any time t can be described by the velocity function
V(t)= 8t (0≤t≤30)
a. Find the position function of the train.
b. Find the position after 3 seconds. (Note: the train starts from the beginning of the track so when t = 0, the integration constant, C = 0.)

Answer 1

what is that pattern in between? :O

Answer 2

I'm not exactly sure what to answer so I'll just do whatever.

a#1: int(0,2) 3x^3-x^2+2 dx =
|(0,2) 3x^4/4 - x^3/3 + 2x.
3(2^4)/4 - (2)^3/3 + 2(2) - 3(0^4)/4 - (0)^3/3 + 2(0)
3(16)/4 - 8/3 + 4 - 0 + 0 - 0
48/4 - 8/3 + 4
12 - 8/3 + 4
16 - 8/3
48/3 - 8/3
40/3

a#2
same question, same answer.

b.
int(0,1) (e^2x - x^2)dx =
|(0,1) e^3x/3 - x^3/3
e^3(1)/3 - (1)^3/3 - e^3(0)/3 - (0)^3/3
e^3/3 - 1/3 - 1/3 + 0
e^3/3 - 2/3

c. int 1/x+3 dx =
ln(x+3) + C


2.
a.
acceleration equals velocity's derivative. a(t) = v'(t)
velocity equals displacement's derivative. v(t) = d'(t)
so displacement can be velocity's integral. d(t) = int (v(t)dt)

v(t) = 8t
d(t) = int(v(t)dt)
d(t) = int (8t)dt
d(t) = 8t^2/2 + C
d(t) = 4t^2 + C
d(0) = 4(0)^2 + C [step 1 of 2 for showing you why C = 0.]
C=0 [step 2 of 2 for showing you why C = 0.]
d(t) = 4t^2

b. int(0,3) v(t)dt =
d(3) - d(0) =
4(3^2) - 4(0^2) =
4(9) - 0 =
36m

I stepped it out for you because clearly you're either completely lost (these are forms of the most basic integrals) or you're just starting this unit, which can be difficult to comprehend.

Also i did this as fast as possible, so please don't hold me to any silly errors. good luck.

Answer 3

Your evaluations match what I have so far. Thanks a lot, I was checking it all and it really helped me out.