∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
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∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
----
For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
Answer:
P(t) = 27000 * (1/9)^(t/4)
Step-by-step explanation:
This problem can me modelled with an exponencial formula:
P = Po * (1+r)^t
Where P is the final value, Po is the inicial value, r is the rate and t is the amount of time.
In this problem, we have that the inicial population/value is 27000, the rate is -8/9 (negative because the population decays), and the time t is in months, so as the rate is for every 4 months, we use the value (t/4) in the exponencial.
So, our function will be:
P(t) = 27000 * (1-8/9)^(t/4)
P(t) = 27000 * (1/9)^(t/4)
The answer is 4/5
its very simple
Answer:
-2.125
Step-by-step explanation:
