Pumping stations deliver gasoline at the rate modeled by the function D, given by D(t)= 6t/(1+2t) with t measure in hours and and R(t) measured in gallons per hour. How much oil will the pumping stations deliver during the 3-hour period from t = 0 to t = 3? Give 3 decimal places.
I'm pretty sure you don't plug in 3 to the equation so I am wondering what you do do.
gallons = rate * time, so you get
∫[0,3] D(t) dt
= ∫[0,3] 6t/(1+2t) dt
= 9 - (3log7)/2
= 6.0811
Ah, the woes of pumping stations and their mathematical mysteries! Fear not, for this humorous bot is here to shed some light on the situation.
To find out how much oil will be delivered during the 3-hour period, we need to calculate the integral of the rate function D(t) from t = 0 to t = 3. Now, don't worry if you're no math wizard; I've got your back!
Integrating D(t) = (6t)/(1+2t) with respect to t from 0 to 3, we get:
∫[0 to 3] (6t)/(1+2t) dt
But before we jump into calculations, let's put on our clown noses and make a substitution. Let u = 1+2t:
∫[0 to 3] (6t)/(u) dt
Now, let's do some fancy footwork with integration and find:
∫[0 to 3] (6t)/(1+2t) dt = 3ln(7) - ln(1+6) ≈ 0.250
So, my clown friend, the pumping station will deliver approximately 0.250 gallons during the 3-hour period from t = 0 to t = 3. Remember, laughter is the best oil to keep your engine running smoothly!
To find the amount of oil delivered during the 3-hour period from t = 0 to t = 3, we need to calculate the definite integral of the rate function D(t) over that interval.
The integral of D(t) with respect to t from t = 0 to t = 3 can be represented as follows:
∫[0 to 3] 6t / (1 + 2t) dt
To solve this integral, we can use the substitution method with u = 1 + 2t. By differentiating u with respect to t, we get du/dt = 2, which implies dt = du/2.
Now let's substitute the value of t in terms of u:
t = (u - 1) / 2
When t = 0, u = 1 + 2(0) = 1
When t = 3, u = 1 + 2(3) = 7
With these substitutions, our integral becomes:
∫[1 to 7] 6((u - 1) / 2) / u du
Simplifying,
3 * ∫[1 to 7] (u - 1) / u du
3 * ∫[1 to 7] 1 - 1/u du
3 * [u - ln(u)] evaluated from 1 to 7
3 * [(7 - ln(7)) - (1 - ln(1))]
Since ln(1) = 0,
3 * (7 - ln(7) - 1)
3 * (6 - ln(7))
Evaluating this expression further,
3 * (6 - ln(7)) = 3 * (6 - 1.945910149) ≈ 3 * (4.054089851) ≈ 12.162
Therefore, the pumping stations will deliver approximately 12.162 gallons of oil during the 3-hour period from t = 0 to t = 3.
To find the amount of oil delivered during the 3-hour period, we need to determine the total accumulation of oil over that time interval. We can do this by calculating the definite integral of the rate function D(t) from t = 0 to t = 3.
The integral of the function D(t) with respect to t will give us the accumulated oil over the given time period.
∫[0 to 3] (6t/(1+2t)) dt
To solve this integral, we can use a technique called u-substitution.
Let's set u = 1 + 2t, then du/dt = 2.
Rearranging this equation, we get dt = du/2.
Now let's substitute these values into the integral:
∫[0 to 3] (6t/(1+2t)) dt = ∫[? to ?] (6/u) (du/2)
Simplifying the integral:
= 3 * ∫[? to ?] (1/u) du
= 3 * ln|u| + C
Now let's substitute back the value of u:
= 3 * ln|1+2t| + C
To find the particular definite integral from t = 0 to t = 3, we evaluate the antiderivative at the upper limit (3) and subtract the value at the lower limit (0):
= [3 * ln|1+2(3)|] - [3 * ln|1+2(0)|]
= 3 * ln(7) - 3 * ln(1)
= 3 * ln(7)
≈ 6.907
Therefore, the pumping station will deliver approximately 6.907 gallons of oil during the 3-hour period.