If a factory continuously pumps pollutants into the air at the rate of √(t)/15 tons per day, then the amount dumped after 3 days is
a)3.464
b)0.019
c)0.115
d)0.231
Note: I have integral from 0 to 3 of √(t)/15 dt but I guess something is wrong. Can you please, show me your work. Thanks
the answer is .231
Oh boy, let me see if I can clown my way through this math problem for you!
We want to find the amount of pollutants dumped after 3 days, which means we need to evaluate the integral of √(t)/15 from 0 to 3.
Let's start by finding the anti-derivative of √(t)/15. The derivative of √(t) is (1/2√(t)), so the integral becomes:
(1/2) integral of (1/√(t)) / 15 dt
Using the power rule for integration, we can simplify this to:
(1/30) integral of (1/√(t)) dt
The integral of (1/√(t)) is 2√(t), so our expression becomes:
(1/30) * 2√(t)
Now, we can evaluate this expression from 0 to 3:
(1/30) * 2√3 - (1/30) * 2√0
Since the square root of 0 is 0, the second term drops out and we're left with:
(2/30) * √3
Wait a sec, let me simplify that for you:
(1/15) * √3
Now, let's plug in 3:
(1/15) * √3 ≈ 0.115
And there you have it, option c) 0.115 is the approximate amount of pollutants dumped after 3 days. Who knew math could be this fun, huh?
To find the amount of pollutants dumped after 3 days, we need to calculate the definite integral of the expression √(t)/15 from 0 to 3.
First, let's evaluate the definite integral:
∫(0 to 3) √(t)/15 dt
To find the antiderivative of √(t)/15, we can use the power rule for integration. The power rule states that the integral of x^n dx is (1/(n+1)) * x^(n+1), where n is any real number except -1.
Applying the power rule, we get:
(1/15) * ∫(0 to 3) t^(1/2) dt
Integrating t^(1/2) using the power rule, we add 1 to the exponent and divide by the new exponent:
(1/15) * (2/3) * t^(3/2)| from 0 to 3
Evaluating the expression at the upper and lower limits:
(1/15) * (2/3) * (3^(3/2)) - (1/15) * (2/3) * (0^(3/2))
Simplifying further:
(2/45) * (3^(3/2)) - (1/15) * (2/3) * (0)
Notice that the term involving 0 is 0, so we can ignore it:
(2/45) * (3^(3/2))
Now, let's compute this expression using a calculator:
(2/45) * (3^(3/2)) ≈ 0.115
Therefore, the amount of pollutants dumped after 3 days is approximately 0.115 tons.
So the correct answer is c) 0.115.
Stop disliking the comment. Reiny has it correct. Just solve the integral with the range being 0 to 3 and you'll get the right answer.
Did you get this?
∫ √(t)/15 dt
= [(2/45 t^(3/2) ] from 0 to 3
I get one of the choices as the answer