Answer:
The total amount of pollution discharged during the first 7 years of operation is 1.955 tons
Step-by-step explanation:
Given
[tex]r(t) = \frac{t}{t^2 + 1}[/tex]
Required
The total amount in the first 7 years
This implies that:
[tex]r(t) = \frac{t}{t^2 + 1}; [0,7][/tex]
The total amount is calculated by integrating r(t) i.e.
[tex]v = \int\limits^a_b {r(t)} \, dt[/tex]
So:
[tex]v = \int\limits^7_0 {\frac{t}{t^2 + 1}} \, dt[/tex]
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We have:
[tex]t^2 + 1[/tex]
Differentiate
[tex]d(t^2 + 1) = 2t[/tex]
Rewrite as:
[tex]2t = d(t^2 + 1)[/tex]
Solve for t
[tex]t = \frac{1}{2}d(t^2 + 1)[/tex]
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So:
Make t the subject
[tex]v = \int\limits^7_0 {\frac{t}{t^2 + 1}} \, dt[/tex]
[tex]v = \int\limits^7_0\frac{1}{2}* {\frac{d(t^2 + 1)}{t^2 + 1}} \, dt[/tex]
[tex]v = \frac{1}{2}\int\limits^7_0 {\frac{d(t^2 + 1)}{t^2 + 1}} \, dt[/tex]
Integrate
[tex]v = \frac{1}{2}\ln(t^2 +1)|\limits^7_0[/tex]
Expand
[tex]v = \frac{1}{2}[\ln(7^2 +1) - \ln(0^2 +1)][/tex]
[tex]v = \frac{1}{2}[\ln(50) - \ln(1)][/tex]
[tex]v = \frac{1}{2}[3.91 - 0][/tex]
[tex]v = \frac{1}{2}[3.91][/tex]
[tex]v = 1.955[/tex]
