A factory is discharging pollution into a lake at the rate of r(t) tons per year given below, where t is the number of years that the factory has been in operation. Find the total amount of pollution discharged during the first 7 years of operation. (Round your answer to two decimal places.)

Sagot :

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]