The level of air pollution a distance x miles from a tire factory is given by

L(x) = e-0.1x + 1/x2
What is the average level of pollution between 10 and 20 miles from the factory?

0.144

To find the average level of pollution between 10 and 20 miles from the factory, we need to compute the definite integral of the pollution function, L(x), over the interval [10, 20] and then divide by the width of the interval.

We can start by finding the antiderivative of L(x).

The antiderivative of e^(-0.1x) is -10e^(-0.1x), and the antiderivative of 1/x^2 is -1/x.

So the antiderivative of L(x) is -10e^(-0.1x) - 1/x.

To evaluate the definite integral of L(x) over the interval [10, 20], we can use the antiderivative:

∫[10, 20] L(x) dx = (-10e^(-0.1x) - 1/x)│[10, 20]

Let's plug in the upper and lower limits:

[-10e^(-0.1 * 20) - 1/20] - [-10e^(-0.1 * 10) - 1/10]

Simplifying further:

[-10e^(-2) - 1/20] - [-10e^(-1) - 1/10]

Now, we can evaluate this expression to find the average level of pollution:

(-10e^(-2) - 1/20) - (-10e^(-1) - 1/10) ≈ -10.04

Therefore, the average level of pollution between 10 and 20 miles from the factory is approximately -10.04.

To find the average level of pollution between 10 and 20 miles from the factory, we need to calculate the definite integral of the function L(x) over the interval [10, 20] and then divide it by the length of the interval.

The definite integral of a function f(x) over the interval [a, b] is denoted as ∫[a, b] f(x) dx and represents the net area under the curve of the function between x = a and x = b.

In this case, the function we need to integrate is L(x) = e^(-0.1x) + 1/x^2, and the interval is [10, 20]. Therefore, we need to calculate the following integral:

∫[10, 20] (e^(-0.1x) + 1/x^2) dx

To evaluate this integral, we can split it into two separate integrals:

∫[10, 20] e^(-0.1x) dx + ∫[10, 20] 1/x^2 dx

For the first integral, we can use a substitution to simplify the integration. Let u = -0.1x, then du/dx = -0.1, or dx = du/-0.1. Substituting these values in the integral, we have:

∫[10, 20] e^(-0.1x) dx = -1/0.1 ∫[10, 20] e^u du

= -10 ∫[10, 20] e^u du

= -10 [e^u] from 10 to 20

= -10 (e^(20*(-0.1)) - e^(10*(-0.1)))

For the second integral, we can simply apply the power rule of integration:

∫[10, 20] 1/x^2 dx = [-1/x] from 10 to 20

= [-1/20 - (-1/10)]

Now we can add the two integrals together and divide by the length of the interval:

Average level of pollution = ( -10 (e^(20*(-0.1)) - e^(10*(-0.1))) + (-1/20 - (-1/10)) ) / (20 - 10)

Simplifying the expression further will give you the average level of pollution between 10 and 20 miles from the factory.