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The daily production level for a product is given by N(x)=120−120e−0.2x𝑁(𝑥)=120−120𝑒−0.2𝑥 units, where x is the time in hours after production begins. Find the average production during the first 77 hours.Enter your answer in exact form or rounded to the nearest whole number.

Question

The daily production level for a product is given by N(x)=120−120e−0.2x𝑁(𝑥)=120−120𝑒−0.2𝑥 units, where x is the time in hours after production begins. Find the average production during the first 77 hours.Enter your answer in exact form or rounded to the nearest whole number.

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Solution

The average production level over a time period can be found by integrating the production function over that time period and then dividing by the length of the time period.

The integral of the production function N(x) from 0 to 77 is:

∫ from 0 to 77 of (120 - 120e^-0.2x) dx

This integral can be split into two parts:

= 120*∫ from 0 to 77 dx - 120*∫ from 0 to 77 e^-0.2x dx

The first integral is easy to compute:

= 120*[x] from 0 to 77 = 12077 - 1200 = 9240

The second integral requires the use of the formula for the integral of e^ax, which is (1/a)e^ax. Here, a = -0.2, so the integral is:

= -120*[-5e^-0.2x] from 0 to 77

= -120*(-5e^-0.277 - -5e^-0.20)

= -120*(-5e^-15.4 - -5)

= -120*(-5/447491.8474397752 - -5)

= -120*(-0.000011182 - -5)

= -120*(-4.999988818)

= 599998.5856

So, the integral of the production function from 0 to 77 is:

= 9240 - 599998.5856 = -591758.5856

The average production level over the first 77 hours is then this value divided by 77:

= -591758.5856 / 77 = -7682.5711 units

This is a negative value, which doesn't make sense in the context of the problem. There may have been a mistake in the calculation or the problem may be incorrectly stated. Please check the problem and try again.

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