# The sales of Universal Instruments in the first t years of its operation are approximated by the...

## Question:

The sales of Universal Instruments in the first {eq}t {/eq} years of its operation are approximated by the function {eq}\displaystyle S (t) = t \sqrt {0.8 t^2 + 16} {/eq}, where {eq}S (t) {/eq} is measured in millions of dollars. What were Universal's average yearly sales over its first {eq}5 {/eq} years of operation?

## Average Value of a Function

The average value of a function {eq}f(x) {/eq} over the interval {eq}[a,b] {/eq} is given by {eq}\frac{1}{b-a}\int_a^b f(x)dx {/eq}. This idea is very similar to what most people think of when thinking of an average - add up everything and divide by the number of things you added up. Remember that an integral is really a sum - integration comes from taking the limit of a Riemann sum - and so we are dividing the integral (the sum) by the length of the interval (how many things we added up). Thinking in this way can help make the formula for the average value of a function easier to memorize

## Answer and Explanation:

We need to find the average value of the function {eq}S(t)=t\sqrt{0.8t^2+16} {/eq} over the interval {eq}[0,5] {/eq}

{eq}\frac{1}{5-0}\int_0^5 t\sqrt{0.8t^2+16} dt {/eq}

Using substitution with {eq}u=0.8t^2 +16 , du=1.6tdt {/eq} and remembering to change the bounds of integration, we have

{eq}\frac{1}{5}\cdot\frac{1}{1.6}\int_{16}^{36}u^{\frac{1}{2}}du = \frac{1}{8}\cdot\frac{2}{3}u^{\frac{3}{2}} |_{16}^{36} = \frac{1}{12}(36^{\frac{3}{2}}-16^{\frac{3}{2}}) = \frac{1}{12}(216-64) = \frac{38}{3}\approx 12.67 {/eq}million dollars

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