# The electric current i (in A) as a function of the time t (in s) for a certain circuit is given...

## Question:

The electric current {eq}i {/eq} (in {eq}A {/eq}) as a function of the time {eq}t {/eq} (in {eq}s {/eq}) for a certain circuit is given by {eq}i = 2 t - t^2 {/eq}. Find the average value of the current with respect to time for the first {eq}5.0\ s {/eq}.

## Average Value of a Function

If we wish to find the average value of a function, we can't use the same technique as we learned for taking the average of a data set. Instead, there's an integral that will find this average.

{eq}\frac{1}{b-a} \int_a^b f(x) dx {/eq}

## Answer and Explanation:

Since we wish to find the average value of this function over the first five seconds, that means finding a numerical value for the following integral.

{eq}\frac{1}{5-0} \int_0^5 (2 t - t^2) dt {/eq}

Since this integrand has an antiderivative that can be found easily, we can apply the Fundamental Theorem of Calculus.

{eq}\begin{align*} \frac{1}{5} \int_0^5 (2 t - t^2) dt &= \frac{1}{5} ( t^2 - \frac{1}{3}t^3)|_0^5\\ &= \frac{1}{5} ( ( (5)^2 - \frac{1}{3}(5)^3) - ( (0)^2 - \frac{1}{3}(0)^3))\\ &= \frac{1}{5} (-\frac{50}{3})\\ &= - \frac{10}{3} \end{align*} {/eq}

Therefore, the average value of this function is {eq}- \frac{10}{3} {/eq}.

#### Learn more about this topic:

Average Value Theorem

from Math 104: Calculus

Chapter 12 / Lesson 9
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