Find the average value of the function f(x)=x^2 on the interval x=1 to x=4.

Question:

Find the average value of the function {eq}f(x)=x^2{/eq} on the interval {eq}x=1{/eq} to {eq}x=4{/eq}.

Average Value of a Function

Let {eq}f(x) {/eq} be an integrable function over the domain {eq}[a,b], {/eq} where {eq}a \neq b . {/eq} Then the average value of the function over the domain is given by the definite integral {eq}\dfrac{1}{b-a} \displaystyle \int_b^a f(x) dx. {/eq}

Answer and Explanation:

Consider the function {eq}f(x) = x^2 {/eq} over the interval {eq}1 \leq x \leq 4. {/eq} The average value is given by the following integral:

{eq}\begin{eqnarray*} f_{avg} &=& \dfrac{1}{4 -1} \displaystyle \int_1^4 x^2 dx \\ &=& \dfrac13 \left( \dfrac13 x^3 \right|_1^4 \\ &=& \dfrac19 \left( 64 - 1 \right) \\ &=& \dfrac{63}{9} = 7. \end{eqnarray*} {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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