What is the average value of the function $f (x) = \frac{1}{x^{2}}$ $f(x) = 1/x^2$ on the interval $[1, 3]$ $[1,3]$ ?

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What is the average value of the function $f (x) = \frac{1}{x^{2}}$ $f(x) = 1/x^2$ on the interval $[1, 3]$ $[1,3]$ ?

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Answer 1 · 2016-03-23T02:32:23

$\frac{1}{3}$ $1/3$

Explanation:

The average value of the function $f (x)$ $f(x)$ on the continuous and differentiable interval $[a, b]$ $[a,b]$ can be found through:

$\frac{1}{b - a} \int_{a}^{b} f (x) d x$ $1/(b-a)int_a^bf(x)dx$

Applying this to the given situation gives:

$= \frac{1}{3 - 1} \int_{1}^{3} \frac{1}{x^{2}} d x = \frac{1}{2} \int_{1}^{3} x^{- 2} d x$ $=1/(3-1)int_1^3 1/x^2dx=1/2int_1^3 x^-2dx$

Using the integration rule

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx=x^(n+1)/(n+1)+C$

So, we want to evaluate

$\frac{x^{- 2 + 1}}{- 2 + 1} = \frac{x^{- 1}}{- 1} = - \frac{1}{x}$ $x^(-2+1)/(-2+1)=x^-1/(-1)=-1/x$

From $1$ $1$ to $3$ $3$ , all multiplied by $\frac{1}{2}$ $1/2$ .

$= \frac{1}{2} {[- \frac{1}{x}]}_{1}^{3} = \frac{1}{2} (- \frac{1}{3} - (- \frac{1}{1})) = \frac{1}{2} (- \frac{1}{3} + 1)$ $=1/2[-1/x]_1^3=1/2(-1/3-(-1/1))=1/2(-1/3+1)$

$= \frac{1}{2} (\frac{2}{3}) = \frac{1}{3}$ $=1/2(2/3)=1/3$

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What is the average value of the function $f (x) = \frac{1}{x^{2}}$ $f(x) = 1/x^2$ on the interval $[1, 3]$ $[1,3]$ ?

1 Answer

Explanation:

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What is the average value of the function $f (x) = \frac{1}{x^{2}}$ $f(x) = 1/x^2$ on the interval $[1, 3]$ $[1,3]$ ?