Best solutions to - "Find the average value fave of the function..."

Khadija Wells

Answered question

2021-09-28

Find the average value fave of the function ff on the given interval.

f (x) = \frac{x^{2}}{{(x^{3} + 30)}^{2}}, x \in [- 3, 3]

Clara Reese

Skilled2021-09-29Added 120 answers

The average value of the function f on the interval [a,b] is equal to:

f_{a v g} = \frac{1}{b - a} \int_{a}^{b} f (x) d x

The given function is

f (x) = \frac{x^{2}}{{(x^{3} + 30)}^{2}}, x \in [- 3, 3]

Thus, the average value of f is given by

f_{a v g} = \frac{1}{3 - (- 3)} \int_{- 3}^{3} \frac{x^{2}}{{(x^{3} + 30)}^{2}} d x = \frac{1}{6} \int_{- 3}^{3} \frac{x^{2}}{{(x^{3} + 30)}^{2}} d x

Now first calculate the indefinite integral

\int \frac{x^{2}}{{(x^{3} + 30)}^{2}} d x

\int_{- 3}^{3} \frac{x^{2}}{{(x^{3} + 30)}^{2}} d x = \int \frac{1}{3 u^{2}} d u

[Apply u - substitution:

u = x^{3} + 30

]

= \frac{1}{3} \cdot \frac{u^{- 2 + 1}}{- 2 + 1}

= - \frac{1}{3 u}

= - \frac{1}{3 (x^{3} + 30)}

[Substitute back:

u = x^{3} + 30

]
Therefore we get

f_{a v g} = \frac{1}{6} \int_{- 3}^{3} \frac{x^{2}}{{(x^{3} + 30)}^{2}} d x

= \frac{1}{6} ({[- \frac{1}{3 (x^{3} + 30)}]}_{x = 3} - {[- \frac{1}{3 (x^{3} + 30)}]}_{x = 3})

= \frac{1}{6} [- \frac{1}{171} - (- \frac{1}{9})]

= \frac{1}{6} \cdot \frac{2}{19}

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