Average value of function

Apr 2010
156
0
What is the average value of the function f(x) = x^2(x^3 +1)^1/2 on the interval x = -1 to x = 2?

I tried using substitution on:

1/3 ∫ -1 to 2 x^2(x^3 + 1)^1/2 dx

which ended up to be 2/6 ∫ 0 to 3 u^2 du (after changing limits of integrals to u)

but I didnt get the right answer.

The correct answer was:

1/3 [ 2/9 (x^3 + 1)^3/2] x= -1 to x = 2 = 2

and it doesnt look like they used substitution.
 

mr fantastic

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Dec 2007
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Zeitgeist
What is the average value of the function f(x) = x^2(x^3 +1)^1/2 on the interval x = -1 to x = 2?

I tried using substitution on:

1/3 ∫ -1 to 2 x^2(x^3 + 1)^1/2 dx

which ended up to be 2/6 ∫ 0 to 3 u^2 du (after changing limits of integrals to u)

but I didnt get the right answer.

The correct answer was:

1/3 [ 2/9 (x^3 + 1)^3/2] x= -1 to x = 2 = 2

and it doesnt look like they used substitution.
You have not applied the correct substitution.

Substitute \(\displaystyle u = x^3 + 1\). You get \(\displaystyle \frac{1}{9} \int_{0}^9 u^{1/2} \, du\).
 
Dec 2009
411
131
What I think they did, although I'm not sure how 'average value' is defined for a function.

But it seems they took the length of the interval of \(\displaystyle [-1,2]\) wich is 3. And then calculated \(\displaystyle \frac{1}{3}\int_{-1}^{2}x^2(x^3+1)^{\frac{3}{2}}dx\)

You can use the substitution \(\displaystyle u= x^3+1, du=3x^2\)