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How do you find the average value of #f(x)=x-2sqrtx# as x varies between #[0,2]#?

Answer 1

Ezekiel Alexander

Average value is given by

#A = 1/(b - a) int_a^b f(x)dx#

Therefore

#A = 1/2int_0^2 x - 2sqrt(x)dx#

#A = 1/2[1/2x^2 - 4/3x^(3/2)]_0^2#

#A = 1/2[2 - 4/3sqrt(8)]#

#A = 1 - 2/3(2)sqrt(2)#

#A = 1 - 4/3sqrt(2) ~~-0.89#

Hopefully this helps!

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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