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Susie's question at Yahoo! Answers regarding minimizing a definite integral

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MarkFL

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Feb 24, 2012
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Here is the question:

Calculus question, please, please answer.?

Find the value of a >0 that minimizes Int_{a}^{a^2} dx/(x_sqrt(x))

I am asking for what value of a the integral from a to a^2 dx/(x*sqrt(x)) will generate the smallest number.

Additional Details

I'm sorry, I meant to type:

The integral from a to a^2 of dx/(x+sqrt(x))
Here is a link to the question:

Calculus question, please, please answer.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

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Feb 24, 2012
13,775
Re: Susie's question at Yahoo! Answers regarding minimzing a definite integral

Hello Susie,

We are given to minimize:

\(\displaystyle g(a)=\int_a^{a^2}\frac{dx}{x+\sqrt{x}}\)

Obviously, we want to equate the derivative to zero and find the critical value(s). So we may utilize the anti-derivative form of the FTOC, and differentiate using the chain rule:

\(\displaystyle \frac{d}{da}\left(\int_a^{a^2}\frac{dx}{x+\sqrt{x}} \right)=0\)

\(\displaystyle \frac{1}{a^2+a}\cdot2a-\frac{1}{a+\sqrt{a}}=0\)

\(\displaystyle \frac{2}{a+1}=\frac{1}{a+\sqrt{a}}\)

\(\displaystyle a+2\sqrt{a}-1=0\)

Using the quadratic formula (and discarding the negative root), we find:

\(\displaystyle \sqrt{a}=\sqrt{2}-1\)

\(\displaystyle a=3-2\sqrt{2}\)

Use of the first derivative test shows the derivative of the function is negative to the left of the critical number and positive to the right, hence the extremum associated with the critical number we found is a minimum. As it is the only extremum in the given domain, we may conclude it is a global minimum.

To Susie, and any other guests viewing this topic, I invite and encourage you to post other calculus problem in our Calculus forum.

Best Regards,

Mark.