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

In summary, the question asks for the value of a that minimizes the integral from a to a^2 of dx/(x+sqrt(x)). After finding the derivative and using the quadratic formula, the critical value is found to be a = 3-2sqrt(2), which is a global minimum as it is the only extremum in the given domain. The link to the question is also provided for reference.
  • #1
MarkFL
Gold Member
MHB
13,288
12
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.
 
Mathematics news on Phys.org
  • #2
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 http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.
 

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

1. How can I minimize a definite integral?

To minimize a definite integral, you need to use calculus techniques such as differentiation and optimization. First, take the derivative of the integrand and set it equal to zero. Then, solve for the variable and substitute it back into the original integral to find the minimum value.

2. Can I use any integration method to minimize a definite integral?

Yes, you can use any integration method to minimize a definite integral as long as you follow the steps mentioned above. Some commonly used integration methods are substitution, integration by parts, and trigonometric substitution.

3. Is there a shortcut or easier way to minimize a definite integral?

Unfortunately, there is no shortcut or easier way to minimize a definite integral. It requires a good understanding of calculus and integration techniques to find the minimum value accurately.

4. What are the applications of minimizing a definite integral?

Minimizing a definite integral has various real-world applications, especially in the fields of physics and engineering. It is used to find the minimum value of a function, which can represent the most efficient or optimal solution in a given scenario.

5. Are there any online resources or tools that can help with minimizing a definite integral?

Yes, there are several online resources and tools available that can help with minimizing a definite integral. Some popular ones include WolframAlpha, Symbolab, and Mathway. However, it is always recommended to have a good understanding of the concepts and techniques before using these tools.

Similar threads

Replies
1
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
  • General Math
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
874
  • Calculus and Beyond Homework Help
Replies
10
Views
674
Back
Top