Integral and its largest value

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Homework Help Overview

The problem involves finding the curve from the family y=x^n along which the integral of the expression (25xy - 8y^2)dx attains its largest value, with specified boundaries from (0,0) to (1,1).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss evaluating the integral along the curve x^n and question the meaning of the original problem statement regarding the intersection of the curve and the integral.

Discussion Status

The discussion includes attempts to clarify the problem and explore potential interpretations. Some participants suggest evaluating the integral directly, while others express uncertainty about the original poster's intent. There is an exploration of the relationship between the curve and the integral.

Contextual Notes

There is a mention of the boundaries for the integral and a consideration of whether 'n' can be a rational number, indicating constraints on the problem setup.

nepenthe
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hello..could you please help me to solve this problem?

Along what curve of the family y=x^n does the integral
int{(25xy-8y^2)dx} attain its largest value? and the boundaries for the integral is from (0,0) to (1,1)

thank you..
 
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Any thoughts on this problem? Maybe just evaluating the integral along the curve x^n?
 
Along what curve of the family y=x^n does the integral
int{(25xy-8y^2)dx} attain its largest value?
\ doesn't mean a whole lot to me. Do you mean "where does x^n intersect the integral of 26xy- 8y^2 dx with the largest y value?
 
nepenthe said:
hello..could you please help me to solve this problem?
Along what curve of the family y=x^n does the integral
int{(25xy-8y^2)dx} attain its largest value? and the boundaries for the integral is from (0,0) to (1,1)
thank you..

Is this what you're looking for ??

[tex]\begin{gathered}<br /> y = x^n \Rightarrow 25xy - 8y^2 = 25x^{n + 1} - 8x^{2n} \hfill \\<br /> \frac{d}<br /> {{dn}}\left[ {\int\limits_0^1 {\left( {25x^{n + 1} - 8x^{2n} } \right)dx} } \right] = \frac{{16}}{{\left( {2n + 1} \right)^2 }} - \frac{{25}}{{\left( {n + 2} \right)^2 }} = 0 \Rightarrow \frac{4}{{2n + 1}} = \frac{5}{{n + 2}} \Rightarrow n = \frac{1}<br /> {2} \hfill \\<br /> \therefore {\text{Curve is }}y = \sqrt x \hfill \\ <br /> \end{gathered}[/tex]

(If you allow 'n' to be a rational number, that is :wink:)

---?Though I'm not sure this is what you're looking for :frown: ?
 
Last edited:

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