Integral and its largest value

  • Thread starter Thread starter nepenthe
  • Start date Start date
  • Tags Tags
    Integral Value
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 4K views
nepenthe
Messages
4
Reaction score
0
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..
 
Physics news on Phys.org
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: