Interval of the Riemann integral value

3 replies · 2K views · last reply
goody1
Messages
16
Reaction score
0
Hello everyone, I have to find an interval of this Riemann integral. Does anybody know the easiest way how to do it? I think we need to do something with denominator, enlarge it somehow. My another guess is the integral is always larger than 0 (A=0) because the whole function is still larger than 0 on interval from 0 to 1. Thank you in advance.
 
Attachments
  • Inequality A ≤ integral from 0 to 1 of dx / sqrt(4 - x^2 - x^3) ≤ B, with A and B as unspecified bounds.
    integral.png
    1 KB · Views: 171
The function $\tfrac1{\sqrt{4-x^2-x^3}}$ increases on the interval $0\leqslant x\leqslant 1$. The minimum value occurs when $x=0$, and the maximum at $x=1$. You can use that to get estimates for $A$ and $B$.
 
Opalg said:
The function $\tfrac1{\sqrt{4-x^2-x^3}}$ increases on the interval $0\leqslant x\leqslant 1$. The minimum value occurs when $x=0$, and the maximum at $x=1$. You can use that to get estimates for $A$ and $B$.
I tried this. Is that corect solution?
 
Attachments
  • Handwritten chain of inequalities: A = 1/2 = 1/sqrt(4 - 0^2 - 0^3) ≤ 1/sqrt(4 - x^2 - x^3) ≤ 1/sqrt(4 - 1^2 - 1^3) = 1/sqrt(2) = sqrt(2)/2 = B.
    inn.png
    1.2 KB · Views: 190
goody said:
I tried this. Is that corect solution?
Yes, that is what I had in mind. :)