Interval of the Riemann integral value

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SUMMARY

The discussion focuses on finding the interval of the Riemann integral for the function $\tfrac1{\sqrt{4-x^2-x^3}}$ over the range $0 \leq x \leq 1$. Participants confirm that the function is increasing within this interval, with a minimum value at $x=0$ and a maximum at $x=1$. The integral is established to be greater than 0, reinforcing the conclusion that the estimates for the integral's bounds, A and B, can be derived from these values.

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goody1
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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.
 

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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?
 

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goody said:
I tried this. Is that corect solution?
Yes, that is what I had in mind. :)
 

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