# Interval of the Riemann integral value

• MHB
• goody1
In summary, the conversation discusses finding an interval for a Riemann integral and suggests using the function $\tfrac1{\sqrt{4-x^2-x^3}}$, which is known to increase on the interval $0\leqslant x\leqslant 1$, to estimate values for $A$ and $B$.

#### goody1

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. :)