MHB Interval of the Riemann integral value

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To find the interval of the Riemann integral for the function $\tfrac1{\sqrt{4-x^2-x^3}}$, it is important to note that the function is increasing on the interval from 0 to 1. The minimum value occurs at x=0, while the maximum is at x=1, which helps in estimating the integral's bounds. The integral is expected to be greater than 0 since the function remains positive within this interval. Enlarging the denominator can also aid in finding the integral's value. Overall, the approach discussed aligns with the correct method for solving the problem.
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. :)
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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