Estimate Maximum and Minimum of Integral $$I=\int_{0}^{1} \frac{dx}{\sqrt{4-x^2-x^3}}$$
My try: we have By Mean value Theorem of Integration , if $f(x)$ is Continuous with Infimum and Supremum as $m$ and $M$ Then
$$m(b-a) \lt \int_{a}^{b} f(x)dx \lt M(b-a)$$
It Means we need to find Infimum and Supremum of $f(x)=\frac{1}{\sqrt{4-x^2-x^3}}$ in $\left[0 \:\:, 1\: \right]$
Now should we use differentiation to find $m$ and $M$ or is there any other way to estimate?
Since the function $f(x)=\frac{1}{\sqrt{4-x^2-x^3}}$ is increasing for $0\leq x\leq 1$, we have $$\frac 12=f(0)\leq f(x)\leq f(1)=\frac{\sqrt 2}2$$ for all $x\in [0,1]$. Consequently, $$\frac 12\leq\int_0^1\frac{1}{\sqrt{4-x^2-x^3}}\mathrm dx\leq\frac{\sqrt 2}2$$