Question

P.T. $\pi/2 < \int_{0}^{1} \frac{dx}{\sqrt{1-x^{3/2}}} < 2$.

Answer 1

The statement $\frac{\pi}{2} < \int_{0}^{1} \frac{dx}{\sqrt{1-x^{3/2}}} < 2$ is true.

Answer 2

1. Lower Bound: For $x \in (0,1)$, establish $x^2 < x^{3/2}$. This implies $1-x^2 > 1-x^{3/2}$. Consequently, $\frac{1}{\sqrt{1-x^2}} < \frac{1}{\sqrt{1-x^{3/2}}}$. Integrating from 0 to 1 yields $\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}} < \int_{0}^{1} \frac{dx}{\sqrt{1-x^{3/2}}}$. Since $\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{2}$, we get $\frac{\pi}{2} < \int_{0}^{1} \frac{dx}{\sqrt{1-x^{3/2}}}$.
2. Upper Bound: Use the substitution $x = \sin^2 \theta$, so $dx = 2 \sin \theta \cos \theta d\theta$. The integral transforms to $\int_{0}^{\pi/2} \frac{2 \sin \theta \cos \theta}{\sqrt{1-\sin^3 \theta}} d\theta$. For $\theta \in (0, \pi/2)$, establish $\sin^3 \theta < \sin^2 \theta$. This implies $1-\sin^3 \theta > 1-\sin^2 \theta = \cos^2 \theta$. Thus, $\sqrt{1-\sin^3 \theta} > \cos \theta$, leading to $\frac{1}{\sqrt{1-\sin^3 \theta}} < \frac{1}{\cos \theta}$. Multiplying by $2 \sin \theta \cos \theta$ and integrating from 0 to $\pi/2$ yields $\int_{0}^{\pi/2} \frac{2 \sin \theta \cos \theta}{\sqrt{1-\sin^3 \theta}} d\theta < \int_{0}^{\pi/2} 2 \sin \theta d\theta = 2$. Therefore, $\int_{0}^{1} \frac{dx}{\sqrt{1-x^{3/2}}} < 2$.