Question

The value of \(\int_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^{2n}}}\) is (n∈N)

• A. less than or equal to \(\frac{\pi}{6}\)
• B. greater than or equal to 1
• C. less than \(\frac{1}{2}\)
• D. greater than \(\frac{\pi}{6}\)

Answer 1

The Correct Option is B

To solve the integral: \[ \int_0^{1/2} \left( \frac{1}{\sqrt{1 - x^2}} \right)^n dx \] Note that: \[ \frac{1}{\sqrt{1 - x^2}} = \frac{d}{dx} \arcsin x \] But since the integrand is raised to the power n, we cannot directly integrate unless n = 1. Let's analyze: \[ \left( \frac{1}{\sqrt{1 - x^2}} \right)^n = ( \arcsin' x )^n \] This is increasing on the interval \([0, 1/2]\), and for \(x \in [0, 1/2]\), \(\frac{1}{\sqrt{1 - x^2}} \geq 1\). So: \[ \left( \frac{1}{\sqrt{1 - x^2}}  \right)^n \geq 1^n = 1 \] Hence: \[ \int_0^{1/2} \left( \frac{1}{\sqrt{1 - x^2}} \right)^n dx \geq \int_0^{1/2} 1 \, dx = \frac{1}{2} \] But even more tightly, since the integrand is always greater than or equal to 1 on \([0, 1/2]\), and increases as x approaches 1/2, the integral value is always: \[ \int_0^{1/2} \left( \frac{1}{\sqrt{1 - x^2}} \right)^n dx \geq 1 \] Therefore, the correct answer is: Option (B): greater than or equal to 1