Integrals - Definite Integral

Evaluate $\int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$.

Let $I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ (Equation 1). Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, we get: $I = \int_{0}^{\pi/2} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} dx$ $I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ (Equation 2). Adding (1) and (2): $2I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ $2I = \int_{0}^{\pi/2} 1 dx = [x]_0^{\pi/2} = \frac{\pi}{2}$ Therefore, $I = \frac{\pi}{4}$.

This example demonstrates the 'King's Property'. By substituting $x$ with $(\pi/2 - x)$, the denominator remains the same while the numerator changes from $\sin$ to $\cos$, allowing the two integrals to be added easily.

Evaluate $\int_{-1}^{2} |x^3 - x| dx$.

First, analyze the sign of $f(x) = x^3 - x = x(x-1)(x+1)$. In $[-1, 0]$, $f(x) \geq 0$. In $[0, 1]$, $f(x) \leq 0$. In $[1, 2]$, $f(x) \geq 0$. Thus, split the integral: $I = \int_{-1}^{0} (x^3 - x) dx + \int_{0}^{1} -(x^3 - x) dx + \int_{1}^{2} (x^3 - x) dx$. Calculating each: $\int (x^3 - x) dx = \frac{x^4}{4} - \frac{x^2}{2}$. $[\frac{x^4}{4} - \frac{x^2}{2}]_{-1}^{0} = 0 - (\frac{1}{4} - \frac{1}{2}) = \frac{1}{4}$. $-[\frac{x^4}{4} - \frac{x^2}{2}]_{0}^{1} = -[(\frac{1}{4} - \frac{1}{2}) - 0] = \frac{1}{4}$. $[\frac{x^4}{4} - \frac{x^2}{2}]_{1}^{2} = (\frac{16}{4} - \frac{4}{2}) - (\frac{1}{4} - \frac{1}{2}) = (4-2) + \frac{1}{4} = 2 + \frac{1}{4} = \frac{9}{4}$. Total $I = \frac{1}{4} + \frac{1}{4} + \frac{9}{4} = \frac{11}{4}$.

This demonstrates the splitting property of definite integrals based on the critical points of a modulus function. The integral is evaluated separately for regions where the expression inside the modulus is positive or negative.