Question

(int_{0}^{8} \frac{dx}{sqrt{1 + x}})

Explanation:

Step1: Identify antiderivative of $\frac{1}{\sqrt{x}}$

The antiderivative of $x^{-\frac{1}{2}}$ is $2x^{\frac{1}{2}} = 2\sqrt{x}$

Step2: Apply Fundamental Theorem of Calculus

$$\int_{1}^{8} \frac{dx}{\sqrt{x}} = 2\sqrt{x}\bigg|_{1}^{8}$$

Step3: Evaluate bounds

$$2\sqrt{8} - 2\sqrt{1} = 2(2\sqrt{2}) - 2(1) = 4\sqrt{2} - 2$$

Answer:

$2(\sqrt{8} - 1)$ or simplified as $4\sqrt{2} - 2$