Question

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

Explanation:

Step1: Simplify the integrand

$\frac{\sqrt{x}}{x} = x^{\frac{1}{2}} \cdot x^{-1} = x^{-\frac{1}{2}}$

Step2: Apply power rule for integration

The power rule for integration is $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ for $n
eq -1$. Here $n=-\frac{1}{2}$, so:
$\int x^{-\frac{1}{2}} dx = \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = 2x^{\frac{1}{2}} + C = 2\sqrt{x} + C$

Step3: Evaluate definite integral

Compute $2\sqrt{x}$ at the bounds 8 and 0:
$2\sqrt{8} - 2\sqrt{0} = 2 \cdot 2\sqrt{2} - 0 = 4\sqrt{2}$

Answer:

$4\sqrt{2}$