Question

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

Explanation:

Step1: Use substitution method

Let \( u = 1 + x \), then \( du = dx \). When \( x = 0 \), \( u = 1 \); when \( x = 8 \), \( u = 9 \).
The integral becomes \( \int_{1}^{9} \frac{du}{\sqrt{u}} \)

Step2: Integrate the function

Rewrite \( \frac{1}{\sqrt{u}} \) as \( u^{-\frac{1}{2}} \). The integral of \( u^{n} \) is \( \frac{u^{n + 1}}{n + 1} + C \) (for \( n
eq - 1 \)). So the integral of \( u^{-\frac{1}{2}} \) is \( \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=\frac{u^{\frac{1}{2}}}{\frac{1}{2}} = 2u^{\frac{1}{2}}+C \)
Evaluating from 1 to 9: \( 2u^{\frac{1}{2}}\big|_{1}^{9}=2\sqrt{9}-2\sqrt{1} \)

Step3: Calculate the result

\( 2\times3 - 2\times1=6 - 2 = 4 \)

Answer:

\( 4 \)