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

We know that \(\int u^n du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). For \(\int\frac{du}{\sqrt{u}}=\int u^{-\frac{1}{2}}du\), here \(n=-\frac{1}{2}\).
So \(\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C=\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C = 2\sqrt{u}+C\)

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus \(\int_{a}^{b}F^\prime(x)dx=F(b)-F(a)\), we have:
\(2\sqrt{u}\big|_{1}^{9}=2\sqrt{9}-2\sqrt{1}\)

Step4: Calculate the result

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

Answer:

\(\boxed{4}\)