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 \( \int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C = 2u^{\frac{1}{2}}+C=2\sqrt{u}+C\).

Step3: Evaluate the definite integral

Evaluate \( 2\sqrt{u}\big|_{1}^{9}=2\sqrt{9}-2\sqrt{1}=2\times3 - 2\times1=6 - 2 = 4\).

Answer:

\(4\)