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

Answer:

\(4\)