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 the integral of \( u^{-\frac{1}{2}} \) with respect to \( u \) is \( 2u^{\frac{1}{2}}+C \) (by the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\), here \( n=-\frac{1}{2}\), so \(\frac{u^{-\frac{1}{2}+ 1}}{-\frac{1}{2}+1}=2u^{\frac{1}{2}}\)). Now we evaluate the definite integral: \( 2u^{\frac{1}{2}}\big|_{1}^{9}=2\times(9^{\frac{1}{2}}-1^{\frac{1}{2}}) \)

Step3: Calculate the result

Since \( 9^{\frac{1}{2}} = 3 \) and \( 1^{\frac{1}{2}}=1 \), we have \( 2\times(3 - 1)=2\times2 = 4 \)

Answer:

\( 4 \)