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 \( u^{-\frac{1}{2}} \) with respect to \( u \) is \( 2u^{\frac{1}{2}} + C \) (where \( C \) is the constant of integration, but we don't need it for definite integral).

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus, we have \( 2u^{\frac{1}{2}}\big|_{1}^{9} = 2(9^{\frac{1}{2}} - 1^{\frac{1}{2}}) \).

Step4: Simplify the expression

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

Answer:

\( 4 \)