Question

(int_{0}^{8} \frac{dx}{sqrt{1 + x}})

Explanation:

Step1: Substitute \( u = 1 + x \)

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: Rewrite the integrand

Rewrite \( \frac{1}{\sqrt{u}} \) as \( u^{-\frac{1}{2}} \). So the integral is \( \int_{1}^{9} u^{-\frac{1}{2}} du \).

Step3: Integrate using the power rule

The power rule for integration is \( \int x^n dx=\frac{x^{n + 1}}{n+1}+C \) (\( n
eq - 1 \)). For \( n=-\frac{1}{2} \), we have \( \int u^{-\frac{1}{2}} du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C=\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C = 2\sqrt{u}+C \).

Step4: Evaluate the definite integral

Evaluate \( 2\sqrt{u} \) from 1 to 9: \( 2\sqrt{9}-2\sqrt{1}=2\times3 - 2\times1=6 - 2 = 4 \).

Answer:

\( 4 \)