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: Rewrite the integrand. \( \frac{1}{\sqrt{u}} = u^{-\frac{1}{2}} \), so the integral is \( \int_{1}^{9} u^{-\frac{1}{2}} du \)

Step3: Integrate using the power rule. The antiderivative of \( u^{n} \) is \( \frac{u^{n + 1}}{n+1} \) (for \( n

eq - 1 \)). Here \( n=-\frac{1}{2} \), so the antiderivative is \( \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=2u^{\frac{1}{2}} \)

Step4: Evaluate the definite integral. \( 2u^{\frac{1}{2}}\big|_{1}^{9}=2\times(9^{\frac{1}{2}} - 1^{\frac{1}{2}})=2\times(3 - 1) \)

Step5: Calculate the result. \( 2\times2 = 4 \)

Answer:

\( 4 \)