$\\int_{0}^{8} \\frac{dx}{\\sqrt{1 + x}} = $

$\\int_{0}^{8} \\frac{dx}{\\sqrt{1 + x}} = $

$\\int_{0}^{8} \\frac{dx}{\\sqrt{1 + x}} = $

Answer

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^{n} ) is ( \frac{u^{n + 1}}{n+1}+C ) (for ( n\neq - 1 )). So ( \int u^{-\frac{1}{2}}du=2u^{\frac{1}{2}}+C = 2\sqrt{u}+C )

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus, ( \left[2\sqrt{u}\right]_{1}^{9}=2\sqrt{9}-2\sqrt{1} )

Step4: Calculate the result

( 2\times3 - 2\times1=6 - 2 = 4 )

Answer:

( 4 )