$\\int_{0}^{8}\\frac{dx}{\\sqrt{1 + x}} =$
$\\int_{0}^{8}\\frac{dx}{\\sqrt{1 + x}} =$
Answer
Explanation:
Step1: Use substitution
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}}).
Step3: Integrate
Using the power rule (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 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 = 2u^{\frac{1}{2}}+C).
Step4: Evaluate the definite integral
(\left[2\sqrt{u}\right]_{1}^{9}=2\sqrt{9}-2\sqrt{1}).
Step5: Calculate the result
(2\times3 - 2\times1=6 - 2).
Answer:
(4)