$$\\int_{0}^{8} \\frac{dx}{\\sqrt{1 + x}}$$
$$\\int_{0}^{8} \\frac{dx}{\\sqrt{1 + x}}$$
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
We know that ( \int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)). For ( \int\frac{du}{\sqrt{u}}=\int u^{-\frac{1}{2}}du), here (n=-\frac{1}{2}), so ( \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) Now evaluate the definite integral: ( \left[2\sqrt{u}\right]_{1}^{9}=2\sqrt{9}-2\sqrt{1})
Step3: Calculate the result
Since ( \sqrt{9} = 3) and ( \sqrt{1}=1), we have (2\times3 - 2\times1=6 - 2 = 4)
Answer:
(4)