$$\\int_{0}^{8} \\frac{dx}{\\sqrt{1+x}} =$$
$$\\int_{0}^{8} \\frac{dx}{\\sqrt{1+x}} =$$
Answer
Explanation:
Step1: Identify the integral form
The given expression is a definite integral of the form $\int_{a}^{b} (1+x)^{-1/2} dx$.
Step2: Apply the power rule for integration
The antiderivative of $(1+x)^{n}$ is $\frac{(1+x)^{n+1}}{n+1}$. Here $n = -1/2$. $$ \int (1+x)^{-1/2} dx = \frac{(1+x)^{1/2}}{1/2} = 2\sqrt{1+x} $$
Step3: Evaluate at the upper and lower limits
Substitute the upper limit $x=8$ and the lower limit $x=0$ into the antiderivative. $$ \left[ 2\sqrt{1+x} \right]_{0}^{8} = 2\sqrt{1+8} - 2\sqrt{1+0} $$
Step4: Simplify the numerical expression
Calculate the square roots and perform the subtraction. $$ 2\sqrt{9} - 2\sqrt{1} = 2(3) - 2(1) = 6 - 2 = 4 $$
Answer:
4