the integral in this exercise converges. evaluate the integral without using a table.\n int_{0}^{4} \frac{d…

the integral in this exercise converges. evaluate the integral without using a table.\n int_{0}^{4} \frac{d x}{sqrt{4-x}} \n int_{0}^{4} \frac{d x}{sqrt{4-x}}=square

the integral in this exercise converges. evaluate the integral without using a table.\n int_{0}^{4} \frac{d x}{sqrt{4-x}} \n int_{0}^{4} \frac{d x}{sqrt{4-x}}=square

Answer

Explanation:

Step1: Use substitution

Let (u = 4 - x), then (du=-dx). When (x = 0), (u = 4); when (x = 4), (u = 0). The integral (\int_{0}^{4}\frac{dx}{\sqrt{4 - x}}) becomes (-\int_{4}^{0}\frac{du}{\sqrt{u}}).

Step2: Integrate

We know that (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)). For (y=\frac{1}{\sqrt{u}}=u^{-\frac{1}{2}}), (\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C = 2u^{\frac{1}{2}}+C). So (-\int_{4}^{0}\frac{du}{\sqrt{u}}=\int_{0}^{4}\frac{du}{\sqrt{u}}). Evaluating (2u^{\frac{1}{2}}\big|_{0}^{4}). Substitute the upper - and lower - limits: (2\sqrt{4}-2\sqrt{0}).

Answer:

(4)