3. $\\int_{\\infty}^{3}\\frac{1}{(x - 2)^{3/2}}dx$ 4. $\\int_{\\infty}^{0}\\frac{1}{\\sqrt4{1 + x}}dx$ 5…

3. $\\int_{\\infty}^{3}\\frac{1}{(x - 2)^{3/2}}dx$ 4. $\\int_{\\infty}^{0}\\frac{1}{\\sqrt4{1 + x}}dx$ 5. $\\int_{-1}^{-\\infty}\\frac{1}{\\sqrt{2 - w}}dw$ 6. $\\int_{\\infty}^{0}\\frac{x}{(x^{2}+2)^{2}}dx$ 7. $\\int_{\\infty}^{4}e^{-y/2}dy$

3. $\\int_{\\infty}^{3}\\frac{1}{(x - 2)^{3/2}}dx$ 4. $\\int_{\\infty}^{0}\\frac{1}{\\sqrt4{1 + x}}dx$ 5. $\\int_{-1}^{-\\infty}\\frac{1}{\\sqrt{2 - w}}dw$ 6. $\\int_{\\infty}^{0}\\frac{x}{(x^{2}+2)^{2}}dx$ 7. $\\int_{\\infty}^{4}e^{-y/2}dy$

Answer

Explanation:

Step1: Substitute ( t = x - 2 )

Let ( t=x - 2 ), then ( dt=dx ). When ( x = 3 ), ( t = 1 ); as ( x\to\infty ), ( t\to\infty ). The integral becomes ( \int_{1}^{\infty}\frac{1}{t^{3/2}}dt )

Step2: Use the power - rule for integration

The power - rule for integration is ( \int t^{n}dt=\frac{t^{n + 1}}{n+1}+C) ((n\neq - 1)). For (n=-\frac{3}{2}), we have ( \int t^{-\frac{3}{2}}dt=\frac{t^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+C=\frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}+C=-2t^{-\frac{1}{2}}+C)

Step3: Evaluate the improper integral

(\int_{1}^{\infty}\frac{1}{t^{3/2}}dt=\lim_{b\to\infty}\int_{1}^{b}t^{-\frac{3}{2}}dt=\lim_{b\to\infty}\left[-2t^{-\frac{1}{2}}\right]{1}^{b}) [ \begin{align*} &=\lim{b\to\infty}\left(-2b^{-\frac{1}{2}}+2\times1^{-\frac{1}{2}}\right)\ &=\lim_{b\to\infty}\left(-\frac{2}{\sqrt{b}} + 2\right) \end{align*} ] Since (\lim_{b\to\infty}\frac{2}{\sqrt{b}}=0), the value of the integral is (2)

Answer:

The integral ( \int_{3}^{\infty}\frac{1}{(x - 2)^{3/2}}dx) is convergent and its value is (2)