12. $$ \\int _ { 1 } ^ { \\infty } \\frac { e ^ { - \\sqrt { x } } } { \\sqrt { x } } d x $$
12. $$ \\int _ { 1 } ^ { \\infty } \\frac { e ^ { - \\sqrt { x } } } { \\sqrt { x } } d x $$
Answer
Explanation:
Step1: Substitution
Let ( t = \sqrt{x}), then (x=t^{2}) and (dx = 2t\space dt). When (x = 1), (t=1); as (x\rightarrow\infty), (t\rightarrow\infty). The integral becomes (\int_{1}^{\infty}\frac{e^{-t}}{t}\cdot2t\space dt=2\int_{1}^{\infty}e^{-t}\space dt).
Step2: Integrate (e^{-t})
The antiderivative of (e^{-t}) is (-e^{-t}). Using the fundamental theorem of calculus for improper integrals: (\lim_{b\rightarrow\infty}2\int_{1}^{b}e^{-t}\space dt=\lim_{b\rightarrow\infty}2\left[-e^{-t}\right]_{1}^{b}).
Step3: Evaluate the limit
(\lim_{b\rightarrow\infty}2\left(-e^{-b}+e^{-1}\right)). Since (\lim_{b\rightarrow\infty}e^{-b}=0), we have (2\left(0 + e^{-1}\right)).
Answer:
(\frac{2}{e})