$$ int _ { 0 } ^ { infty } e ^ { - x } d x $$
$$ int _ { 0 } ^ { infty } e ^ { - x } d x $$
Answer
Explanation:
Step1: Find the antiderivative
The antiderivative of (e^{-x}) is (-e^{-x}).
Step2: Apply the fundamental theorem of calculus
[ \begin{align*} \int_{0}^{\infty}e^{-x}dx&=\lim_{b\rightarrow\infty}\int_{0}^{b}e^{-x}dx\ &=\lim_{b\rightarrow\infty}\left[-e^{-x}\right]{0}^{b}\ &=\lim{b\rightarrow\infty}\left(-e^{-b}+e^{0}\right) \end{align*} ]
Step3: Evaluate the limit
As (b\rightarrow\infty), (e^{-b}=\frac{1}{e^{b}}\rightarrow0). So (\lim_{b\rightarrow\infty}\left(-e^{-b}+e^{0}\right)= 0 + 1=1)
Answer:
(1)