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Question: The improper integral \(\int_{0}^{\infty}{e^{- x}dx}\) is …… and the value is…....

The improper integral ∫0∞e−xdx\int_{0}^{\infty}{e^{- x}dx} is …… and the value is….

A

Convergent, 1

B

Divergent, 1

C

Convergent, 0

D

Divergent, 0

Answer

Convergent, 1

Explanation

Solution

I=∫0∞e−xdx=lim⁡k→∞∫0ke−xdxI = \int_{0}^{\infty}{e^{- x}dx = \lim_{k \rightarrow \infty}\int_{0}^{k}{e^{- x}dx}} ⇒ I=lim⁡k→∞[−e−x]0k=lim⁡k→∞[−e−k+e0]I = \lim_{k \rightarrow \infty}\lbrack - e^{- x}\rbrack_{0}^{k} = \lim_{k \rightarrow \infty}\lbrack - e^{- k} + e^{0}\rbrack ⇒I=lim⁡k→∞(1−e−k)=1−0=1[∵lim⁡k→∞e−k=e−∞=0]I = \lim_{k \rightarrow \infty}(1 - e^{- k}) = 1 - 0 = 1\lbrack\because\lim_{k \rightarrow \infty}e^{- k} = e^{- \infty} = 0\rbrack Thus, lim⁡k→∞∫0ke−xdx\lim_{k \rightarrow \infty}\int_{0}^{k}{e^{- x}dx} exists and is finite. Hence the given integral is convergent.