To solve the integral [tex]\(\int_0^{\infty} e^{-x} \, dx\)[/tex], we can proceed as follows:
1. Identify the integrand and the limits of integration:
- The integrand is [tex]\(e^{-x}\)[/tex].
- The limits of integration are from 0 to [tex]\(\infty\)[/tex].
2. Recognize that this is an improper integral:
- The upper limit is [tex]\(\infty\)[/tex], so we need to interpret this as a limit.
3. Set up the limit to handle the improper integral:
- We will rewrite the integral as a limit of a definite integral with an upper bound [tex]\(b\)[/tex] that approaches [tex]\(\infty\)[/tex]:
[tex]\[
\int_0^{\infty} e^{-x} \, dx = \lim_{b \to \infty} \int_0^b e^{-x} \, dx
\][/tex]
4. Evaluate the definite integral:
- To find [tex]\(\int_0^b e^{-x} \, dx\)[/tex], we need to find the antiderivative of [tex]\(e^{-x}\)[/tex].
- The antiderivative of [tex]\(e^{-x}\)[/tex] is [tex]\(-e^{-x}\)[/tex].
- So, we evaluate:
[tex]\[
\int_0^b e^{-x} \, dx = \left[ -e^{-x} \right]_0^b
\][/tex]
5. Apply the bounds to the antiderivative:
- Substitute the upper bound [tex]\(b\)[/tex] and the lower bound 0 into the antiderivative:
[tex]\[
\left[ -e^{-x} \right]_0^b = -e^{-b} - (-e^0)
\][/tex]
- Simplify this expression:
[tex]\[
-e^{-b} + 1
\][/tex]
6. Take the limit as [tex]\(b\)[/tex] approaches [tex]\(\infty\)[/tex]:
- As [tex]\(b\)[/tex] approaches [tex]\(\infty\)[/tex], [tex]\(e^{-b}\)[/tex] approaches 0 because [tex]\(e^{-b}\)[/tex] becomes very small.
[tex]\[
\lim_{b \to \infty} \left( -e^{-b} + 1 \right) = 0 + 1
\][/tex]
- Therefore,
[tex]\[
\lim_{b \to \infty} \left( -e^{-b} + 1 \right) = 1
\][/tex]
Thus, the value of the integral [tex]\(\int_0^{\infty} e^{-x} \, dx\)[/tex] is [tex]\(1.0000000000000002\)[/tex].
