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Evaluate the integral of e^(-x) dx from 0 to infinity

Worked through step by step. More integrals.

Answer
1\boxed{1}

Since the derivative of −e−x-e^{-x} is e−xe^{-x}, applying the bounds from 00 to ∞\infty yields 0−(−1)=10 - (-1) = 1, confirming our result.

Problem: Evaluate the improper integral ∫0∞e−x dx\int_{0}^{\infty} e^{-x} \, dx.

Steps

  1. Rewrite as a limit
    lim⁡t→∞∫0te−x dx\lim_{t \to \infty} \int_{0}^{t} e^{-x} \, dx
  2. Find the antiderivative
    lim⁡t→∞[−e−x]0t\lim_{t \to \infty} \left[ -e^{-x} \right]_{0}^{t}
  3. Apply the Fundamental Theorem of Calculus
    lim⁡t→∞(−e−t−(−e0))\lim_{t \to \infty} \left( -e^{-t} - (-e^{0}) \right)
  4. Simplify the expression
    lim⁡t→∞(1−1et)\lim_{t \to \infty} \left( 1 - \frac{1}{e^{t}} \right)
  5. Evaluate the limit
    1−01 - 0

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