Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{0} e^{-x} d x $$

Short answer

The given improper integral converges and the value of the integral is 1.

Step-by-step solution

Rewrite the Integral

Rewrite the improper integral as a limit. This is done because the concept of integration from negative infinity isn't directly defined. So, we use the concept of a limit to provide a meaning. Rewrite as \(\lim_{a \to -\infty} \int_{a}^{0} e^{-x} dx \). Where 'a' tends to negative infinity.

Evaluate the Integral

Evaluate the integral within the limit as you normally would. The integral of \(e^{-x}\) is \(-e^{-x}\). Therefore, substitute the integral in the limit by its value to give: \(\lim_{a \to -\infty} [-e^{-x}]_{a}^{0}\).

Substitution of the Limits

Substitute the limits of the integral into the expression. Remember the property \(-[f(b) - f(a)] = f(a) - f(b)\), so the integral becomes: \(\lim_{a \to -\infty} [-e^{0} - (-e^{-a})]\) which in turn simplifies to \(\lim_{a \to -\infty} [1 - e^{-a}]\).

Apply the Limit

Apply the limit and simplify. As \(a\) tends to \(-\infty\), \(e^{-a}\) tends to 0 (since \(-a\) tends to infinity). Thus, the limit simplifies to \(1 - 0 = 1\).