QuestionJuly 16, 2025

Note: You may need to assume the fact that lim _(Marrow infty )M^ne^-M=0 for all n. Decide whether or not the given integral converges. int _(0)^infty e^-5xdx The integral converges The integral diverges. If the integral converges compute its value. (If the integral diverges enter DNE.) square

Note: You may need to assume the fact that lim _(Marrow infty )M^ne^-M=0 for all n. Decide whether or not the given integral converges. int _(0)^infty e^-5xdx The integral converges The integral diverges. If the integral converges compute its value. (If the integral diverges enter DNE.) square
Note: You may need to assume the fact that lim _(Marrow infty )M^ne^-M=0 for all n. Decide whether or not the given integral converges. int _(0)^infty e^-5xdx The integral converges The integral diverges. If the integral converges compute its value. (If the integral diverges enter DNE.) square

Solution
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Answer

The integral converges and its value is \frac{1}{5}. Explanation 1. Identify the Type of Integral The integral is an improper integral from 0 to \infty. 2. Determine Convergence Evaluate \int_{0}^{\infty} e^{-5x} \, dx. Since e^{-5x} decays exponentially as x \to \infty, the integral converges. 3. Compute the Integral Use the formula for the integral of an exponential function: \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C. Here, a = -5. \int e^{-5x} \, dx = -\frac{1}{5} e^{-5x} + C. 4. Evaluate the Definite Integral Evaluate from 0 to \infty: \lim_{b \to \infty} \left[-\frac{1}{5} e^{-5b}\right] + \frac{1}{5} e^{0}. As b \to \infty, e^{-5b} \to 0. Thus, the expression becomes 0 + \frac{1}{5} = \frac{1}{5}.

Explanation

1. Identify the Type of Integral<br /> The integral is an improper integral from 0 to $\infty$.<br /><br />2. Determine Convergence<br /> Evaluate $\int_{0}^{\infty} e^{-5x} \, dx$. Since $e^{-5x}$ decays exponentially as $x \to \infty$, the integral converges.<br /><br />3. Compute the Integral<br /> Use the formula for the integral of an exponential function: $\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C$. Here, $a = -5$.<br /> $\int e^{-5x} \, dx = -\frac{1}{5} e^{-5x} + C$.<br /><br />4. Evaluate the Definite Integral<br /> Evaluate from 0 to $\infty$: <br /> $\lim_{b \to \infty} \left[-\frac{1}{5} e^{-5b}\right] + \frac{1}{5} e^{0}$.<br /> As $b \to \infty$, $e^{-5b} \to 0$. Thus, the expression becomes $0 + \frac{1}{5} = \frac{1}{5}$.
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