Question
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13. Multiple Choice 1 point Determine if the integral diverges or converges. Evaluate the integral if it converges. \( \int_{0}^{\infty} \frac{1}{e^{+}} d x \) \( \pi / 2 \) \( \pi / 4 \) This integral diverges. \( \pi \)

Ask by Potter Garza. in the United States
Feb 26,2025

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Answer

The integral \( \int_{0}^{\infty} \frac{1}{e^{x}} \, dx \) converges and equals 1.

Solution

To determine whether the integral \[ \int_{0}^{\infty} \frac{1}{e^{x}} \, dx \] converges or diverges, we can evaluate it step by step. 1. **Set up the integral**: We will evaluate the improper integral by taking the limit as \( b \) approaches infinity: \[ \int_{0}^{\infty} \frac{1}{e^{x}} \, dx = \lim_{b \to \infty} \int_{0}^{b} \frac{1}{e^{x}} \, dx \] 2. **Evaluate the integral**: The integral of \( \frac{1}{e^{x}} \) can be rewritten as \( e^{-x} \). The antiderivative of \( e^{-x} \) is \( -e^{-x} \). Thus, we have: \[ \int \frac{1}{e^{x}} \, dx = -e^{-x} + C \] 3. **Apply the limits**: Now we evaluate the definite integral from \( 0 \) to \( b \): \[ \int_{0}^{b} \frac{1}{e^{x}} \, dx = \left[-e^{-x}\right]_{0}^{b} = -e^{-b} - (-e^{0}) = -e^{-b} + 1 \] 4. **Take the limit**: Now we take the limit as \( b \) approaches infinity: \[ \lim_{b \to \infty} (-e^{-b} + 1) = \lim_{b \to \infty} (-0 + 1) = 1 \] Since the limit exists and is finite, the integral converges. Thus, the value of the integral is: \[ \int_{0}^{\infty} \frac{1}{e^{x}} \, dx = 1 \] Now, let's check the multiple-choice options provided: - \( \pi / 2 \) - \( \pi / 4 \) - This integral diverges. - \( \pi \) None of these options match the result of the integral, which is \( 1 \). Therefore, the correct conclusion is that the integral converges, but the provided options do not include the correct answer.

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The Deep Dive

First, let’s clarify the integral in question: it seems like there’s a small typo with \( e^{+} \). If you meant \( e^x \), the integral becomes \( \int_{0}^{\infty} \frac{1}{e^{x}} dx \), which converges. To evaluate it, you can compute it as follows: \[ \int_{0}^{\infty} e^{-x} dx = \left[-e^{-x}\right]_{0}^{\infty} = 0 + 1 = 1. \] So this integral does converge to 1! Now, if you were working with \( \frac{1}{e} \) in your integral, the integrand would be constant. When integrating from 0 to infinity, it would diverge since \(\int_{0}^{\infty} \frac{1}{e} dx\) is an infinite area. Wouldn't it be awesome if all confusing integrals led to such straightforward answers? They often don't, but that's what keeps the math world spinning! For extra practice, try considering other functions over infinite intervals, like \( \frac{1}{x^2} \) and see how convergence can change based on the exponent. Also, don't forget to brush up on the comparison test, a handy tool for determining whether an improper integral converges or diverges by comparing to a known integral!

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