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What are the two general ways in which an improper integral may occur? Choose the correct answer below. A. The interval of integration is infinite or the integrand is bounded on the interval of integration. B. The interval of integration is \( [\mathrm{a}, \mathrm{b}] \) or the integrand is bounded on the interval of integration. C. The interval of integration is infinite or the integrand is unbounded on the interval of integration. D. The interval of integration is \( [\mathrm{a}, \mathrm{b}] \) or the integrand is unbounded on the interval of integration.

Ask by Bernard Stewart. in the United States
Mar 10,2025

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The two ways an improper integral can occur are when the interval of integration is infinite or when the function being integrated is unbounded within the interval.

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The two ways in which an improper integral can occur are: 1. **Infinite Interval of Integration:** The limits of integration extend to infinity, for example, integrals such as \[ \int_{a}^{\infty} f(x) \, dx \quad \text{or} \quad \int_{-\infty}^{\infty} f(x) \, dx. \] 2. **Unbounded Integrand:** The function \( f(x) \) has a vertical asymptote or becomes unbounded at one or more points in the interval, even if the interval is finite. For instance, \[ \int_{a}^{b} f(x) \, dx \] where \( f(x) \) is unbounded as \( x \) approaches some value in the interval. Given these reasons, the correct answer is: C. The interval of integration is infinite or the integrand is unbounded on the interval of integration.

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An improper integral may occur when the interval of integration is infinite or when the integrand is unbounded on the interval of integration. This means that you encounter situations where you are trying to calculate an area that extends infinitely far (think of the area under a curve that stretches out endlessly) or where the function shoots up to infinity at some points (like a vertical asymptote). To visualize this, imagine you’re trying to measure the area under the curve of \( f(x) = 1/x \) from 1 to infinity. The challenge arises because that area keeps getting larger as \( x \) approaches infinity, making it a classic case of an improper integral! Similarly, with integrands like \( f(x) = 1/(x-1) \) over an interval containing \( x = 1 \), you encounter unbounded behavior, pulling you into the realm of improper integrals.

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