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Chapter 17: Problem 1

Find \(\int_{1}^{\infty} \mathrm{e}^{-x} \mathrm{~d} x\).

Short Answer

Expert verified
Answer: The value of the improper integral is \(\frac{1}{\mathrm{e}}\).

Step by step solution

01

Identify the antiderivative

We need to find the antiderivative of \(e^{-x}\). The antiderivative of \(e^{ax}\), where \(a\) is a constant, is \(\frac{e^{ax}}{a}\). In this case, \(a = -1\), so the antiderivative of \(e^{-x}\) is \(-\frac{e^{-x}}{1}\) or simply \(-e^{-x}\).
02

Apply the limits of integration

We will now apply the limits of integration to the antiderivative we found in Step 1. This means we will evaluate the antiderivative at the upper limit (\(\infty\)) and the lower limit (1), and then subtract the lower limit result from the upper limit result: \(\left[-e^{-x}\right]_1^\infty = -e^{-\infty} - (-e^{-1})\)
03

Evaluate the antiderivative at the limits

\(-e^{-\infty}\) becomes \(0\) because as \(x\) approaches infinity, \(e^{-x}\) approaches 0, and \(-e^{-1}\) can be rewritten as \(-e^{-1} = -\frac{1}{\mathrm{e}}\). So the result is \(0 - \left(-\frac{1}{\mathrm{e}}\right)\)
04

Calculate the result

Now, we need to simplify the expression: \(0 - \left(-\frac{1}{\mathrm{e}}\right) = \frac{1}{\mathrm{e}}\) Hence, the value of \(\int_{1}^{\infty} \mathrm{e}^{-x} \mathrm{~d} x\) is \(\frac{1}{\mathrm{e}}\).

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Most popular questions from this chapter

Use Simpson's rule with the number of strips specified to approximate the following definite integrals: (a) \(\int_{0}^{0.8} \tan ^{2} x \mathrm{~d} x, 8\) strips (b) \(\int_{1}^{2} \sqrt{1+x^{3}} \mathrm{~d} x, 10\) strips

Find \(\int\left(\cos ^{2} \theta+\sin ^{2} \theta\right) \mathrm{d} \theta\).

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