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Chapter 6: Problem 10

Explain why the integral is improper and determine whether it diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{0} e^{2 x} d x $$

Short Answer

Expert verified
The integral \(\int_{-\infty}^{0} e^{2x} dx\) is improper due to its lower limit being \(-\infty\). It converges and evaluates to \(\frac{1}{2}\).

Step by step solution

01

Identify the Improper Integral

The given integral, \(\int_{-\infty}^{0} e^{2x} dx\), is an improper integral because it has \(-\infty\) as the lower limit of integration.
02

Rewrite the Integral

Rewrite the improper integral as a limit as \(t\) approaches \(-\infty\) to deal with the infinity bound in the integral: \(\lim_{t\to -\infty} \int_{t}^{0} e^{2x} dx\)
03

Evaluate the Integral

Now integrate \(e^{2x}\) within the bounds of \(t\) to 0. The antiderivative of \(e^{2x}\) is \(\frac{1}{2}e^{2x}\), so: \(\lim_{t\to -\infty} [\frac{1}{2}e^{2x}]_{t}^{0} = \lim_{t\to -\infty} (\frac{1}{2}e^{2(0)} - \frac{1}{2}e^{2t})\). Substituting in the values of 0 and \(t\) gives: \(\lim_{t\to -\infty} (\frac{1}{2} - \frac{1}{2}e^{2t})\)
04

Evaluate the Limit

As \(t\) approaches \(-\infty\),\(e^{2t}\) approaches 0. Therefore, the limit of the integral is: \lim_{t\to -\infty} (\frac{1}{2} - \frac{1}{2}e^{2t}) = \frac{1}{2} - 0 = \frac{1}{2}\)
05

Determine Convergence/Divergence and Evaluate the Integral

Since the limit of the integral as \(t\) approaches \(-\infty\) is a finite number \(\frac{1}{2}\), the integral \(\int_{-\infty}^{0} e^{2x} dx\) converges and its value is \(\frac{1}{2}\).

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Most popular questions from this chapter

Decide whether the integral is improper. Explain your reasoning. $$ \int_{0}^{1} \frac{2 x-5}{x^{2}-5 x+6} d x $$

Use the Trapezoidal Rule and simpson's Rule to approximate the value of the definite integral for the indicated value of \(n\). Compare these results with the exact value of the definite integral. Round your answers to four decimal places. $$ \int_{1}^{2} \frac{1}{x} d x, n=8 $$

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