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Chapter 6: Problem 8

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 given integral is improper because it has an infinite interval of integration. After analysis, it can be concluded that the integral \(\int_{-\infty}^{0} e^{2x} dx\) converges, and the value of the integral is \(\frac{1}{2}\).

Step by step solution

01

Identify the integral

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

Replace \(-\infty\) with a variable

Let's replace \(-\infty\) with a variable, say \(a\). Now, the integral becomes \(\int_{a}^{0} e^{2x} dx\) and we will consider the limit as \(a\) approaches \(-\infty\). This modification transforms our problem into one that involves a proper integral, which we know how to handle.
03

Evaluate the integral

We evaluate the antiderivative of \(e^{2x}\) which is \(\frac{1}{2}e^{2x}\). Using the Fundamental Theorem of Calculus, we evaluate \(\frac{1}{2} e^{2x}\) between 0 and \(a\), yielding \(\frac{1}{2}e^{2*0} - \frac{1}{2}e^{2a} = \frac{1}{2} - \frac{1}{2} e^{2a}\).
04

Evaluate limit

Now we need to evaluate \(\lim_{a \to -\infty} (\frac{1}{2} - \frac{1}{2} e^{2a})\). As \(a\) goes to \(-\infty\), \(e^{2a}\) goes to 0, because the exponential function approaches 0 faster than any polynomial when \(x\) tends to \(-\infty\). Thus, the whole expression approaches \(\frac{1}{2}\).

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Most popular questions from this chapter

Show that the indeterminate forms \(0^{\circ}\) \(\infty^{0},\) and \(1^{\infty}\) do not always have a value of 1 by evaluating each limit. (a) \(\lim _{x \rightarrow 0^{+}} x^{\ln 2 /(1+\ln x)}\) (b) \(\lim _{x \rightarrow \infty} x^{\ln 2 /(1+\ln x)}\) (c) \(\lim _{x \rightarrow 0}(x+1)^{(\ln 2) / x}\)

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