/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 13

Evaluate the following integrals or state that they diverge. $$\int_{0}^{\infty} e^{-a x} d x, a>0$$

Short Answer

Expert verified
Answer: The improper integral converges and has the value \(\frac{1}{a}\).

Step by step solution

01

Identify the integral as an improper integral

Since the upper limit of integration is infinity, the integral is an improper integral.
02

Rewrite the improper integral as a limit

We will rewrite the improper integral as a limit to evaluate it: $$\int_{0}^{\infty} e^{-ax} dx = \lim_{b \to \infty} \int_{0}^{b} e^{-ax} dx$$
03

Evaluate the integral

Now, we will evaluate the integral before taking the limit: $$\int{e^{-ax}} dx = -\frac{1}{a}e^{-ax} + C$$
04

Apply the limits of integration

Next, we will apply the limits of integration: $$\lim_{b \to \infty} \left( -\frac{1}{a}e^{-ab} + \frac{1}{a}e^{-0} \right)$$
05

Evaluate the limit

Now, we will evaluate the limit as b goes to infinity: $$-\frac{1}{a} \lim_{b \to \infty} (e^{-ab}) + \frac{1}{a}e^{0}$$ Since \(a > 0\), as \(b \to \infty\), we have \(ab \to \infty\) and thus \(e^{-ab} \to 0\). Thus, we can simplify the limit: $$-\frac{1}{a}(0) + \frac{1}{a}(1) = \frac{1}{a}$$
06

State the result

We have found a finite value for the integral, so it converges: $$\int_{0}^{\infty} e^{-ax} dx = \frac{1}{a}, \ a > 0$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the volume of the following solids. The region bounded by \(y=\frac{1}{\sqrt{4-x^{2}}}, y=0, x=-1,\) ar \(x=1\) is revolved about the \(x\) -axis.

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1-\cos x}$$

When is the volume finite? Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(0

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{d t}{2+e^{-t}}$$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ Verify relation \(A\) by differentiating \(x=2 \tan ^{-1} u\). Verify relations \(B\) and \(C\) using a right-triangle diagram and the double-angle formulas $$\sin x=2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) \text { and } \cos x=2 \cos ^{2}\left(\frac{x}{2}\right)-1$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.