/*! 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 103 Evaluate the definite integral. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 103

Evaluate the definite integral. Use a graphing utility to verify your result.\(\int_{1}^{3} \frac{e^{3 / x}}{x^{2}} d x\)

Short Answer

Expert verified
The definite integral of the function \( \frac{e^{3 / x}}{x^{2}} \) over the interval from 1 to 3 is \( -(e^3 - e) \).

Step by step solution

01

Substitute u

Let \( u = 3/x \). Then \( du = -3/x^2 \, dx \). Now replace \( 3/x \) with \( u \) in the integral and \( dx \) with \( -du/3 \). The new integral becomes \( -\int e^u \, du \).
02

Solve the new integral

The antiderivative of \( e^u \) is just \( e^u \). So the integral is \( -e^u + C \).
03

Substitute back the original term

Now substitute back for \( u \). We get \( -e^{3/x} + C \).
04

Evaluate the definite integral

As we are solving a definite integral, we substitute the limits of integration. So we get \( -e^{3/1} - ( - e^{3/3}) = - e^3 + e = -(e^3 - e) \).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Techniques
Integration is an essential tool in calculus used to solve problems involving areas under curves, among other applications. There are various techniques to evaluate integrals, each suited to different types of functions. Some of the main techniques include:
  • Basic Integration: Finding the antiderivative of simple functions.
  • Integration by Substitution: Simplifying complex integrals by changing variables.
  • Integration by Parts: Useful for products of functions.
  • Partial Fraction Decomposition: Breaking down complex fractions into simpler parts.
  • Trigonometric Substitution: Suitable for integrals involving radical expressions.
In the given exercise, the substitution method is used due to the complexity of the integrand. By transforming the variable, we simplify the process and make the integral easier to evaluate.
Substitution Method in Calculus
Substitution is one of the most effective methods to evaluate an integral by making a variable change. The goal is to transform the original integral into a simpler form. Here's how it works:
  • Select an appropriate substitution, often represented as \( u \), that reduces the complexity of the integrand.
  • Calculate \( du \) in terms of the original variable \( dx \). This process typically involves differentiation of the substitution \( u \).
  • Rewrite the integral in terms of \( u \), replacing both the original function and the differential \( dx \).
  • Evaluate the new integral, which should now be simpler to solve.
  • Finally, substitute back the original expression for \( u \) to express the result in terms of the initial variable.
In our exercise, \( u = 3/x \) is chosen, converting the integral into a simpler form, \( -\int e^u \, du \). This approach significantly facilitates finding the antiderivative.
Definite Integration Limits
When dealing with definite integrals, the integration limits define the start and end points on the x-axis. These are the bounds within which the integration is performed. Here are the key steps to using definite integration limits:
  • Identify the integral's limits of integration, denoted as the lower and upper bounds \( a \) and \( b \).
  • Once the indefinite integral is found, apply the limits. This involves calculating the antiderivative at the upper limit and subtracting the antiderivative evaluated at the lower limit.
  • This process gives the net area between the curve of the function and the x-axis over the specified interval.
In the solution presented, after substituting the variable and simplifying the integral, the limits \( x = 1 \) and \( x = 3 \) are applied to finalize the evaluation. This results in substituting these limits into the expression \( -e^{3/x} \) and calculating the difference, which yields the value of the definite integral.

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

In Exercises 81 and 82, find \(d y / d x\) at the given point for the equation. \(x=y^{3}-7 y^{2}+2, \quad(-4,1)\)

Use the functions \(f(x)=x+4\) and \(g(x)=2 x-5\) to find the given function. \((f \circ g)^{-1}\)

Find, to three decimal places, the value of \(x\) such that \(e^{-x}=x\). (Use Newton's Method or the zero or root feature of a graphing utility.)

Find the integral. $$ \int x \operatorname{csch}^{2} \frac{x^{2}}{2} d x $$

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

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.