/*! 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 96 Compute \(\int_{0}^{1} \ln x d x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 96

Compute \(\int_{0}^{1} \ln x d x\) using integration by parts. Then explain why \(-\int_{0}^{\infty} e^{-x} d x\) (an easier integral) gives the same result.

Short Answer

Expert verified
Question: Show that the integral of ln(x) on the interval [0,1] using integration by parts is equivalent in magnitude but opposite in sign to the integral \(-\int_{0}^{\infty} e^{-x} \, dx\). Solution: Based on our calculation using integration by parts, we found that $$\int_{0}^{1} \ln(x) \, dx = -1$$ and $$-\int_{0}^{\infty} e^{-x} \, dx = 1$$ These two integral results are equal in magnitude but have opposite signs.

Step by step solution

01

Choose u(x) and v'(x)

To use integration by parts, we first pick u(x) and v'(x). We want to choose these functions such that the integral of u'(x)v(x) is easier to compute than the original integral. It's common to choose u(x) = ln(x) and v'(x) = 1.
02

Compute u'(x) and v(x)

Next, we need to compute the derivatives and antiderivatives for our chosen functions u(x) = ln(x) and v'(x) = 1. For u(x) = ln(x), the derivative u'(x) is: $$u'(x) = \frac{1}{x}$$ For v'(x) = 1, the antiderivative v(x) is: $$v(x) = x$$
03

Apply the integration by parts formula and evaluate on the interval [0,1]

Now, we will apply the integration by parts formula using our chosen functions and their derivatives: $$\int_{0}^{1} \ln(x) \, dx = \int_{0}^{1} u(x) v'(x) \, dx = \bigg[u(x)v(x) - \int_{0}^{1} u'(x)v(x) \, dx\bigg] $$ Substitute u(x), u'(x), and v(x) from Step 2: $$= \bigg[\ln(x)x - \int_{0}^{1} \frac{1}{x}x \, dx\bigg]$$ Now, evaluate the integral on the interval [0,1]: $$\bigg[\ln(1)(1) - \ln(0)(0) - \int_{0}^{1} 1 \, dx\bigg] = -\int_{0}^{1} 1 \, dx$$ The integral of 1 is x, so we get: $$-\Bigg[x\Bigg]_0^1 = -(1 - 0) = -1\text.$$ Thus, $$\int_{0}^{1} \ln(x) \, dx = -1\text.$$
04

Compute the second integral and compare results

Now we need to compute the given integral \(-\int_{0}^{\infty} e^{-x} \, dx\). We find the antiderivative of \(e^{-x}\): $$\int e^{-x} \, dx = -e^{-x}$$ Evaluate the integral on the interval [0,鈭瀅: $$-\Bigg[-e^{-x}\Bigg]_0^\infty = -\Big(-e^{-\infty} + e^0\Big) = 1-0 = 1$$ Now we have the results for both integrals: $$\int_{0}^{1} \ln(x) \, dx = -1$$ $$-\int_{0}^{\infty} e^{-x} \, dx = 1$$ As the results are negative of each other, we can conclude that they are equivalent in magnitude, but opposite in sign.

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

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{d x}{(x+1)\left(x^{2}+2 x+2\right)^{2}}$$

Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b],\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 2 and an illustration.

A long, straight wire of length \(2 L\) on the \(y\) -axis carries a current \(I\). According to the Biot-Savart Law, the magnitude of the magnetic field due to the current at a point \((a, 0)\) is given by $$B(a)=\frac{\mu_{0} I}{4 \pi} \int_{-L}^{L} \frac{\sin \theta}{r^{2}} d y$$ where \(\mu_{0}\) is a physical constant, \(a>0,\) and \(\theta, r,\) and \(y\) are related as shown in the figure. a. Show that the magnitude of the magnetic field at \((a, 0)\) is $$B(a)=\frac{\mu_{0} I L}{2 \pi a \sqrt{a^{2}+L^{2}}}$$ b. What is the magnitude of the magnetic field at \((a, 0)\) due to an infinitely long wire \((L \rightarrow \infty) ?\)

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}} ; x=u^{6}$$

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{2 x^{3}+x^{2}-6 x+7}{x^{2}+x-6} d x$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

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.