/*! 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 32 Evaluate the following definite ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 32

Evaluate the following definite integrals. $$\int_{1}^{e} \ln 2 x d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral $$\int_{1}^{e} \ln(2x) dx$$. Answer: The definite integral equals $$e\ln(2e) - e - \ln(2) + 1$$.

Step by step solution

01

Determine the parts for Integration by parts

To apply integration by parts, we need to choose a part for u and a part for dv. Since the derivative of ln(2x) is simpler, we choose our parts as follows:$$ u = \ln(2x)\, , \, du = \frac{2}{2x} dx = \frac{1}{x} dx\\ dv = dx \, , \, v = x $$
02

Apply integration by parts formula

Apply the integration by parts formula:$$ \int u dv = uv - \int v du $$Substitute the parts determined in step 1 into the formula:$$ \int \ln(2x) dx = x \ln(2x) - \int x \cdot \frac{1}{x} dx $$Notice that \(x\) in the integral on the right-hand side can be canceled out:$$ \int \ln(2x) dx = x \ln(2x) - \int dx $$Now we can integrate the remaining part:$$ \int \ln(2x) dx = x \ln(2x) - x + C $$where C is the constant of integration. Since we are evaluating a definite integral, we don't need to worry about the constant.
03

Evaluate the definite integral using the Fundamental Theorem of Calculus

Now, we will use the Fundamental Theorem of Calculus to find the definite integral between the limits 1 and e:$$ \int_{1}^{e} \ln(2x) dx = \Big[x \ln(2x) - x\Big]_1^e $$Apply the limits:$$ \Big[(e \ln(2e) - e) - (1 \ln(2) - 1)\Big] $$Simplify the expression:$$ \Big[(e \ln(2e) - e) - (\ln(2) - 1)\Big] = e\ln(2e) - e - \ln(2) + 1 $$This is the value of the definite integral:$$ \int_{1}^{e} \ln(2x) dx = e\ln(2e) - e - \ln(2) + 1

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

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \sin 3 x \sin 2 x d x$$

Circumference of a circle Use calculus to find the circumference of a circle with radius \(a.\)

Prove that the Trapezoid Rule is exact (no error) when approximating the definite integral of a linear function.

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.

Recall that the substitution \(x=a \sec \theta\) implies that \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0\) ) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). $$ \begin{array}{l} \text { Show that } \int \frac{d x}{x \sqrt{x^{2}-1}}= \\ \qquad\left\\{\begin{array}{ll} \sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1 \end{array}\right. \end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Geometry

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.