Problem

Evaluate the integral. $ \displaystyle \int_0^…

04:26

Problem 27 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_1^5 \frac{\ln R}{R^2} dR $

Answer

$\int_{1}^{5} \frac{\ln R}{R^{2}} d R=\left[-\frac{1}{R} \ln R\right]_{1}^{5}-\int_{1}^{5}-\frac{1}{R^{2}} d R=-\frac{1}{5} \ln 5-0-\left[\frac{1}{R}\right]_{1}^{5}=-\frac{1}{5} \ln 5-\left(\frac{1}{5}-1\right)=\frac{4}{5}-\frac{1}{5} \ln 5$

Topics

Integration Techniques

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Discussion

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Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

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Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

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Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

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Problem 49

Problem 50

Problem 51

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Video Transcript

The problem is you wanted the integral Integral from one to five. L and R or R squared D r For this problem, we can use my third of the integration by parts common. It is into girl from A to B. You have released from yaks. It's a call to you'd have screamed roommate being minus into girl from a to B you prom? Absolutely. Yaks. Now for our problem, we can like you. Is the coat too? How in and re prime is O'Toole one over. Uh, square. Then you prom this Rico to one over our arm. And we It's the code to negative. I wanna know where r now this into girl is you go to about this formula to city photo. You damn Swede! This's negative one or we are ums when are from one to five minus integral from one toe. Five. You promised me that this is negative one over our square. We are This is the coach. So for the first two term line five to one from crying foul and want Teo, are we have This is negative, Alan, Five oh five and minus zero. Second term is plus into peril of one of our squire's. This's negative one over R from one to five. This is a country negative. Our style over five class negative one over five minus Meg to Juan. The author is negative now and over a five class whom you are.

Wen Z.

Topics

Integration Techniques

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Problem

Evaluate the integral. $ \displaystyle \int_0^…

Problem 27 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_1^5 \frac{\ln R}{R^2} dR $

Answer

$\int_{1}^{5} \frac{\ln R}{R^{2}} d R=\left[-\frac{1}{R} \ln R\right]_{1}^{5}-\int_{1}^{5}-\frac{1}{R^{2}} d R=-\frac{1}{5} \ln 5-0-\left[\frac{1}{R}\right]_{1}^{5}=-\frac{1}{5} \ln 5-\left(\frac{1}{5}-1\right)=\frac{4}{5}-\frac{1}{5} \ln 5$

Topics

Calculus: Early Transcendentals

Discussion

Top Calculus 2 / BC Educators

Catherine R.

Heather Z.

Kristen K.

Samuel H.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 7

Video Transcript

Topics

Calculus: Early Transcendentals

Top Calculus 2 / BC Educators

Catherine R.

Heather Z.

Kristen K.

Samuel H.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Problem

Evaluate the integral. $ \displaystyle \int_0^…

Problem 27 Medium Difficulty

Evaluate the integral. $ \displaystyle \int_1^5 \frac{\ln R}{R^2} dR $

Answer

$\int_{1}^{5} \frac{\ln R}{R^{2}} d R=\left[-\frac{1}{R} \ln R\right]_{1}^{5}-\int_{1}^{5}-\frac{1}{R^{2}} d R=-\frac{1}{5} \ln 5-0-\left[\frac{1}{R}\right]_{1}^{5}=-\frac{1}{5} \ln 5-\left(\frac{1}{5}-1\right)=\frac{4}{5}-\frac{1}{5} \ln 5$

Topics

Calculus: Early Transcendentals

Discussion

Top Calculus 2 / BC Educators

Catherine R.

Heather Z.

Kristen K.

Samuel H.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 7

Video Transcript

Topics

Calculus: Early Transcendentals

Top Calculus 2 / BC Educators

Catherine R.

Heather Z.

Kristen K.

Samuel H.

Calculus 2 / BC Bootcamp

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Evaluate the integral.

$ \displaystyle \int_1^5 \frac{\ln R}{R^2} dR $