Learn How to Integrate xlnx Using the Integration by Parts Method

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Homework Help Overview

The discussion revolves around the integration of the function xlnx using the integration by parts method. Participants are exploring different approaches to set up the integration and evaluate it, particularly focusing on the correct assignment of variables and limits of integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants suggest various assignments for u and dv, with some proposing u = ln x and others sticking with u = x. There are questions about the evaluation of the integral, especially concerning the limits of integration and the correctness of the calculations.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts and calculations. Some guidance has been offered regarding the evaluation of the integral, and there is an ongoing exploration of different interpretations of the integration process.

Contextual Notes

There is a specific focus on integrating the function from 1 to 2, and participants are addressing potential errors in their evaluations and the setup of the integration by parts method.

ricekrispie
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little help? simple question...

Homework Statement




Integrate:
xlnx dx

use integration by parts


Homework Equations





The Attempt at a Solution



let u = x du = dx
dv = ln x v = ??
 
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If you don't know the integral of lnx...try letting u=lnx and dx=x dx instead of the way you wrote it
 

Homework Statement




Integrate:
xlnx dx



Homework Equations





The Attempt at a Solution



let u = x du = dx
dv = ln x v = ??
 
How about

Let u = lnx
du = 1/x
dv = xdx
v = x^2 / 2
 
if I have to integrate it from 1 to 2 where are these plugged in?
 
\int_a^b udv = uv|_a^b - \int_a^b vdu
Or in other words
\int_a^b udv = u(b)v(b) - u(a)v(a) - \int_a^b vdu
 
my attempt

i wonder if this worked...

\int_1^2 x lnx dx = lnx (x^2/2)|_1^2 - \int_1^2 (x^2/2)* (1/x) dx
= 2ln2 - ((1/4)(x^2))|_1^2

= 2ln2 - (5/4)
??
 
ricekrispie said:
i wonder if this worked...

\int_1^2 x lnx dx = lnx (x^2/2)|_1^2 - \int_1^2 (x^2/2)* (1/x) dx
= 2ln2 - ((1/4)(x^2))|_1^2

= 2ln2 - (5/4)
??
Almost! (1/4)(x^2) when x= 2 is 1. when x= 1, it is 1/4 so the last part is -(1- 1/4)= -3/4, not -5/4.
 
check your evaluation for the last part

it might be me but you did the integration correctly, gj.
 
  • #10
Almost...

x^2|^2_1 Is 3 not 5.
 
  • #11
hehe i noticed a lack of parenthesis... thanks for helping :smile:
 

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