Question

integration by parts

Answer 1

\[7x^{2}\ln(3x)-(7/2)x ^{2}\]

Answer 2

343-ln(2)-(6160/36)

Answer 3

no minus between 343 and ln(2)

Answer 4

\[\int\limits_{1/3}^{7}14xln(3)dx\]

Answer 5

\[\int\limits_{1/3}^{7}14x\ln(3x)dx\]?

Answer 6

yes

Answer 7

are you just asking us to check your work?
if so the indefinite is right, let me check your sub

Answer 8

yes, check please

Answer 9

eh the evaluation is annoying let me just write it out here

Answer 10

I think that should be ln(21)

Answer 11

what am i doing i will let wolfram do it

Answer 12

yeah you are off somewhere
I don't see how you got log(2) either

Answer 13

I would just change the bounds and simplify the integral with a u-sub to avoid these kinds of errors

Answer 14

I get \[343 \ln 21 - \frac{ 1540 }{ 9 }\]

Answer 15

you were right. how did you get that. where do you think i messed up

Answer 16

\[\int\limits_{1/3}^{7}14x\ln(3x)dx\\u=3x\implies dx=\frac13du\\\frac{14}3\int_1^{21}u\ln udu=\frac{14}3\left(u^2\ln u-\int_1^{21}udu\right)=\frac{14}3\left.\left(u^2\ln u-\frac12u^2\right)\right|_1^{21}\]evaluation is easier now

Answer 17

thanks, everyone give medals

Answer 18

I have enough, Noel can have mine

Answer 19

\[(7*7^{2}\ln (3\times7)-\frac{ 7 }{ 2 }7^{2}\]
Gee, I was still typing into the equation editor. By the way, Mr. Turing, what editor are you using?

Answer 20

I use embedded latex
do inline latex by enclosing commands with \.()\
do regular latex by enclosing in \.[]\
(take out the . in both cases)
If you don't know what latex is, I'll try to find you a tutorial site

Answer 21

we also have a latex practice group that can be found by clicking \(\href{http:///openstudy.com/study#/groups/LaTeX%20Practicing!%20%3A)}{here}\)
(me showing off fancy tricks you can do with latex :P )