Question

show that (the integral of x/(3x^2+3) from x=5 to x=7 actually equals (1/6)[ln(150)-ln(78)]...the closest I can get is to show that the integral is equal to (1/2)[ln(150)-ln(78)] and I'm on the verge of thinking my teacher is wrong, however when I put both of these into wolfram, they came out to be o.11108 and 0.10898, which is very close, so please anyone...HELP ME BEFORE I GO CRAZY!

Answer 1

U can use the substitution u = 3x²+3. Then du = 6x. Now write the integral as follows:\[\int\limits_{5}^{7}\frac{ x }{ 3x^2+3 }dx=\frac{ 1 }{ 6 } \int\limits_{5}^{7}\frac{ 6x }{ 3x^2+3 }dx=\int\limits_{?}^{?}\frac{ 1 }{ u }du\]
This one is much easier to find! Just one problem left: calculate the new boundaries for u.
So: if x=5 then u=?
if x=7 then u=?

Answer 2

Ah, you probably did all this stuff already ;)
So:\[\frac{ 1 }{ 6 }\int\limits_{78}^{150}\frac{ 1 }{ u }du=\frac{ 1 }{ 6 }\left( \ln 150 - \ln 78 \right)\], the same as your result!

Answer 3

Was it the boundaries?

Answer 4

hi...the computer is taking your formulas and making them very hard to read

Answer 5

OK, I will upload it as an image...

Answer 6

when i did it, i made u=x^2+1, which made 1/2du=x dx
but i kept the "3" that i factored out of the deominator on the bottom...then i took the integral of ln(1/(3u) du

Answer 7

then that made 1/2[3(x^2+1)...then i replaced x=50 and x=26

Answer 8

i saw your image...i need a few minutes to look at it and understand it...

Answer 9

So ZeHanz...were you actually able to get the left to equal the right?

Answer 10

But if you factor out the 3, it can be put in front of the integral sign as 1/3.
You then would have:\[\frac{ 1 }{ 3 }\int\limits_{5}^{7}\frac{ x }{ x^2+1 }dx=\frac{ 1 }{ 6 }\int\limits_{26}^{50}\frac{ 1 }{ u }du\] where u=x²+1 and du=2xdx, so the result would be: 1/6(ln(50)-ln(26))

Answer 11

No matter what you do, you always get the same answer:
1/6(ln(150-ln(78))=1/6(ln(50)-ln(26))=1/6(ln(25)-ln(13)),
because 150/78=50/26=25/13.

Answer 12

hi ZeHanz...i really appreciate your help! Is there any way you could make an image of the entire problem? I can't understand what the computer has made of your entries

Answer 13

OK, I'll do the same image-trick again!

Answer 14

Here it is:

Answer 15

oh wait a second...1/6[ln(150)-ln(78)] is equal to 1/6(ln(50)-ln(26))???...i think i had that second part as an answer but didn't know that it equaled what i was looking for

Answer 16

yeah...i can get 1/6 out in front, but then i am removing the "3" that i need to multiply with both "50" and "26", and i then cannot get the 150 and 78 that i need

Answer 17

would you do me a huge favor and show it all worked out for me? i still can't get it to work unfortunately...i have spent nearly 3 hours on this problem and i am going insane!...please help if you can

Answer 18

Hey ZeHanz...I finally got it!!!!!!!!!!! THANK YOU THANK YOU THANK YOU!

Answer 19

Here is my calculation...

Answer 20

Hi ZeHanz, I just thought you would be interested to know that my teacher will only accept 1/6[ln(150)-ln(78)] as an answer...when I made u=x^2+1, i could only get the answer 1/6[ln(50)-ln(26)]...and even though that is mathematically equal to 1/6[ln(150)-ln(78)], he would have marked me wrong...my teacher is very tricky because the only way you can get 1/6[ln(150)-ln(78)], is to set u=3x^2+3...this is very tricky problem but I wanted to say thank you again...you saved my sanity!!! Danke!