Question

have two problems (evaluating definite integrals by substitution) they are fractions and I gone know what to do with the numerator.. I have a picture of it if anyone wants to accept the challenge

Answer 1

numbers 29 AND 36

Answer 2

For 29 \[\Large \int\limits_{0}^{2}\frac{ x }{ (5x^2 + 1)^2 } dx = \]Notice the numerator x is similar to the derivative of 5x^2+1 (which would be 10x).. so substitute \[\large u = 5x^2+1\] \[\large du = 10x dx \text{ ...so... } \frac{ du }{ 10 } = x dx\]
\[\Large \int\limits\limits_{0}^{2}\frac{ x }{ (5x^2 + 1)^2 } dx = \int\limits\limits_{0}^{2}\frac{1 }{ (5x^2 + 1)^2 } x dx =\]

Answer 3

Now you can replace the xdx with du/10, and replace 5x^2+1 with u, and find the new limits using u = 5x^2+1. When x=0, u=1. When x=2, u=21
\[\Large \int\limits\limits\limits_{?}^{?}\frac{1 }{ u^2 } \frac{ du }{10 } = \frac{ 1 }{10 }\int\limits\limits\limits_{1}^{21}\frac{1 }{ u^2 }du\]

Answer 4

how did you find the 21?

Answer 5

and I have the du=2x.. would I make it 2x/10, and the plug in for x?

Answer 6

my mistake.. Im looking at number 36.. sorry about that

Answer 7

Find u=1 and u=21 by using: u=5x^2+1 and plugging in x=0 and x=2.

Answer 8

For 36, you'll have to factor the denominator first... x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2. So...
\[\Large \int\limits_{1}^{2} \frac{ 1 }{ x^2 - 6x +9 } dx = \int\limits_{1}^{2} \frac{ 1 }{ (x-3)^2 } dx \]Now see if you can find a u-sub that'll work.

Answer 9

you can see my earlier confusion

Answer 10

Really hard to follow your work with it being that small... I can't really tell what you did.

Answer 11

I was plugging in the o,and the 2 where the x is on the numerator of the problem.. and getting 1/10-0/1-2/441

Answer 12

Yeah, you definitely can not do that. That's not how integration works.

Answer 13

its the fractions throwing me off. I had no trouble with the earlier problems, just dont know what to do with the numerators

Answer 14

yea, I can tell by me not getting the correct answers.. this is a better photo of me doing incorrect work, because I dont understand how to do it properly.

Answer 15

Follow the steps I showed above (you still have to integrate what I gave), you can use a similar process to finish 36.

For 36, use the substitution u=x-3, so the derivative du = dx.

Your new limits of integration in 36: using u=x-3 when x=1, u=-2. When x=2, u=-1.

Answer 16

thanks

Answer 17

what is u2?

Answer 18

for number 29

Answer 19

What do you mean by u2? u^2?
\[\Large \frac{ 1 }{10 }\int\limits\limits\limits\limits_{1}^{21}\frac{1 }{ u^2 }du = \frac{ 1 }{10 }\int\limits\limits\limits\limits_{1}^{21} u^{-2} du\]that might make it easier to integrate