Question

Evaluate the definite integral.
I am using the equation writer to enter the integral below.

Answer 1

\[\int\limits_{0}^{4}\frac{ x }{ \sqrt{1+2x} }dx\]

Answer 2

you know how to integrate ?

Answer 3

yes this particular question is under the heading "The substitution rule" and is my last question for this section but I do not see a place to use the substitution rule and I do not think it has symmetry like some of the questions before it.

Answer 4

Ok its gotta be easy then

let

y=1+2x

and diffrentiate it

Answer 5

then how do I deal with the x in the numerator? because the dy/dx piece for what you posted would just be 2 wouldn't it?

Answer 6

I am trying to remember,what i can...

Answer 7

It's often a good idea to substitute terms that are in the denominator, or under a root signs just as @hba has done. Doing this may make more terms in the numerator but those can be split up into separate fractions.

Answer 8

@Mathmuse It's been a long time since I've done this but i am gonna try to help,I just want you to be here,so you can watch if i go wrong.

Answer 9

Okay but how this is the step that I do not understand. I cannot see how you get rid of the numerator through substitution.

Answer 10

I appreciate the help though

Answer 11

Just diffrentiate it and tell me what you get ?

Answer 12

The y=1+2x equation?
I get that y'=2

Answer 13

dy=2dx

Answer 14

so,

\[4\int\limits_{0}^{4}\frac{ dy }{ y }\]

Answer 15

Looks cool to me :D

Answer 16

Okay I did not do that because that is not how my textbook taught me substitution

Answer 17

Now integrate.

Answer 18

y'=dy/dx=2

so

dy=2dx

Answer 19

^ thats how i got it.

Answer 20

i better show some steps

Answer 21

isn't it
\[\int\limits_{0}^{4}\frac{ 2x \times dy }{ y}\]

Answer 22

I guess I am just missing what happened to the x somehow

Answer 23

@Mathmuse Please show some step i also wanna revise.

Answer 24

I am sorry but I just do not see it

Answer 25

@hba I hate to bring it up but in that discussion yesterday I was saying why do her teachers need to complicate things not you guys. She did not understand me, unfortunately I was not around to correct the misunderstanding.

Answer 26

@whatisthequestion I do not even remember what your'e talking about,link pleasee

Answer 27

Starting with
\[\int\limits\limits_{0}^{4}\frac{ x }{ \sqrt{1+2x} }dx\]

If you want to substitute, you must get rid of any terms that are related to 'x'

so we start with:
\[u=1+2x\]

which leads to:
\[\int\limits\limits_{}^{}\frac{ x }{ \sqrt{u} }dx\]

At this point there is still a dx left, so lets take the derivative of our substitution to solve for dx so we can replace it.
\[u = 1+2x\]
\[\frac{du}{dx} = 2\]
\[\frac{du}{2} = dx\]

Now we have:
\[\int\limits\limits_{}^{}\frac{ x }{ \sqrt{u} }\frac{1}{2}du\]

Answer 28

http://openstudy.com/study#/updates/50b1b9f0e4b09749ccac7ef1

Answer 29

I think i deal with many problems like this daily but we have to refrain users from cheating.

Answer 30

the x term in the numerator needs to go so lets rearrange our original substitution to find what x is equal to:
\[u=1+2x\]
\[2x=u-1\]
\[x=\frac{u-1}{2}\]

making this substitution and pulling the constants in front of the integral we get:
\[\frac{1}{2}\frac{1}{2}\int\limits\limits_{}^{}\frac{ u-1 }{ \sqrt{u} }du\]

Answer 31

Okay I never even though of going back to rewrite the 2x+1=u equation

Answer 32

I just wanted to make it clear that I was not saying that you were complicating things but that her teachers were

Answer 33

now we can clean it up:
\[\frac{1}{4}[\int\limits\limits\limits_{}^{}\frac{ u }{ \sqrt{u} }du - \int\limits\limits\limits_{}^{}\frac{ 1 }{ \sqrt{u} }du]\]
\[\frac{1}{4}[\int\limits\limits\limits_{}^{}u^{\frac{1}{2}}du - \int\limits\limits\limits_{}^{}u^{-\frac{1}{2}}du]\]

Answer 34

@whatisthequestion Yeah don't worry i know you were helping :)

@Mathmuse Let him do some work lol :D

Answer 35

and I have the formulas for how to differentiate those thank you besides my final (and studying for it) that was my last math question for calculus this semester and I just could not see how to solve it

Answer 36

don't forget the bounds of integration. They used to go from 0 to 4 in the x domain, but now you're taking the integral in the u domain. you have to sub those out too.

Answer 37

Yeah you have to sub them back after your'e done .

Answer 38

yes with G(x) which is =y

Answer 39

expect at those points

Answer 40

you don't have to sub anything after you're done. the definite integral gives you a strictly numeric answer. once you've calculated it your finished.
For this question I would follow only what I've written. there is no y, there is only u.

Answer 41

sorry my textbook always uses the notation u for the piece they sub in, I accidentally slipped back into that notation

Answer 42

y is normally reserved for the dependent variable of a function. you will see u's and v's for substitutions and other calculus methods