Question

have a question about integration by substitution

Answer 1

yeah.I think I can help

Answer 2

when yu use substitution it is my understanding that you need to get the integral in the form \[\int\limits u du\]and du=f'(x)dx.if you have a problem such as\[\int\limits 2x(4(x)^2-10)^2\] wouldn't you say that f'(x) = 8x, when you use\[u=4x^2 -10\] then because of the fact that you have 2xdx in the original problem you then divide du=8xdx by four to get the 2x dx that was in the original problem?

Answer 3

And that means that 2xdx is equal to \[(1/4) du\]

Answer 4

so I am confused in my text book it gives one problem as \[\int\limits 2x(4x^2 -10)^2 \] with the answer as \[{[(4x^2 -10)^3]/12}+c\]

Answer 5

w8
gmme 2minutes

Answer 6

the next part shows \[\int\limits (x^2 -1)/\sqrt{2x-1} dx\] and an answer of u=2x+1, x=(1/2)(u+1) dx=(1/2)du

Answer 7

how do they get the (1/2)du? This is where my confusion comes in

Answer 8

If du=f'(x) dx in this case du=2dx I can't relate how these concepts are the same

Answer 9

I think you have this strict thought that there is suppose to be a specific form that u substitution needs to look.

Answer 10

let 2x - 1 = u, du = 2dx
and other other x => (u+1)/2

Answer 11

The thing is that u substitution is done to make an impossible integration possible. So you might still have to do a lot of work when you u substitute.

Answer 12

Romero .... I was told that any integration could be done without substitution I am now even more confused

Answer 13

Well impossible for me means doing an integration for several minutes or an hour or even more.

Answer 14

if you choose to do u substitution then how do you deal with du? do you put du equal to what you already have?

Answer 15

So you can do it without substitution but it's going to take you far more time.

Answer 16

yes the answer is true..lets do this step by step. let u=4x^2,the du=8xdx, therefore 1/8du=xdx then take this 1/8du and substitue on the original integral for xdx. ushould have 1/8( integralsign) 2u^2du. u'll get the answer

Answer 17

@ekim2012 please don't do the problem yet we are trying to explain how u substitution work

Answer 18

cool

Answer 19

Ok first thing you do it find what u is going to be: Your book said 2x - 1 = u

Answer 20

From there you need to find a way to replace all the x and the dx in your integration and you can do it from
2x - 1 = u

Answer 21

then how do you apply that to the other problem where dx=(1/2)du?

Answer 22

yep

Answer 23

ok from 2x - 1 = u we want to know what dx is right?

well what you can do is integrate u in respect to x (dx)
but you also have to do that to both sides.

du/dx= d(2x-1)/dx

Answer 24

that isn't making sense to me

Answer 25

du/dx= d(2x-1)/dx

the right hand side just means find the derivative of 2x-1.
which is just 2

du/dx=2

we want to isolate the dx so we switch dx and 2

du/2= dx

Answer 26

let 2x - 1 = u
du = 2 dx
dx = du/2

Answer 27

and change x's in to u's

Answer 28

x = (u+1)/2
x^2 = (u+1)^2/4

Answer 29

2x - 1 = u
well we also want to change x into u right

so from 2x - 1 = u we want to get what x equal.
Simple just solve for x

2x - 1 = u
2x - 1+1= u+1 ----we added 1 to both sides
2x = u+1
(2x)/2 = (u+1)/2---- we divided 2 to both sides
x = (u+1)/2 ----- yay now we have x

Answer 30

experimentX .... the du=2dx then, du/2=dx makes sense. How can that be applied to the first problem? where you set du=f'(x)dx then solve for dx?

Answer 31

I get the part solving for x to replace x with somethin in terms of u.

Answer 32

du=f'(x)dx is the same as

du/2=dx

he just skipped some steps

Answer 33

u=x
du/dx= f'(x)

where x =2x-1 in this case.

Answer 34

du/dx= f'(2x-1)
du/dx=2
du = 2 dx
du(1/2)=dx

Answer 35

Hey you should practice on more problems. some easier ones here. I think you go it you just need to work on easier problems just so you get a hang of it.