substitution? i cant figure it out?

Question

Im supposed to use 'u' to integrate this

integrate from [0,4] x / sqrt(1-2x)

Im so sorry- its 1+2x under the sqrt. not minus

Answer 1

Let u = sqrt(1-2x)
Then u^2 = 1-2x
x = (1-u^2)/2
dx = -u du
Make the above substitutions and you are on your way

Answer 2

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

And put the bounds in terms of u:
u=1-2x
x=0: u=1-2*0=1
x=4: u=1-2*4=-7
Now substitute these back in:

int[((1-u)dx/(2u^.5)]1,-7
to get du into the equation, we need an extra -2 in the numerator, so multiply top and bottom by -2:
int[(1-u)(-2dx)/(2u^.5)]1,-7
we can substitute du for (-2dx), and we can move the 1/2 outside the integral:
.5 * int[(du-udu)/(u^.5)]1,-7

This looks ugly in keyboard font, but it's actually easy to solve now, because everything is in plain terms of u. If you want (I want to but you could do it without), separate the fraction at the subtraction sign and make 2 integrals:
.5 * int[du/u^.5]1,-7 - .5*int[udu/u^.5]1,-7
Now integrate:
.5*[2u^.5]1,-7 - .5*[2/3u^1.5]1,-7
Calculate across and solve.
But, as the previous poster astutely observes, you can't integrate from the initial bounds because they include areas where the function isn't defined. So check the bounds (perhaps 0->4 is the bound for the integration after you substitute u?). I won't calc it out since it would be wrong with the wrong bounds, but if you fix the bounds then this should work fine.

Answer 3

put sqrt(1-2x)= u so -1/sqrt(1-2x)*dx =du
and 1-2x=u^2 so x = 1/2((1-u^2)
The integral becomes Int ......
but pay attention 1-2x>=0 so x<=1/2 so you cant integrate from 0 to 4

Answer 4

Take 1 + 2x = u² and find the limits for u accordingly. The rest should be easy for you - I hope!

Answer 5

I = ∫ x / [ (1 - 2x)^(1/2) ] dx
Let u = (1 - 2x)^(1/2)
du / dx = (1/2)(1 - 2x)^(-1/2)(- 2)
du / dx = (- 1) (1 - 2x)^(-1/2)
du = (-1) (1 - 2x)^(-1/2) dx
du = [ (-1) / (1 - 2x)^(1/2) ] dx
- du = 1 / [(1 - 2x)^(1/2)] dx
u² = 1 - 2x
2x = 1 - u ²
x = (1/2) (1 - u ²)
I = - (1/2) ∫ (1 - u ²) / u du
I = - (1/2) ∫1 / u - u du
I= - (1/2) [ log u - u ² / 2 ] + C
I = (- 1/2) log [(1 - 2x)^(1/2)] - (1 - 2x) / 2 + C

Answer 6

correctly your quiz is integrate from [0,4] x / sqrt(1-2x) dx...dx is very impotent here

use sqrt(1-2x) = y

then X = (1-Y^2)
remember now dx = -2YdY

then problem is int[1,-7] {(1-Y^2)/Y} *(-2YdY)
==> int[1,-7] {(1-Y^2)/} *(-2dY)
==> int[1,-7]{-2}dY +2Y^2dY

Answer 7

santmann is right, you cannot integrate from 0 to 4. The function is not defined from 0.5 to 4.