conflicting solutions to this q (1 Viewer)

[mecramarathon]

how do you integrate by substitution the following q?

x√(1-x) dx ?

i get conflicting answers from fitzpatrick and cambridge! (in terms of positive and negative signs)

thnx

 

[ilikebeeef]

let u=1-x
x=1-u

Sum[x(1-x)^1/2]dx
=Sum[(1-u)u^1/2]du
=Sum[u^1/2 - u^3/2]du
=(2/3)u^3/2 - (2/5)u^5/2 + c
=(2/3)(1-x)^3/2 - (2/5)(1-x)^5/2 + c

Hopefully that's correct

 

[xV1P3R]

One issue
u=1-x
x=1-u
du = -dx

 

[Trebla]

 

[jet]

Just to clarify Trebla's post, he realised that to get rid of the square root, you could say u² = 1 - x instead of u = 1 - x (which will still leave in the square root).

Then he rearranged to get x = 1 - u² and could differentiate from there.

That is a very useful substitution method for these sorts of questions

 

[The Nomad]

Just to clarify Trebla's post, he realised that to get rid of the square root, you could say u² = 1 - x instead of u = 1 - x (which will still leave in the square root).

Then he rearranged to get x = 1 - u² and could differentiate from there.

That is a very useful substitution method for these sorts of questions

I would just like to point out that u² = 1 - x is not a clear-cut definition of the value of u.

That particular solution suggests that (u²)^1/2 = u, when in actuality, (u²)^1/2 = |u|.

Hence, there are 2 cases:

• If u > 0, |u| = u, and (u²)^1/2 = u.
• If u < 0, |u| = -u, and (u²)^1/2 = -u.

 

[jet]

I would just like to point out that u² = 1 - x is not a clear-cut definition of the value of u.

That particular solution suggests that (u²)^1/2 = u, when in actuality, (u²)^1/2 = |u|.

Hence, there are 2 cases:

• If u > 0, |u| = u, and (u²)^1/2 = u.
• If u < 0, |u| = -u, and (u²)^1/2 = -u.

That's a true point, though I've always been under the impression that 3 unit doesn't require such a large amount of rigour. I do always feel guilty when I have to resubstitute back in though.

 

[The Nomad]

That's a true point, though I've always been under the impression that 3 unit doesn't require such a large amount of rigour. I do always feel guilty when I have to resubstitute back in though.

I guess the rigour point is fair enough.

Just wondering though, if 10 people get 84/84 in the HSC exam, how do they choose who is ranked first, and who is second...and so on? Do they not check working out? Just a little query I have.

 

[jet]

I guess the rigour point is fair enough.

Just wondering though, if 10 people get 84/84 in the HSC exam, how do they choose who is ranked first, and who is second...and so on? Do they not check working out? Just a little query I have.

They look at 2 unit marks. Then they combine it with assessment marks, to 1 decimal place. That usually sorts it out.

 

[Trebla]

I would just like to point out that u² = 1 - x is not a clear-cut definition of the value of u.

That particular solution suggests that (u²)^1/2 = u, when in actuality, (u²)^1/2 = |u|.

Hence, there are 2 cases:

• If u > 0, |u| = u, and (u²)^1/2 = u.
• If u < 0, |u| = -u, and (u²)^1/2 = -u.

Despite this technical issue, it just takes just one extra piece of information to clear it up since the substitution in this case is of a free choice (i.e. just state u > 0 lol)