Integration Question (1 Viewer)

nrlwinner · Jan 6, 2010

Hi. I'm not sure how to do integration by substitution when one the substitution they give us is a square.

Eg.

$u^2=x+2$

Find $\int \frac{x-2}{(\sqrt(x+2)}$

gurmies · Jan 6, 2010

jet · Jan 6, 2010

NRL winner, is this an extension 1 or extension 2 question?

If only extension 1, you need to rearrange it like this:

$u^2 = x + 2 \Rightarrow x = u^2 - 2 \\ \frac{dx}{du} = 2u \Rightarrow dx = 2u \,du \\ \therefore \int \frac{x - 2}{\sqrt{x + 2}} \, dx = \int \frac{u^2 - 2 - 2}{\sqrt{u^2 + 2 - 2}} \times 2u \, du \\ = \int \frac{u^2 - 4}{\sqrt{u^2}} \times 2u \, du \\ = 2\int u^2 - 4 \, du \\ = \frac{2u^3}{3} - 8u + c \\ = \frac{2\sqrt{\left (x + 2 \right )^3}}{3} - 8\sqrt{x + 2} + C$

Now I'm not sure who made a mistake; me or gurmies

xV1P3R · Jan 6, 2010

Gurmies forgot to sub dx = 2udu entirely; forgot to multiply by "u" when subbing in causing the denominator to be carried through.

Aquawhite · Jan 6, 2010

I always miss out on answering these >_<. I could have done this one

And it looks like a harder 3U question or just the general 4U questions I've seen... <3 4U integration. ^_^

cutemouse · Jan 6, 2010

Knowing implicit differentiation I reckon can help in 3U, especially if you get something nasty like that in this year's 3U HSC.

jet · Jan 6, 2010

jm01 said:
Knowing implicit differentiation I reckon can help in 3U, especially if you get something nasty like that in this year's 3U HSC.

Yeah, but you have to be completely explicit that you are differentiating implicitly in 3 unit work. Even then, I wouldn't.

4 unit not so much.

Also, NRL winner, I avoided taking the square root of the substitution because it would introduce a ± into the equation. On second thought, I may have overlooked that when re-substituting back in :/

The Nomad · Jan 6, 2010

Just for kick's sake, no need for a substitution:

$\\ \int \frac{x - 2}{\sqrt{x + 2}} dx = \\ \int \frac{(x + 2) - 4}{\sqrt{x + 2}}dx = \\ \int (\sqrt{x + 2} - \frac{4}{\sqrt{x + 2}}) dx = \\ \frac{2(x+2)^{3/2}}{3} - 8\sqrt{x+2} + C$

jetblack2007 said:
Yeah, but you have to be completely explicit that you are differentiating implicitly in 3 unit work. Even then, I wouldn't.

4 unit not so much.

Also, NRL winner, I avoided taking the square root of the substitution because it would introduce a ± into the equation. On second thought, I may have overlooked that when re-substituting back in :/

I don't think this is true. I did it all the time.

jet · Jan 6, 2010

The Nomad said:
Just for kick's sake, no need for a substitution:

$\\ \int \frac{x - 2}{\sqrt{x + 2}} dx = \\ \int \frac{(x + 2) - 4}{\sqrt{x + 2}}dx = \\ \int (\sqrt{x + 2} - \frac{4}{\sqrt{x + 2}}) dx = \\ \frac{2(x+2)^{3/2}}{3} - 8\sqrt{x+2} + C$

Would have been my method if not asked for substiution.

gurmies · Jan 6, 2010

xV1P3R said:
Gurmies forgot to sub dx = 2udu entirely; forgot to multiply by "u" when subbing in causing the denominator to be carried through.

This is why I got the marks I did in HSC Mathematics.

Aquawhite · Jan 6, 2010

gurmies said:
This is why I got the marks I did in HSC Mathematics.

Naww, we all make mistakes

jet · Jan 7, 2010

gurmies said:
This is why I got the marks I did in HSC Mathematics.

What, the excellent ones which do you proud?

shaon0 · Jan 7, 2010

nrlwinner said:
Hi. I'm not sure how to do integration by substitution when one the substitution they give us is a square.

Eg.

$u^2=x+2$

Find $\int \frac{x-2}{(\sqrt(x+2)}$

Shit, I always miss out on integration Q's

gurmies · Jan 7, 2010

jetblack2007 said:
What, the excellent ones which do you proud?

Come on man, I was blessed with piss-easy exams and I only did decently in them =/ Yours were far harder!

Integration Question (1 Viewer)

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