what should my thought process be when evaluating this integtral, where I have a fairly large x in front of a third root? Where do I start?

$\displaystyle \int x^{11}\sqrt[3]{2x^4+5}$

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- Feb 21st 2009, 12:15 PMgammamanintegral thought process?
what should my thought process be when evaluating this integtral, where I have a fairly large x in front of a third root? Where do I start?

$\displaystyle \int x^{11}\sqrt[3]{2x^4+5}$ - Feb 21st 2009, 12:18 PMKrizalid
Think a bit: $\displaystyle x^{11}\sqrt[3]{2x^{4}+5}=x^{8}\cdot \sqrt[3]{2x^{4}+5}\cdot x^{3},$ what is the proper substitution?

- Feb 21st 2009, 12:32 PMgammaman
did the x=8 come from taking x^11 and subtracting it from the 3 outside the root?

next, can I get rid of the third root by multiplying (1/3) by x^3 - Feb 21st 2009, 12:34 PMKrizalid
Tell me what's the proper substitution first. :)

- Feb 21st 2009, 12:36 PMgammaman
well 2x^4+5 will be the u

and 8x^3 will be du - Feb 21st 2009, 12:39 PMKrizalid
That's not actually the proper substitution, but it's a good start.

What I call as a "proper substitution," it's such that you can get rid of the cube root. How to do that? Easy, put $\displaystyle u^3=2x^4+5$ and $\displaystyle 3u^2\,du=8x^3\,dx$ and $\displaystyle x^4=\frac{u^3-5}2$ so from there you can get the $\displaystyle x^8.$ Finally $\displaystyle \sqrt[3]{2x^4+5}$ becomes $\displaystyle u.$ - Feb 21st 2009, 12:43 PMgammaman
The only part I do not understand is where this came from

$\displaystyle x^4=\frac{u^3-5}2$ - Feb 21st 2009, 12:44 PMKrizalid
Because of $\displaystyle u^3=2x^4+5$!!

- Feb 21st 2009, 12:47 PMgammaman
and how does this give us x^8??

- Feb 21st 2009, 12:49 PMKrizalid
$\displaystyle x^{8}=\left( x^{4} \right)^{2}=\left( \frac{u^{3}-5}{2} \right)^{2}.$

- Feb 21st 2009, 12:52 PMgammaman
sorry for all the questions, but why are you squaring those terms??

- Feb 21st 2009, 12:53 PMKrizalid
I want you to please tell me how does the integral like before putting $\displaystyle u^3=2x^4+5.$

- Feb 21st 2009, 12:54 PMgammaman
it was the third root?

- Feb 22nd 2009, 05:16 AMtom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/asy/f...ootQuadSub.png

...where straight continuous lines differentiate down / integrate up with respect to the explicit variable, and the staight dashed line with respect to the dashed balloon expression, so that the triangular network satisfies the chain rule. Integrating up the dashed is like finding F(u), and for this we can do...

http://www.ballooncalculus.org/asy/f...adSubPart2.png

This still contains many of the algebra connections that were mystifying you, but you may find it helpful to get this kind of overview.

(Krizalid - respect to the policy of dealing with the cube root in the substitution - but does it really help in this case? The OP's sub of $\displaystyle u = 2x^4 + 5$ seems to work fine. I did a similar picture for $\displaystyle u = (2x^4 + 5)^{\frac{1}{3}}$ and it's alot messier. My written version, too, though maybe not your's. Anyway, hope you don't mind me giving the pot another stir.)

Don't integrate - balloontegrate!

Balloon Calculus: worked examples from past papers - Feb 22nd 2009, 09:58 AMHallsofIvy
My "thought process" is "I sure would like to get rid of that $\displaystyle 2x^4+ 5$ inside the cube root. What would happen if I tried $\displaystyle u= 2x^4+ 4$". Then I would find that $\displaystyle du= 8x^3 dx$ or $\displaystyle (1/8)du= x^3dx$. The "1/8" is a constant so I can alway put in $\displaystyle 8(1/8)$ but where am I going to get that $\displaystyle x^3$ from?

That's when I look at the $\displaystyle x^{11}$ and write it as [tex](x^8)(x^3}= (x^4)^2 (x^3)[tex].

Now I recognize that since $\displaystyle u= 2x^4+ 5$, $\displaystyle x^4= \frac{u- 5}{2}$ and so $\displaystyle (x^4)^2= \left(\frac{u-5}{2}\right)^2$.

Putting that all together, $\displaystyle \int x^11\sqrt[3]{2x^4+ 5}dx= \int (x^4)^2\sqrt{2x^4+ 5}(x^3dx)= \int\frac{(u-5)^2}{4}\sqrt[3]{u} (\frac{1}{8}du)$

Now write that $\displaystyle \sqrt[3]{u}$ as $\displaystyle u^{1/3}$ and you have

$\displaystyle \frac{1}{32}\int (u^2- 10u+ 25)u^{1/3} du= \frac{1}{32}\int (u^{7/3}- 10u^{4/3}+ 25u^{1/3})du$

which should be easy.