integral thought process?

**gammaman** · February 21st 2009, 12:15 PM

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?

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

**Krizalid** · February 21st 2009, 12:18 PM

Think a bit: $x^{11}\sqrt[3]{2x^{4}+5}=x^{8}\cdot \sqrt[3]{2x^{4}+5}\cdot x^{3},$ what is the proper substitution?

**gammaman** · February 21st 2009, 12:32 PM

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

**Krizalid** · February 21st 2009, 12:34 PM

Tell me what's the proper substitution first.

**gammaman** · February 21st 2009, 12:36 PM

well 2x^4+5 will be the u
and 8x^3 will be du

**Krizalid** · February 21st 2009, 12:39 PM

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 $u^3=2x^4+5$ and $3u^2\,du=8x^3\,dx$ and $x^4=\frac{u^3-5}2$ so from there you can get the $x^8.$ Finally $\sqrt[3]{2x^4+5}$ becomes $u.$

**gammaman** · February 21st 2009, 12:43 PM

The only part I do not understand is where this came from

$x^4=\frac{u^3-5}2$

**Krizalid** · February 21st 2009, 12:44 PM

Because of $u^3=2x^4+5$ !!

**gammaman** · February 21st 2009, 12:47 PM

and how does this give us x^8??

**Krizalid** · February 21st 2009, 12:49 PM

$x^{8}=\left( x^{4} \right)^{2}=\left( \frac{u^{3}-5}{2} \right)^{2}.$

**gammaman** · February 21st 2009, 12:52 PM

sorry for all the questions, but why are you squaring those terms??

**Krizalid** · February 21st 2009, 12:53 PM

I want you to please tell me how does the integral like before putting $u^3=2x^4+5.$

**gammaman** · February 21st 2009, 12:54 PM

it was the third root?

**tom@ballooncalculus** · February 22nd 2009, 05:16 AM

Just in case a picture helps...

...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...

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 $u = 2x^4 + 5$ seems to work fine. I did a similar picture for $u = (2x^4 + 5)^{\frac{1}{3}}$ 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

**HallsofIvy** · February 22nd 2009, 09:58 AM

Originally Posted by gammaman

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?

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

My "thought process" is "I sure would like to get rid of that $2x^4+ 5$ inside the cube root. What would happen if I tried $u= 2x^4+ 4$ ". Then I would find that $du= 8x^3 dx$ or $(1/8)du= x^3dx$ . The "1/8" is a constant so I can alway put in $8(1/8)$ but where am I going to get that $x^3$ from?

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

Now I recognize that since $u= 2x^4+ 5$ , $x^4= \frac{u- 5}{2}$ and so $(x^4)^2= \left(\frac{u-5}{2}\right)^2$ .
Putting that all together, $\int x^11\sqrt[3]{2x^4+ 5}dx= \int (x^4)^2\sqrt{2x^4+ 5}(x^3dx)= \int\frac{(u-5)^2}{4}\sqrt[3]{u} (\frac{1}{8}du)$
Now write that $\sqrt[3]{u}$ as $u^{1/3}$ and you have
$\frac{1}{32}\int (u^2- 10u+ 25)u^{1/3} du= \frac{1}{32}\int (u^{7/3}- 10u^{4/3}+ 25u^{1/3})du$
which should be easy.

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