Question

I need help evaluating this integral.

Answer 1

\[\int\limits_{0}^{2}x ^{3} \sqrt (x^2+4) dx\]

Answer 2

The whole 2nd part is supposed to be under the square root.. (x^2+4)

Answer 3

\[\int_{0}^{2}x^3\sqrt{x^2+4}dx\]\[u=x^2+4\]\[du=2xdx\]at first it may not seem like it but this sub will work
just follow it through

Answer 4

Are you sure? I think my professor wants us to use trig substitution because that way till not work

Answer 5

I am just not sure how to use the trig substiution with the damn x^3 out front

Answer 6

that way will work, but if they want you to use a trig sub so be it...

Answer 7

Well would I use x = 2tan(t)? but then i have 8tan^3(t) out front?

Answer 8

That's too bad...I liked the first way you were doing it TT

Answer 9

here that is anyway zarkon, just for completeness :D
I will show you first how my way works\[u=x^2+4\]\[du=2xdx\]\[\int x^3\sqrt{x^2+4}dx=\frac12\int(u-4)udu=\frac12\int (u-4)\sqrt udu\]\[=\frac12\int u^{3/2}-4u^{1/2}du\]ok now the trig way...
yes you will usex=tan(hteta)

Answer 10

sorry x=2tan(theta)

Answer 11

\[\int x^3\sqrt{x^2+4}dx\]\[u=2\tan\theta\]\[du=2\sec^2\theta d\theta\]what do you get after the substitution paintball?

Answer 12

Dont you get \[\int\limits_{}^{}8\tan^3\Theta(2\sec^2\Theta)d \Theta\]

Answer 13

what happened to all that business under the square root?

Answer 14

did you even do the first way right?how did you get that( u-4)udu?what happened to the square root?

Answer 15

sorry so its ∫8tan^3Θ(2secΘ)dΘ

Answer 16

\[u=x^2+4\to x^2=u-4\]then I subbed that in for x^2 in the integral
I assure you it's right...

Answer 17

anyway, now you forgot to put it dx above...

Answer 18

oh..

Answer 19

but its not x^2, its x^3

Answer 20

which way do you want to do it?
I will be happy to explain all the details of my way if you think that is suficient

Answer 21

lets just do the first one I guess if its right. Sorry. I dont see how it works since its x^3, yet you solve for x^2?

Answer 22

it's okay, look...

Answer 23

let me break it up so it is a little more clear what exactly I'm gonna do\[\int x^3\sqrt{x^2+4}dx=\int x^2\sqrt{x^2+4}(xdx)\]now look at the expression we get for the above terms from our sub:\[u=x^2+4\to x^2=u-4\]\[du=2xdx\to xdx=\frac{du}2\]this implies that\[x^3dx=x^2(xdx)=\frac12(u-4)du\]sub these expressions in above for x^2 and xdx and we get\[\frac12\int(u-4)\sqrt udu\]

Answer 24

ohh that make senses then. The du will get rid of one of the x's so it becomes easier. Thank you very much

Answer 25

you're welcome
many people doubt the power of the u-sub in these situations
don't be one of them ;)

Answer 26

for the record here is the trig way\[\int(2\tan\theta)^3\sqrt{4\tan^2\theta+4}(2\sec^2\theta) d\theta=2^5\int\tan^3\theta\sec^3\theta d\theta\]now you would strip out a secxtanx and convert the remaining tan^2 to 1-sec^2, then use a u-sub, then turn it all back into x
much more of a pain!

Answer 27

I did the first way out and get a negative number?Do I just take the absolute value?

Answer 28

hmm... you should not have gotten a negative number
I will check it out, brb

Answer 29

are you evaluating\[\frac15(x^2+4)^{5/2}-\frac43(x^2+4)^{3/2}|_{0}^{2}\]?

because that is how you do it, and the answer is non-negative

Answer 30

sorry I will try it now. just go back

Answer 31

i got \[(152\sqrt{8}+64)/15\]

Answer 32

It seems you are making an algebra error
try to get this answer
http://www.wolframalpha.com/input/?i=integral+from+0+to+2+of+x%5E3*sqrt%28x%5E2%2B4%29dx