# Integral - Pls check my work

• Dec 3rd 2007, 06:23 PM
angel.white
Integral - Pls check my work
$\int_{-1}^{1} (\sqrt[3]{u} + 1)^{2} du$

This is what I did:
(FOIL)
$\int_{-1}^{1} ({u}^{2/3} + 2u^{1/3} + 1) du$

$\left[ \frac{3}{5}u^{5/3} + \frac{3}{2}u^{4/3} + u + C\right]_{-1}^{1}$

$\frac{3}{5}(1)^{5/3} + \frac{3}{2}(1)^{4/3} + 1 + C - \left(\frac{3}{5}(-1)^{5/3} + \frac{3}{2}(-1)^{4/3} - 1 + C \right)$

$\frac{3}{5} + 1 + \frac{3}{5} + 1$

$\frac{16}{5}$

But when I check it here: Function calculator it gives me what appears to be an imaginary number, it says the answer is: $4.53546396068 + 1.80971485451I_{3}$. But I can't understand why, the negative ones should all work because they are never taken to an even root.
• Dec 3rd 2007, 07:15 PM
topsquark
Quote:

Originally Posted by angel.white
$\int_{-1}^{1} (\sqrt[3]{u} + 1)^{2} du$

This is what I did:
(FOIL)
$\int_{-1}^{1} ({u}^{2/3} + 2u^{1/3} + 1) du$

$\left[ \frac{3}{5}u^{5/3} + \frac{3}{2}u^{4/3} + u + C\right]_{-1}^{1}$

$\frac{3}{5}(1)^{5/3} + \frac{3}{2}(1)^{4/3} + 1 + C - \left(\frac{3}{5}(-1)^{5/3} + \frac{3}{2}(-1)^{4/3} - 1 + C \right)$

$\frac{3}{5} + 1 + \frac{3}{5} + 1$

$\frac{16}{5}$

But when I check it here: Function calculator it gives me what appears to be an imaginary number, it says the answer is: $4.53546396068 + 1.80971485451I_{3}$. But I can't understand why, the negative ones should all work because they are never taken to an even root.

Funny. It told me that $(u^{1/3} + 1)^2$ didn't exist on the whole interval. (Drunk)

I guess the solution is that Function calculator sucks. (No)

-Dan
• Dec 3rd 2007, 07:16 PM
Jhevon
Quote:

Originally Posted by topsquark
Funny. It told me that $(u^{1/3} + 1)^2$ didn't exist on the whole interval. (Drunk)

I guess the solution is that Function calculator sucks. (No)

-Dan

did you try evaluating the second part of his answer in a regular calculator?

maple gave me a complex answer as well, i can't see why at the moment
• Dec 3rd 2007, 07:20 PM
topsquark
Quote:

Originally Posted by Jhevon
did you try evaluating the second part of his answer in a regular calculator?

maple gave me a complex answer as well, i can't see why at the moment

I looked at his solution (and found no errors) and after seeing how bad the function calculator was, I ran it through my TI-92.

My guess is that the programs are having trouble with the cube root for negative x. (Though you would think that Maple at least wouldn't have that problem.)

-Dan
• Dec 3rd 2007, 07:24 PM
Jhevon
Quote:

Originally Posted by topsquark
I looked at his solution (and found no errors) and after seeing how bad the function calculator was, I ran it through my TI-92.

My guess is that the programs are having trouble with the cube root for negative x. (Though you would think that Maple at least wouldn't have that problem.)

-Dan

which brings up another interesting--maybe related--point. when you try to graph x^(1/3) with Graph, it only graphs the function for positive x's. why is it that programs have trouble evaluating the odd roots of negative numbers? is it a glich, or do they know something that we don't?
• Dec 3rd 2007, 10:56 PM
DivideBy0
Mine works perfectly fine for negative x
• Dec 3rd 2007, 11:04 PM
angel.white
Quote:

Originally Posted by DivideBy0
Mine works perfectly fine for negative x

Which do you use? & is it free?
• Dec 4th 2007, 12:03 AM
Jhevon
Quote:

Originally Posted by angel.white
Which do you use? & is it free?

Graph is free, and it works great (except, perhaps, in this situation :D). do you have it?
• Dec 4th 2007, 12:17 AM
angel.white
Quote:

Originally Posted by Jhevon
Graph is free, and it works great (except, perhaps, in this situation :D). do you have it?

I've been using:
Cool math .com - Online Graphing Calculator - Graph It!
and Function Grapher Online
• Dec 4th 2007, 12:27 AM
Jhevon
Quote:

Originally Posted by angel.white

I've been using:
Cool math .com - Online Graphing Calculator - Graph It!
and Function Grapher Online

well, Graph does a lot more than just plot graphs. it does other cool things, like finding the area under curves and stuff
• Dec 4th 2007, 12:42 AM
angel.white
Quote:

Originally Posted by Jhevon
Graph is free, and it works great (except, perhaps, in this situation :D). do you have it?

Wow, it's a pretty neat program :) I like how easy it is to check an answer, click the shady parabola thing :P

But, it also could not verify the answer for negative numbers, said it was imaginary. But I checked my answer anyway by taking it from 0 to 1 instead, and graph agreed with me, so if I got it wrong, then it's only due to an arithmetic error at the end. And that is rather unlikely as topsquark got the same answer.

On a side note, I wish I was a robot.
• Dec 6th 2007, 03:38 AM
angel.white
Quote:

Originally Posted by Jhevon
well, Graph does a lot more than just plot graphs. it does other cool things, like finding the area under curves and stuff

And make smiley faces. (Cool)
• Dec 6th 2007, 07:21 AM
topsquark
(Someone has a lot of spare time on their hands!) (Beer)

-Dan