PDA

View Full Version : A little problem


Sombra
Dec5-04, 06:34 PM
y = (x^5 + 4x^4)^(1/2)

They want me to find the area of this loop, and the boundaries are -4 and 0, but when I integrate it and plug in the -4, I get 0, which is clearly not the case. If you can help, I would love it.

Tide
Dec5-04, 07:08 PM
I think you left out the second limit on the integration!

WaR
Dec5-04, 07:12 PM
Can you post your integrating steps?
That integration is not equal to 0
But
\int_{-4}^{0} \sqrt{x^{5}+4x^{4}} dx = \frac{2048}{105}

Sombra
Dec5-04, 07:22 PM
I did:

Integration of (x^5 +4x^4)^(1/2) is 2/3 (x^5 +4x^4)^(3/2)

So 2/3(0 + 0)^(3/2) = 0

0 - [2/3 (-1024 + 1024)^(3/2) = 0

It doesn't work.

Tide
Dec5-04, 07:30 PM
No, you integrated incorrectly!

Sombra
Dec5-04, 07:35 PM
ok, the answer given in the back of the book is 4096/105. I know that even if that answer is not correct, it is more correct than mine because the area of that loop in the graph is clearly not 0.

Sombra
Dec5-04, 07:38 PM
ok, I think the book is wrong because the calculator said that it was 19.50476233. I still don't know how this works though because.... I still get 0.

Pyrrhus
Dec5-04, 07:48 PM
Little hint

\int_{-4}^{0} \sqrt{x^{5}+4x^{4}} dx

\int_{-4}^{0} x^{2} \sqrt{x+4} dx

Sombra
Dec5-04, 07:57 PM
that's how the original problem was set up. I distributed, thinking it would be easier. In any case, I get 0. Thanks

Pyrrhus
Dec5-04, 08:11 PM
Post the original problem as the book states it.

Sombra
Dec5-04, 08:21 PM
Sketch the graph y^2 = x^4(x+4) and find the area enclosed by the loop.

Tide
Dec5-04, 08:34 PM
That is distinctly different from the problem you first posed - but your integration is still incorrect! :-)

Sombra
Dec5-04, 08:36 PM
the boundaries are still -4 and 0, and my answer still comes out to be 0. What is correct? I guess I am not understanding. Please help!!

Tide
Dec5-04, 08:42 PM
\int x^2 \sqrt {x+4} dx = \frac {2 (x+4)^{3/2} (15 x^2 -48 x + 128)}{105}

Sombra
Dec5-04, 08:54 PM
ok so 2/3(x+4)^3/2, I got that part, but how did you get the other part. I am not familiar with integrating something undistributed unless it follows the f(x) * f'(x) rule. Sorry for all the questions. Thank you so much!!

futb0l
Dec5-04, 09:00 PM
Use cyclovenom's hint

Sombra
Dec5-04, 09:15 PM
Now I am totally lost.... I know I have to integrate only the top half of this loop and multiply it by 2 to find the area or it will equal 0. I just do not know how to set up the problem to make it only the top half of the loop.

Tide
Dec5-04, 11:48 PM
You can integrate by parts. You may get a slightly different form than I did because I did some simplification.