Author Topic: Integration (Read 10246 times) Tweet Share

pHysiX · « **Reply #45 on:** April 11, 2009, 09:46:31 pm »

i love integration by parts.

last year's methods teachers were like u can't anti diff logs and had another approach.

along came integration by parts and voila, proved them wrong. woots!

haha something random XD

d0minicz · « **Reply #46 on:** April 14, 2009, 12:38:55 am »

Find the volume of the region bounded by $y= x^2 -5x +6$ and y=0 , about the y-axis.
thanks.

kamil9876 · « **Reply #47 on:** April 14, 2009, 10:49:33 am »

Quote from: d0minicz on April 14, 2009, 12:38:55 am

Find the volume of the region bounded by $y= x^2 -5x +6$ and y=0 , about the y-axis.
thanks.

Refer to diagram below:

$y=(x-2)(x-3)$

$M=y(2.5)$
$=-0.25$

So 0 and -0.25 are your terminals.

Now you must find the x in terms of y. A way to do this is to turn the parabola into turning point form:
$<br />y=x^2-5x+\frac{25}{4}-\frac{25}{4}+6$
$y=(x-\frac{5}{2})^2 -\frac{1}{4}$
$y+\frac{1}{4}=(x+\frac{5}{2})^2$
$+/-\sqrt{y+\frac{1}{4}}=x-\frac{5}{2}$
$x=+/-\sqrt{y+\frac{1}{4}}+\frac{5}{2}$

This is a volume between two curves. One curve is the positive square root while the other is the negative square root. So now put in those two expression in the formula for volume between two curves with terminals 0 and -0.25.

d0minicz · « **Reply #48 on:** April 14, 2009, 11:59:52 am »

thanks ; but i need help with the integration part
can you plz show me

kamil9876 · « **Reply #49 on:** April 14, 2009, 12:53:26 pm »

first pic, first part. Second pic, second part.

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