Find the area

**333arizona** · May 19th 2008, 05:35 PM

First I need to find the area between the functions 2x^2-8x+11 and x^2-4x+10, then I need to find the area when it's rotated around the y-axis, assistance is much appreciated.

**Mathstud28** · May 19th 2008, 05:40 PM

Originally Posted by 333arizona

First I need to find the area between the functions 2x^2-8x+11 and x^2-4x+10, then I need to find the area when it's rotated around the y-axis, assistance is much appreciated.

This one you should practice. Show your work. You do it by setting the two funcions equal solving setting up an integral based on the intersection coordinates and find within the interval of intersection which function is greater. Once you do that take the integral from the first intersection to the second of the greater function minus the smaller function'

show your work and we will critique

**333arizona** · May 19th 2008, 05:51 PM

ok, i set them equal to each other and ended up with x^2-4x+1=0, and now I feel stupid because I don't remember how to factor that, even though I remember doing it.

**Mathstud28** · May 19th 2008, 05:55 PM

Originally Posted by 333arizona

ok, i set them equal to each other and ended up with x^2-4x+1=0, and now I feel stupid because I don't remember how to factor that, even though I remember doing it.

Well dont feel stupid. You cannot factor it.

You have to used the quadratic formula $x=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt{3}$

Which is the correct answer

**333arizona** · May 19th 2008, 05:58 PM

ok, I did that, and ended up with that but it didn't seem to get me anywhere so I threw that idea out.

**Mathstud28** · May 19th 2008, 06:02 PM

Originally Posted by 333arizona

ok, I did that, and ended up with that but it didn't seem to get me anywhere so I threw that idea out.

Ok so if you graphed your equations you would see they intersect at those points...and throughout that interval $(2-\sqrt{3},2+\sqrt{3})$ one is greater than the other

I instead of graphing will test point an element of the interval

$f(x)=2x^2-8x+11$

$g(x)=x^2-4x+10$

$f(0)=11$

$g(0)=10$

$\therefore,f(x)>g(x)\forall{x}\in(2-\sqrt{3},2+\sqrt{3})$

So your integral for the area would be

$\int_{2-\sqrt{3}}^{2+\sqrt{3}}f(x)-g(x)dx$

**333arizona** · May 19th 2008, 06:16 PM

ok so, i plugged in the functions and collected like terms to get

integral x^2-12x+21

took the integral and got x^3/3-6x^2+21x

plugged in the 2 values and got a headache

**Mathstud28** · May 19th 2008, 06:22 PM

Originally Posted by 333arizona

ok so, i plugged in the functions and collected like terms to get

integral x^2-12x+21

took the integral and got x^3/3-6x^2+21x

plugged in the 2 values and got a headache

$\int_{2-\sqrt{3}}^{2+\sqrt{3}}\bigg[x^2-12x+21\bigg]dx=\bigg[\frac{x^3}{3}-6x^2+21x\bigg]\bigg|_{2-\sqrt{3}}^{2+\sqrt{3}}=4\sqrt{3}$

If you need me to show how just say so

**333arizona** · May 19th 2008, 06:24 PM

please, I'm really in deep on this one now, a little help on the next step would be like getting to take graph theory again.

Thread: Find the area

LinkBack

Thread Tools

Display

Find the area

Similar Math Help Forum Discussions

Given 3 points, find the area of the triangle/find the volume of the parallelepiped

Find the area..

find the area

Find the area...

find area of shaded area (yellow)

Search Tags