Intergration Help!

**looi76** · Feb 28th 2008, 07:10 AM

Question:
Find the total area of the region shaded in the following diagram:

$y = 3x^2 - 12x + 9$

Attempt:
$y = 3x^2 - 12x + 9$
$a = 3, b = -12, c = 9$

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x = \frac{-(-12) \pm \sqrt{(-12)^2-4\times3\times9}}{2\times3}$

$x = \frac{12 \pm \sqrt{36}}{6}$

$x = \frac{12 + 6}{6}$ , $x = \frac{12 - 6}{6}$
$x = 3$ , $x = 1$

$=3x^2-12x+9$

$=\frac{3x^{2+1}}{2+1} - \frac{12x^{1+1}}{1+1} + 9x$

$=\frac{3x^{3}}{3} - \frac{12x^{2}}{2} + 9x$

$=x^3-6x^2+9x$

$= [x^3-6x^2+9x]^3_1$

$= [(3)^3-6(3)^2+9(3)] - [(1)^3-6(1)^2+9(1)]$

$=-4$

**Krizalid** · Feb 28th 2008, 07:21 AM

Try it again, remember that areas can't be negative.

**colby2152** · Feb 28th 2008, 07:29 AM

Originally Posted by Krizalid

Try it again, remember that areas can't be negative.

With these questions, you always want to split the integrals in parts to make sure each segment of area is positive.

**Gauss** · Feb 28th 2008, 11:22 AM

Also have in mind that integrals of graphs UNDER the x-axes are always negative. Therefore, if you look for an AREA (and not just for an integral), you have to switch it to positive.

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