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August 30, 2014

August 30, 2014

Posted by **James** on Monday, June 24, 2013 at 9:18am.

- Math 61 part 2 -
**Steve**, Monday, June 24, 2013 at 10:14amThe area of the region is

A = ∫[-1,3] (2x+3)-x^2 dx

= -1/3 x^3 + x^2 + 3x [-1,3]

= (-9+9+9)-(1/3 + 1 - 3)

= 32/3

For xbar, find

X = ∫[-1,3] x((2x+3)-x^2) dx

= -1/4 x^4 + 2/3 x^3 + 3/2 x^2 [-1,3]

= (-81/4 + 18 + 27/2)-(-1/4 + 2/3 + 3/2)

= 32/3

For ybar, find

Y = ∫[-1,3] 1/2 ((2x+3)^2 - (x^2)^2) dx

= -1/5 x^5 + 4/3 x^3 + 6x^2 + 9x [-1,3]

= 544/15

xbar = X/A = 1

ybar = Y/A = 17/5

- Math 61 part 2 -
**MathMate**, Monday, June 24, 2013 at 10:18amTo solve this category of problems where you need the area/centroid of a region bounded by two curves, you need to first find TWO intersection points of the two curves by equating y1(x)=x^2 and y2(x)=2*x+3.

The intersection points are at x=-1 and x=3, with y2(x) above y1(x).

Then you need to find the area by integrating

A=∫(y2(x)-y1(x))dx between the limits x=-1 and x=3.

=32/3

To find the centroid(xBar,yBar), you need to find the first moments,

xBar=∫x(y2(x)-y1(x))dx / A

and

yBar=∫(1/2)(y2(x)+y1(x))(y2(x)-y1(x))dx /A

from which I get

(xBar,yBar)=(1,17/5)

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