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October 1, 2014

October 1, 2014

Posted by **aimee** on Friday, February 12, 2010 at 1:24pm.

Find the area S of this region if a = 6, b = 3. (Give an exact answer.)

The two graphs intercept at 0 and the other limit is not given. The integral is int(ax-x^2)-(bx. So y=ax+x^2 is the graph on top and y=bx is at the bottom.

- calculus -
**drwls**, Friday, February 12, 2010 at 1:35pmThe upper limit of the integration will be where the curves intersect.

That is where 6x - x^2 = 3x.

x^2 -3x = x(x-3) = 0

x = 0 or 3

Now calculate

int(6x-x^2)-(3x)dx from 0 to 3

= int 3x - x^2 dx from 0 to 3

= [3x^2/2 - x^3/3]@x=3 - 0

(Since the value of the indefinite integral in brackets, at x = 0, is 0)

= 27/2 - 27/3 = 27/6 = 9/2

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