calculus
posted by aimee .
Consider the following shaded region.
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(axx^2)(bx. So y=ax+x^2 is the graph on top and y=bx is at the bottom.

The upper limit of the integration will be where the curves intersect.
That is where 6x  x^2 = 3x.
x^2 3x = x(x3) = 0
x = 0 or 3
Now calculate
int(6xx^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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