The 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
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(ax-x^2)-(bx. So y=ax+x^2 is the graph on top and y=bx is at the bottom.
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