anonymous
  • anonymous
(a) integrate the function over the interval (b) find the area of the region between the graph and the x-axis. y=x^2-6x+8, [0,2]
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
(a)\[\int\limits_{0}^{2} \ \ x^2 - 6x+8 \ dx = \left[ x^3/3 - 3x^2+8x \right] \ from \ 0 \to 2\]\[= 2^3/3 - 3(2)^2+8(2) = 20/3\]
anonymous
  • anonymous
(b)\[\int\limits_{2}^{4} \ \ x^2 - 6x+8 \ dx = 4/3\] (below the x-axis).
anonymous
  • anonymous
On part b...where did the 4 and 2 come from?

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anonymous
  • anonymous
The only definite area between the curve and the x-axis is on the interval [2,4].

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