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anonymous
 5 years ago
(a) integrate the function over the interval
(b) find the area of the region between the graph and the xaxis.
y=x^26x+8, [0,2]
anonymous
 5 years ago
(a) integrate the function over the interval (b) find the area of the region between the graph and the xaxis. y=x^26x+8, [0,2]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(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
 5 years ago
Best ResponseYou've already chosen the best response.0(b)\[\int\limits_{2}^{4} \ \ x^2  6x+8 \ dx = 4/3\] (below the xaxis).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0On part b...where did the 4 and 2 come from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The only definite area between the curve and the xaxis is on the interval [2,4].
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