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anonymous
 5 years ago
Find the area of the region bounded by y = x^3  4x^2 + 3x and the xaxis.
anonymous
 5 years ago
Find the area of the region bounded by y = x^3  4x^2 + 3x and the xaxis.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first, you need to draw the graph to know the area the area is equals to the integral of the graph from x=0 to x=1, the graph is above the x axis from x=1 to x=3, the graph is below the x axis \[Area = \int\limits_{0}^{1}x ^{4}4x ^{3}+3x  \int\limits_{1}^{3}x ^{4}4x ^{3}+3x \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your question needs more information. The area bounded by the Xaxis and the equation is infinite. From 0 to 1 y is positive. From 1 to 3 y is negative and from 3 to infinity y approachs infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If we assume it just the first curves area that we are looking for then. All you need to do is integrate from 0 to 1. Your equation will be as follows. \[A = \int\limits_{0}^{1}(x ^{3}4x ^{2}+3x)dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your question needs more information. The area bounded by the Xaxis and the equation is infinite. From 0 to 1 y is positive. From 1 to 3 y is negative and from 3 to infinity y approachs infinity
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