Find the area of the region bounded by y = x^3 - 4x^2 + 3x and the x-axis.

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Find the area of the region bounded by y = x^3 - 4x^2 + 3x and the x-axis.

Mathematics
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first, 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 \[Area = \int\limits_{0}^{1}x ^{4}-4x ^{3}+3x - \int\limits_{1}^{3}x ^{4}-4x ^{3}+3x \]
here's the graph
1 Attachment
Your question needs more information. The area bounded by the X-axis 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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If 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. \[A = \int\limits_{0}^{1}(x ^{3}-4x ^{2}+3x)dx\]
Your question needs more information. The area bounded by the X-axis 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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