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

1. anonymous

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 $Area = \int\limits_{0}^{1}x ^{4}-4x ^{3}+3x - \int\limits_{1}^{3}x ^{4}-4x ^{3}+3x$

2. anonymous

here's the graph

3. anonymous

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

4. anonymous

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. $A = \int\limits_{0}^{1}(x ^{3}-4x ^{2}+3x)dx$

5. anonymous

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