The area A of the region between the two curves x = y^3–4y^2+3y and x+y=y^2 is 11.8334.
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We have to evaluate the integral needed to determine the area A of the region between the two curves x = y^3–4y^2+3y and x+y=y^2 i.e. x = y^2-y.
Now the area A of the given region is:
A = [tex]\int^1_{y=0}\left[g(y)-f(y)\right]dy+\int^4_{y=1}\left[g(y)-f(y)\right]dy[/tex]
A = [tex]\int^1_{y=0}\left[(y^3-4y^2+3y)-(y^2-y)\right]dy+\int^4_{y=1}\left[(y^2-y)-(y^3-4y^2+3y)\right]dy[/tex]
A = [tex]\int^1_{y=0}\left[y^3-4y^2+3y-y^2+y)\right]dy+\int^4_{y=1}\left[y^2-y-y^3+4y^2-3y)\right]dy[/tex]
A = [tex]\int^1_{y=0}\left[y^3-5y^2+4y\right]dy+\int^4_{y=1}\left[5y^2-y^3-4y)\right]dy[/tex]
Further simplification
A = [tex]\left[\frac{y^4}{4}-\frac{5y^3}{3}-\frac{4y^2}{2}\right]^1_{y=0}+\left[\frac{5y^3}{3}-\frac{y^4}{4}-\frac{4y^2}{2}\right]^4_{y=1}[/tex]
A = [tex]\left[(\frac{1}{4}-\frac{5}{3}-2)\right-(0-0+0)]+\left[(\frac{320}{3}-64-32)\right-(\frac{5}{3}-\frac{1}{4}-2)][/tex]
A = (3-20+24)/12 + 320/3 - 96 - (20-3-24)/12
A = 7/12 + 32/3 + 7/12
A = (7+128+7)/12
A = 142/12
A = 11.8334
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