`y=x/(1+e^(x^3)) , y=0 , x=2` Find the area of the region bounded by the graphs of the equations

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If we have two function f(x) and g(x) we can find the area between the curves with the following formula.

`A=int^b_a (f(x)-g(x))dx`

Where f(x) is the upper function and g(x) is the lower function.

In this case let `f(x)=x/(1+e^x^3), g(x)=0` and integrate from x=0 to x=2.

`A=int^2_0 (x/(1+e^x^3)-0)dx`

`A=int^2_0 x/(1+e^x^3)dx`

This integral may not have an analytical solution. So I'm just going to approximate the integral useing Simpsons rule.

`A=int^2_0 x/(1+e^x^3)dx~~(Delta x)/3(y_0+4y_1+2y_2+4y_3+2y_4...+4y_(n-1)+y_n)`

Where `Delta x=(b-a)/n=2/n`

Ill use 10 intervals, so n=10.

`A=int^2_0 x/(1+e^x^3)dx~~0.287099`

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