Tuesday, July 14, 2015

Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 62

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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