# Calculus II

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### 1 Answer

Simpson's rule approximates an integral `int_a^b f(x) dx` to `(h/3)*(f(x_0) + 4*f(x_1) + 2*f(x+2) +...+f(x_n))` where `x_0 = a` and` x_n = b` and `h = (b - a)/n`

The integral `int_0^3 e^(2x) dx = [e^(2x)/2]_0^3 = (e^6 - e^0)/2 = (e^6 - 1)/2 ~~201.21`

Using Simpson's rule with n = 6, `h = (3 - 0)/6 = 1/2` . The approximate value of the integral is:

`(1/6)*(e^0 + 4*e^1 + 2*e^2 + 4*e^3 + 2*e^4 + 4*e^5 + e^6)`

`~~ 202.211`

The error in using the Simpson's approximation is `(h^4/180)*(b - a)*max_(xi in [a,b]) f^4(xi)`

For any value of n, for the given integral, the error is `((3/n)^4/180)*(3)*16*e^6`

If the error has to be within 0.0001, n should be 55

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did you find the exact error??