Simpson's Rule

Related Calculator: Simpson's (Parabolic) Rule Calculator

Idea of Simpson's rule is in following: approximate curve by parabola and then find area of parabola (it is easy to do because we know antiderivative of quadratic function).

Again we divide `[a,b]` into `n` subintervals of equal length `Delta x=(b-a)/n`, and also require `n` to be even number.simpson's rule

Then on each consecutive pair of intervals we approximate the curve `y=f(x)` by a parabola. If `y_i=f(x_i)`, then `P_i=(x_i,y_i)` is the point on the curve lying above `x_i`.

A typical parabola passes through three consecutive points `P_i` , `P_(i+1)` and `P_(i+2)`.

First we find equation of parabola that passes through points `(x_0,y_0)`, `(x_1,y_1)` and `(x_2,y_2)`.

Also note that `x_1=x_0+Delta x` and `x_2=x_0+2Delta x`.

Equation of any parabola has form `y=Ax^2+Bx+C` and so area under parabola from `x=x_0` to `x=x_2=x_0+2Delta x` is

`S=int_(x_0)^(x_0+2Delta x) (Ax^2+Bx+C)dx=(A/3 x^3+B/2 x^2+Cx)|_(x_0)^(x_0+2Delta x)=`

`=(A/3 (x_0+2Delta x)^3+B/2 (x_0+2Delta x)^2+C(x_0+2Delta x))-(A/3 x_0^3+B/2 x_0^2+Cx_0)=`

`=2ADelta x x_0^2+4A(Delta x)^2x_0+8/3A(Delta x)^3+2B Delta x x_0+2B(Delta x)^2+2CDelta x=`

`=(Delta x)/3(A(6x_0^2+12Delta x x_0+8A(Delta x)^2)+B(6x_0+6Delta x)+6C)`.

Additionally parabola should pass through points `P_0=(x_0,y_0)`, `P_1=(x_1,y_1)` and `P_2=(x_2,y_2)` (recall that `x_1=x_0+h` and `x_2=x_0+2h`)so

`{(y_0=Ax_0^2+Bx_0+C),(y_1=A(x_0+Delta x)^2+B(x_0+Delta x)+C),(y_2=A(x_0+2Delta x)^2+B(x_0+2Delta x)+C):}`

From this we have that

`y_0+4y_1+y_2=`

`=Ax_0^2+Bx_0+C+4(A(x_0+Delta x)^2+B(x_0+Delta x)+C)+A(x_0+2Delta x)^2+B(x_0+2Delta x)+C=`

`=A(6x_0^2+12Deltax x_0+8(Delta x)^2)+B(6x_0+6Delta x)+6C`.

If we know multiply both sides of equality by `(Delta x)/3` we will obtain that

`(Delta x)/3 (y_0+4y_1+y_2)=(Delta x)/3(A(6x_0^2+12Delta x x_0+8(Delta x)^2)+B(6x_0+6Delta x)+6C)`.

But right side is exactly area `S` under parabola.

Therefore, `S=(Delta x)/3(y_0+4y_1+y_2)`.

Similarly, it can be shown that the area under parabola through `P_2`, `P_3` and `P_4` from `x=x_2` to `x=x_4` is `(Delta x)/3(y_2+4y_3+y_4)`.

In general, area under parabola through `P_i`, `P_(i+1)`, `P_(i+2)` from `x=x_i` to `x=x_(i+2)` is `(Delta x)/3(y_i+y_(i+1)+y_(i+2))`.

Summing areas under parabolas on each subinterval we will obtain that `int_a^bf(x)dx~~h/3(y_0+4y_1+y_2)+h/3(y_2+4y_3+y_4)+...+h/3(y_(n-2)+4y_(n-1)+y_n)`.

Simpson's Rule. `int_a^bf(x)dx~~S_n=(Delta x)/3(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_(n-2)+4y_(n-1)+y_n)`.

where `n` is even and `Delta x=(b-a)/n`.

Note the pattern of coeffcients: 1,4,2,4,2,4,2,4,2,...,4,2,4,2,4,2,4,1.

Example 1. Use Simpson's Rule to approximate value of `int_1^2 1/x^2dx` with `n=8`.

Here ` a=1`, `b=2`, `f(x)=1/x^2` and `n=8`. So, `Delta x=(b-a)/n=(2-1)/8=0.125`.

So, `int_1^2 1/x(dx)~~S_n=`

`=(0.125)/3(f(1)+4f(1.125)+2f(1.25)+4f(1.375)+2f(1.5)+4f(1.625)+2f(1.75)+4f(1.875)+f(2))=`

`=0.125/3(1/1^2+4/(1.125)^2+2/(1.25)^2+4/(1.375)^2+2/(1.5)^2+4/(1.625)^2+2/(1.75)^2+4/(1.875)^2+1/2^2)~~0.5000299.`

True value of integral is `I=int_1^2 1/x^2dx=0.5`. As can be seen Simpson's rule gave very good approximation.

When we approximate integral we will always have some error: `E=int_a^bf(x)dx-App` where `App` is approximation and `E` is error.

Error Bound for Simpson's Rule. Suppose `|f^((4))(x)|<=M` for `a<=x<=b` then `|E|<=(M(b-a)^5)/(180n^4)`.

Example 2. How large should we take `n` in order to guarantee that the Simpson's Rule approximation for `int_1^2 1/x^2 dx` are accurate to within 0.0002?

Here `a=1`, `b=2`, `f(x)=1/x^2`.

Then `f'(x)=-2/x^3`, `f''(x)=6/x^4`, `f'''(x)=-24/x^5` and `f^((4))(x)=120/x^6`.

Therefore `|f^((4))(x)|<=120` for `1<=x<=2`.

Thus, `(120(2-1)^5)/(180n^4)<0.0002` or `n^4>120/(180*0.0002)=1/(0.0003)`.

So, `n>1/(root(4)(0.0003))~~7.6`.

So, we need to take `n=8` (remember `n` must be even).

This is much better then `n=36` for Midpoint Rule and `n=51` for Trapezoidal Rule.