1. ## Trapezoidal rule proof

Consider the function y=x^2 for x[0,1]. Show using the trapezoidal rule with n intervals that the integral 0 -> 1 of x^2 dx is equivalent to n(n+1)(2n+1)/6n^3 - 1/2n.

I am having a lot of trouble arranging the trapezoidal rule into a sigma type sequence.

My current working goes as far as
integral of x^2 is equivilant to 1-0/2n [f(x0) +2(f(x1)+f(x2)+...+f(xn-1))+f(n)]

All help is greatly appreciated.

2. Let the width of a strip be $\displaystyle h$ so $\displaystyle nh = 1$.

Then $\displaystyle f(x_1) = h^2,$ $\displaystyle f(x_2) = (2h)^2,$ $\displaystyle f(x_3) = (3h)^2$, and so on.

If you substitute these into your trapezium rule formula and remove $\displaystyle h^2$ as a common factor, the formula for the sum of squares

$\displaystyle 1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n}{6}(n+1)(2n+1),$

can be used to simplify.

Towards the end, remove $\displaystyle h$ by writing it as $\displaystyle 1/n$.

Incidently it's best to sum the series as

$\displaystyle 1^2 + 2^2 + ... + (n-1)^2 = 1^2 + 2^2 + ... + n^2 - n^2$.