Trapezoidal rule

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Section 13.3 Trapezoidal rule

When a function is increasing, the leftpoint rule underestimates its integral, and rightpoint rule overestimates it. So, their average $T_n=(L_n+R_n)/2$ should be a better approximation. This is the Trapezoidal Rule:

\begin{equation*} T_n = \frac{h}{2}(f(a)+f(b)) + h( f(x_1)+\dots+f(x_{n-1})) \end{equation*}

In Matlab, the computation is similar to Example 13.2.1

T = (f(x(1)) + f(x(end))) * h/2 + sum(f(x(2:end-1)))*h;

It may be simpler to compute the vector y = f(x) first, which reduces the number of calls to function f and the number of parentheses:

T = (y(1) + y(end)) * h/2 + sum(y(2:end-1))*h;

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Example 13.3.1. Comparing the accuracy of four rules.

For the function $f(x) = e^x $ on the interval $[-1, 1]\text{,}$ use Matlab to compute the left-, right, and midpoint approximations with $n=10$ and find the error of each approximation (that is, its difference with the actual integral).

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Solution.

f = @(x) exp(x);
a = -1;
b = 1;
n = 10;
h = (b-a)/n;
x = a:h:b;
y = f(x);

L = sum(y(1:end-1))*h;
R = sum(y(2:end))*h;
T = (y(1) + y(end)) * h/2 + sum(y(2:end-1))*h; 
midpoints = (x(1:end-1) + x(2:end))/2;
M = sum(f(midpoints))*h;

exact = exp(1)-exp(-1);
er = abs([L R T M] - exact);
fprintf('Errors: Leftpoint %g, Rightpoint %g, Trapezoidal %g, Midpoint %g\n', er);

Here fprintf “unpacks” the vector er, inserting its entries in the right places.

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