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The Trapezoidal Rule

We can begin to get more creative with our Reimann Sums to get better approximations for area under curves. Instead of using a rectangle approximation, we can place trapezoids on the graph to approximate area.



Trapezoid Rule

abf(x)dx i=1nf(xi1)+f(xi)2=Δx2(f(x0)+2f(x1)+2f(x2)+...+2f(xn1)+f(xn))\boxed{\int_a^bf\left(x\right)dx\ \approx\sum_{i=1}^n\frac{f\left(x_{i-1}\right)+f\left(x_i\right)}{2}=\frac{\Delta x}{2}\left(f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+...+2f\left(x_{n-1}\right)+f\left(x_n\right)\right)}

where Δx=ban,  xi=a+iΔx\displaystyle \Delta x=\frac{b-a}{n} , \ \ x_i=a+i\Delta x , and nn is the number of subintervals.

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Example: The Trapezoid Rule

Use a trapezoid approximation for

04ex2dx\displaystyle \int_0^4e^{x^2}dx

using 4 sub intervals.


abf(x)dx i=1nf(xi1)+f(xi)2=Δx2(f(x0)+2f(x1)+2f(x2)+...+2f(xn1)+f(xn))\displaystyle \int_a^bf\left(x\right)dx\ \approx\sum_{i=1}^n\frac{f\left(x_{i-1}\right)+f\left(x_i\right)}{2}=\frac{\Delta x}{2}\left(f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+...+2f\left(x_{n-1}\right)+f\left(x_n\right)\right)


Δx=ban=404=1\displaystyle \Delta x=\frac{b-a}{n}=\frac{4-0}{4}=1

04ex2dx12(f(0)+2f(1)+2f(2)+2f(3)+f(4)) =12(e02+2e12+2e22+2e32+e42)\displaystyle \int_0^4e^{x^2}dx\approx\frac{1}{2}\left(f\left(0\right)+2f\left(1\right)+2f\left(2\right)+2f\left(3\right)+f\left(4\right)\right)\ =\frac{1}{2}\left(e^{0^2}+2e^{1^2}+2e^{2^2}+2e^{3^2}+e^{4^2}\right)

=12(1+2e+2e4+2e9+e16)\displaystyle =\frac{1}{2}\left(1+2e+2e^4+2e^9+e^{16}\right)
Which of the following is the integral approximation of
13cos(x3)dx\int_1^3 \cos (x^3)dx
using the trapezoid rule with n=3n=3?