Trapezoidal Rule

**omgeezy** · January 11th 2009, 11:03 PM

the 3rd time needing help today...

Problem:
Use the trapezoidal rule to approximate the
$\int (\frac{1}{4}x^3-2x+3) dx$ on the interval $[-4,8]$ using 6 trapezoids of equal height.

I believe the height for each trapezoid will be 2 because 8-(-4) is 12, 12/6=2
I never get the graphing part...
I thought I could get the two main bases of the big trapezoid by plugging in -4 and 8 into the integrand
I got -5 and 115
tried to graph it, didn't go well

are $x=-2, 0, 2, 4, 6$
the other pts that should be plugged into the integrand to find the heights of the rectangles that will be connected to make trapezoids, or directly find the bases of each trapezoid?

**lukaszh** · January 12th 2009, 11:05 AM

Trapezoidal rule:
$\int_a^bf(x)\,\text{d}x\approx\Delta x\sum_{k=1}^{n}\frac{f(x_{k})+f(x_{k-1})}{2}$
So if there is six elements with height $\Delta x=2$
$\int_a^bf(x)\,\text{d}x\approx2\sum_{k=1}^{6}\frac {f(x_{k})+f(x_{k-1})}{2}$
First element:
$\mathcal{A}_1=\frac{f(-4)+f(-2)}{2}=\frac{-5+5}{2}=0$
Secon element:
$\mathcal{A}_2=\frac{f(-2)+f(0)}{2}=\frac{5+3}{2}=4$
Other:
$\mathcal{A}_3=\frac{f(0)+f(2)}{2}=\frac{3+1}{2}=2$
$\mathcal{A}_4=\frac{f(2)+f(4)}{2}=\frac{1+11}{2}=6$
$\mathcal{A}_5=\frac{f(4)+f(6)}{2}=\frac{11+45}{2}= 28$
$\mathcal{A}_6=\frac{f(6)+f(8)}{2}=\frac{45+115}{2} =80$
So the area:
$\int_{-4}^{8}\left(\frac{x^3}{4}-2x+3\right)\,\text{d}x\approx2\sum_{k=1}^{6}\mathc al{A}_k=2\left(0+4+2+6+28+80\right)=240$

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