AP Calc: use Trap. Rule w/ n=4 for int[2,3][1/(x-1)^2]dx

[venialove]

I don't understand how to use this rule. (I have sketched the graph)
This is what I did.
1/(2-1)^2 +1 (3-1)^2 etc... But my answer came out to be 1.7348

Use the Trapezoidal Rule wih n=4 to approximate

3
? 1/(x-1)^2 dx
2

 

[Dr. Flim-Flam]

Re: ap calculus (I have sketched the graph)

Integral[1/(x-1)^2,x,2,3] = (1/8)[f(2)+2f(9/4)+2f(5/2)+2f(11/4)+f(3)] =(1/8)[1+32/16+8/9+32/49+1/4] = about.508.
Trapezoid Rule, n = 4.

 

[galactus]

Re: ap calculus (I have sketched the graph)

The trapezoid method involves the formula:

\(\displaystyle \int_{a}^{b}f(x)dx\approx{\left(\frac{b-a}{2n}\right)\left[y_{0}+2y_{1}+.....+2y_{n-1}+y_{n}\right]\)

So, we have \(\displaystyle \frac{b-a}{2n}=\frac{3-2}{2(4)}=\frac{1}{8}\)

\(\displaystyle \begin{vmatrix}x&y\\2&1\\\frac{9}{4}&\frac{16}{25}\\\frac{5}{2}&\frac{4}{9}\\\frac{11}{4}&\frac{16}{49}\\3&\frac{1}{4}\end{vmatrix}\)

Remember, the terms in the middle are multiplied by 2.

So, we get \(\displaystyle \frac{1}{8}\left[1+\frac{32}{25}+\frac{8}{9}+\frac{32}{49}+\frac{1}{4}\right]=\frac{179573}{352800}\)