# Homework Help: Trapazoidal Approximation help

1. May 2, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

$$\int^{5}_{1} \frac{x}{x+1} dx$$

using dx = .5

2. Relevant equations

$$\sum^{a}_{b} \frac {f(x)+f(x+dx)}{2} dx = [\frac{1}{2}f(x_{0})+f(x_{1})+\cdots+f(x_{n-1})+\frac{1}{2}f(x_{n})]dx$$

3. The attempt at a solution
TAI = Trapazoidal Approximation Input value

$$x_{0}=1 \rightarrow f(x_{0})=\frac {1}{2} \rightarrow TAI = \frac{1}{2}\frac{1}{2} =\frac{1}{4}$$

$$x_{1} = 1.5 \rightarrow f(x_{1}) = \frac{1.5}{2.5} \rightarrow TAI = \frac{1.5}{2.5}$$

$$x_{2} = 2 \rightarrow f(x_{2}) = \frac{2}{3} \rightarrow TAI = \frac{2}{3}$$

$$x_{3} = 2.5 \rightarrow f(x_{3})= \frac{2.5}{3.5} \rightarrow TAI = \frac {2.5}{3.5}$$

$$x_{4} = 3 \rightarrow f(x_{4}) = \frac{4}{5} \rightarrow TAI = \frac {4}{5}$$

$$x_{5} = 3.5 \rightarrow f(x_{5})= \frac{3.5}{4.5} \rightarrow TAI = \frac {3.5}{4.5}$$

$$x_{6} = 4 \rightarrow f(x_{6}) = \frac {4}{5} \rightarrow TAI = \frac {4}{5}$$

$$x_{7} = 4.5 \rightarrow f(x_{7}) = \frac{4.5}{5.5} \rightarrow TAI = \frac {4.5}{5.5}$$

$$x_{8} = 5 \rightarrow f(x_{8}) = \frac {5}{6} \rightarrow TAI = \frac{1}{2}\frac{5}{6} = \frac {5}{12}$$

so now, following the formula I add up all the TAI values, and the 1/2's have already been applied, so:

$$\frac{1}{4} + \frac {1.5}{2.5} + \frac{2}{3} + \frac{2.5}{3.5} + \frac {3}{4}+ \frac{3.5}{4.5} + \frac {4}{5} + \frac {4.5}{5.5} + \frac {5}{12} = 5.843578644 = \frac{20248}{3465}$$

and then I multiply the sum by dx which is .5:

$$\frac {20248}{3465} * \frac {1}{2} = 2.921789$$

the book states that the answer they are looking for is 2.8968

and for E(dx) (error) I take the second derivative:

$$y = \frac {x}{x+1}$$

$$y' = \frac {(x+1)(1)-(x)(1)}{(x+1)^{2}}$$

$$y' = \frac {x+1-x}{(x+1)^{2}}$$

$$y' = \frac {1}{(x+1)^{2}}$$

which the same as:

$$y' = (x+1)^{-2}$$

so:

$$y''= -2(x+1)^{-3}$$

and the formula for error is:

$$|f''(x)| \leq M$$ for $$a \leq x \leq b$$

and:

$$E(dx)= \frac {b-a}{12} M (dx)^{2}$$

so now:

$$|f''(1)| = .25$$

$$|f''(5)| = .00926$$

so f'' has M at 1 so to find error we do:

$$\frac {5-1}{12}(.25)(.25)^{2} = \frac {1}{192}$$

the book states that the correct E(dx) is $$\frac {1}{48}$$

I have done this many times over and can not find where I went wrong, especially with the actually summing in the first part.

2. May 2, 2010

### hgfalling

For x4 you put down 4/5 instead of 3/4. You wrote the right thing in the horizontal addition, but the sum you come up with is the sum including 4/5 instead of 3/4.

In the error part, your dx is 0.5, not 0.25.

3. May 3, 2010

### Asphyxiated

hmm i just mixed up on the post with the 4/5 instead of 3/4, thats what I have written on the paper but i just did the problem again and it all works out, i wonder why i couldnt get it to come out before.... and the error, yeah i got mixed up, thanks for the help!