Trapazoidal Approximation help

[Asphyxiated]

Homework Statement

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

using dx = .5

Homework 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

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.

 

[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.

 

[Asphyxiated]

hmm i just mixed up on the post with the 4/5 instead of 3/4, that's what I have written on the paper but i just did the problem again and it all works out, i wonder why i couldn't get it to come out before... and the error, yeah i got mixed up, thanks for the help!