Trapezium Rule

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November 13th 2010, 03:25 PM
Hellbent

Trapezium Rule

Hi,

A little help required and some checking.

The trapezium rule, with 2 intervals of equal width , is to be used to find an approximate value for $\int_0^1 e^{-x}\,dx$ .
Explain, with the aid of a sketch, why the approximate value will be greater than the exact value of the integral.

My graphs are different from Wolfram's. Here they are: e^-x and -e^-x
EXPLANATION PART: I don't know this part. Help required.

Calculate the approximate and exact values respectively, to 3 d.p.

$h = \frac{1-0}{2} = 0.5$

$\begin{vmatrix} x_r & 0 & 0.5 & 1.0 \\y_r & 1 & 0.607 & 0.368 \end{vmatrix}$

$\int_0^1 e^{-x}\,dx = \frac{1}{2}\cdot 0.5(1+2(0.607)+0.368)<br /> = 0.645$

$\int_0^1 e^{-x}\,dx = \bigg[-e^{-x}\bigg]_0^1$

$=[-0.368] - [-1]$

$= 0.632$

Another approximation to $\int_0^1 e^{-x}$ , is to be calculated by using two trapezia of unequal width. The first trapezium has width $h$ and the second has width $(1 - h)$ , so that the three ordinates are at $x = 0$ , $x = h$ and $x = 1$ . Show that the total area $T$ of these two trapezia is given by $T = \frac{1}{2}(e^{-1} + h(1 - e^{-1})+e^{-k})$ .

Width $= h$ and $(1 - h)$

$x = 0$ , $x = h$ , $x = 1$

$T=\frac{1}{2}h(e^0+e^{-h})+(1-h)(e^{-h}+e^{-1})$

$=\frac{1}{2}{h+e^{-h}+e^{-1} - he^{-1}}$

$T= \frac{1}{2}{e^{-1}+h(1-e^{-1})+e^{-h}}$

Show that the value of $h$ for which $T$ is a minimum is given by $h = \ln\bigg(\frac{e}{e-1}\bigg)$ .

$\frac{dT}{dh} = \frac{1}{2}(1-e^{-1})-e^{-h}$

$0 = 1 - e^{-1} - e^{-h}$

$-1 + e^{-1} = -e^{-h}$

$-\ln\,1- \frac{1}{e} = h$

$-\ln\,\frac{e-1}{e} = h$

$\ln\bigg(\frac{e}{e-1}\bigg) = h$
November 13th 2010, 03:35 PM
Prove It

Quote:

Originally Posted by Hellbent

Hi,

A little help required and some checking.

The trapezium rule, with 2 intervals of equal width , is to be used to find an approximate value for $\int_0^1 e^{-x}\,dx$ .
Explain, with the aid of a sketch, why the approximate value will be greater than the exact value of the integral.

My graphs are different from Wolfram's. Here they are: e^-x and -e^-x
EXPLANATION PART: I don't know this part. Help required.

Calculate the approximate and exact values respectively, to 3 d.p.

$h = \frac{1-0}{2} = 0.5$

$\begin{vmatrix} x_r & 0 & 0.5 & 1.0 \\y_r & 1 & 0.607 & 0.368 \end{vmatrix}$

$\int_0^1 e^{-x}\,dx = \frac{1}{2}\cdot 0.5(1+2(0.607)+0.368)<br /> = 0.645$

$\int_0^1 e^{-x}\,dx = \bigg[-e^{-x}\bigg]_0^1$

$=[-0.368] - [-1]$

$= 0.632$

Another approximation to $\int_0^1 e^{-x}$ , is to be calculated by using two trapezia of unequal width. The first trapezium has width $h$ and the second has width $(1 - h)$ , so that the three ordinates are at $x = 0$ , $x = h$ and $x = 1$ . Show that the total area $T$ of these two trapezia is given by $T = \frac{1}{2}(e^{-1} + h(1 - e^{-1})+e^{-k})$ .

Width $= h$ and $(1 - h)$

$x = 0$ , $x = h$ , $x = 1$

$T=\frac{1}{2}h(e^0+e^{-h})+(1-h)(e^{-h}+e^{-1})$

$=\frac{1}{2}{h+e^{-h}+e^{-1} - he^{-1}}$

$T= \frac{1}{2}{e^{-1}+h(1-e^{-1})+e^{-h}}$

Show that the value of $h$ for which $T$ is a minimum is given by $h = \ln\bigg(\frac{e}{e-1}\bigg)$ .

$\frac{dT}{dh} = \frac{1}{2}(1-e^{-1})-e^{-h}$

$0 = 1 - e^{-1} - e^{-h}$

$-1 + e^{-1} = -e^{-h}$

$-\ln\,1- \frac{1}{e} = h$

$-\ln\,\frac{e-1}{e} = h$

$\ln\bigg(\frac{e}{e-1}\bigg) = h$

1. Have you tried drawing the graph? I would say it's because the function $\displaystyle e^{-x}$ is convex (check by showing that the second derivative is always positive), so all chords will lie above the graph.
November 14th 2010, 01:27 AM
Hellbent

Yes, I did draw the graphs. The ones I gave are Wolfram's.
The graphs I drew are probably okay, so forget that part. Hehe :)
November 14th 2010, 01:30 AM
Hellbent

Delete.
November 14th 2010, 10:24 AM
Hellbent

Is the rest of what I did okay?