
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 $\displaystyle \int_0^1 e^{x}\,dx$.
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.
$\displaystyle h = \frac{10}{2} = 0.5$
$\displaystyle \begin{vmatrix} x_r & 0 & 0.5 & 1.0 \\y_r & 1 & 0.607 & 0.368 \end{vmatrix}$
$\displaystyle \int_0^1 e^{x}\,dx = \frac{1}{2}\cdot 0.5(1+2(0.607)+0.368)
= 0.645$
$\displaystyle \int_0^1 e^{x}\,dx = \bigg[e^{x}\bigg]_0^1$
$\displaystyle =[0.368]  [1]$
$\displaystyle = 0.632$
Another approximation to $\displaystyle \int_0^1 e^{x}$, is to be calculated by using two trapezia of unequal width. The first trapezium has width $\displaystyle h$ and the second has width $\displaystyle (1  h)$, so that the three ordinates are at $\displaystyle x = 0$, $\displaystyle x = h$ and $\displaystyle x = 1$. Show that the total area $\displaystyle T$ of these two trapezia is given by $\displaystyle T = \frac{1}{2}(e^{1} + h(1  e^{1})+e^{k})$.
Width $\displaystyle = h$ and $\displaystyle (1  h)$
$\displaystyle x = 0$, $\displaystyle x = h$, $\displaystyle x = 1$
$\displaystyle T=\frac{1}{2}h(e^0+e^{h})+(1h)(e^{h}+e^{1})$
$\displaystyle =\frac{1}{2}{h+e^{h}+e^{1}  he^{1}}$
$\displaystyle T= \frac{1}{2}{e^{1}+h(1e^{1})+e^{h}}$
Show that the value of $\displaystyle h$ for which $\displaystyle T$ is a minimum is given by $\displaystyle h = \ln\bigg(\frac{e}{e1}\bigg)$.
$\displaystyle \frac{dT}{dh} = \frac{1}{2}(1e^{1})e^{h}$
$\displaystyle 0 = 1  e^{1}  e^{h}$
$\displaystyle 1 + e^{1} = e^{h}$
$\displaystyle \ln\,1 \frac{1}{e} = h$
$\displaystyle \ln\,\frac{e1}{e} = h$
$\displaystyle \ln\bigg(\frac{e}{e1}\bigg) = h$

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 $\displaystyle \int_0^1 e^{x}\,dx$.
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.
$\displaystyle h = \frac{10}{2} = 0.5$
$\displaystyle \begin{vmatrix} x_r & 0 & 0.5 & 1.0 \\y_r & 1 & 0.607 & 0.368 \end{vmatrix}$
$\displaystyle \int_0^1 e^{x}\,dx = \frac{1}{2}\cdot 0.5(1+2(0.607)+0.368)
= 0.645$
$\displaystyle \int_0^1 e^{x}\,dx = \bigg[e^{x}\bigg]_0^1$
$\displaystyle =[0.368]  [1]$
$\displaystyle = 0.632$
Another approximation to $\displaystyle \int_0^1 e^{x}$, is to be calculated by using two trapezia of unequal width. The first trapezium has width $\displaystyle h$ and the second has width $\displaystyle (1  h)$, so that the three ordinates are at $\displaystyle x = 0$, $\displaystyle x = h$ and $\displaystyle x = 1$. Show that the total area $\displaystyle T$ of these two trapezia is given by $\displaystyle T = \frac{1}{2}(e^{1} + h(1  e^{1})+e^{k})$.
Width $\displaystyle = h$ and $\displaystyle (1  h)$
$\displaystyle x = 0$, $\displaystyle x = h$, $\displaystyle x = 1$
$\displaystyle T=\frac{1}{2}h(e^0+e^{h})+(1h)(e^{h}+e^{1})$
$\displaystyle =\frac{1}{2}{h+e^{h}+e^{1}  he^{1}}$
$\displaystyle T= \frac{1}{2}{e^{1}+h(1e^{1})+e^{h}}$
Show that the value of $\displaystyle h$ for which $\displaystyle T$ is a minimum is given by $\displaystyle h = \ln\bigg(\frac{e}{e1}\bigg)$.
$\displaystyle \frac{dT}{dh} = \frac{1}{2}(1e^{1})e^{h}$
$\displaystyle 0 = 1  e^{1}  e^{h}$
$\displaystyle 1 + e^{1} = e^{h}$
$\displaystyle \ln\,1 \frac{1}{e} = h$
$\displaystyle \ln\,\frac{e1}{e} = h$
$\displaystyle \ln\bigg(\frac{e}{e1}\bigg) = h$
1. Have you tried drawing the graph? I would say it's because the function $\displaystyle \displaystyle e^{x}$ is convex (check by showing that the second derivative is always positive), so all chords will lie above the graph.

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 :)


Is the rest of what I did okay?