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.

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

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

$\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 $\displaystyle (1 - h)$

$\displaystyle x = 0$, $\displaystyle x = h$, $\displaystyle x = 1$

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

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

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

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

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

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

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

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

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

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