Compute Riemann sums of a linear function,

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

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Question statement

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be the function $f(x)=\simplify{{a}*x+{b}}$ .

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			Name	Type	Generated Value

a	integer	-4
c	integer	1
b	integer	4
uppersum2	number	-153
d	integer	10
theint	number	-162
monotonic	string	a decreasing
n	integer	9
lowersum	number	-180
lowersum2	number	-171
type	string	decreasing
uppersum	number	-144
side	integer	0

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-4

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Compute the upper and lower sums of $f$ for the partition of the interval $[\var{c},\var{d}]$ into subintervals each of length 1:

\[\Delta = \{\var{c},\var{c+1},\;\ldots,\var{d}\}.\]

Note that $f$ is {monotonic} function over this interval.

Upper sum = ?Gap 0. (Input to 3 decimal places.)

Lower sum = ?Gap 1. (Input to 3 decimal places.)

Advice

{plotf(a,b,c,d)}

$f(x)$ is a linear function which is {monotonic} function. It follows that:

1. The lower sum is given by taking the area over each interval $[x_j,x_{j+1}]$ to be $f(x_{\simplify{j+{1-side}}})\times(x_{j+1}-x_j)$ as $f(x_{\simplify{j+{1-side}}})\lt f(x_{\simplify{j+{side}}})$

2. The upper sum is given by taking the area over each interval $[x_j,x_{j+1}]$ to be $f(x_{\simplify{j+{side}}})\times(x_{j+1}-x_j)$ as $f(x_{\simplify{j+{side}}})\gt f(x_{\simplify{j+{1-side}}})$

a) In this case, $x_j=j,\;\;j=\var{c},\var{c+1},\dots,\var{d}$ and so the interval length $x_{j+1}-x_{j}=1$ and so we have the for the lower sum:

Lower sum.

\[\begin{eqnarray}\mbox{LS}=\sum_{j=\var{c}}^{\var{d-1}}f(x_{\simplify{j+{1-side}}})=\sum_{j=\var{c}}^{\var{d-1}}\left(\simplify[all,!collectNumbers,!noLeadingMinus]{{a}*(j+{1-side})+{b}}\right)&=&{\var{a}}\times\sum_{j=\var{c}}^{\var{d-1}}(\simplify[all,!collectNumbers,!noLeadingMinus]{(j+{1-side}) })+\var{n*b}\\&=&\var{lowersum}\end{eqnarray}\]

We are using the arithmetic series sum: $\displaystyle \sum_{j=a}^b j=\frac{1}{2}(a+b)(b-a+1)$ so that $\sum_{j=\var{c}}^{\var{d-1}}(\simplify[all,!collectNumbers,!noLeadingMinus]{(j+{side}) })=\var{(c+d-1+2*side)*(d-c)/2}$.

Upper sum.

\[\begin{eqnarray}\mbox{US}=\sum_{j=\var{c}}^{\var{d-1}}f(x_{\simplify{j+{side}}})=\sum_{j=\var{c}}^{\var{d-1}}\left(\simplify[all,!collectNumbers,!noLeadingMinus]{{a}*(j+{side})+{b}}\right)&=&{\var{a}}\times\sum_{j=\var{c}}^{\var{d-1}}(\simplify[all,!collectNumbers,!noLeadingMinus]{(j+{side}) })+\var{n*b}\\&=&\var{uppersum}\end{eqnarray}\]

{riemann(a,b,c,d)}

You can use the slider above to see the effect of changing the number, $n$ of partitions between $\var{c}$ and $\var{d}$.

If we double the number of partitions them we have $x_j=\frac{j}{2},\;\;j=\var{2*c},\var{2*c+1},\ldots,\var{2*d}$.

Also we have to remember to multiply by the interval length $x_{j+1}-x_j=1/2$.

We have:

Lower sum.

\[\begin{eqnarray}\mbox{LS}=\sum_{j=\var{2*c}}^{\var{2*d-1}}f(x_{\simplify{j+{1-side}}})\times \frac{1}{2}=\sum_{j=\var{2*c}}^{\var{2*d-1}}\left(\simplify[all,!collectNumbers,!noLeadingMinus]{{a}*((j+{1-side})/2)+{b}}\right)\times \frac{1}{2}&=&\frac{\var{a}}{4}\times\sum_{j=\var{2*c}}^{\var{2*d-1}}\left(\simplify[all,!collectNumbers,!noLeadingMinus]{(j+{(1-side)}) }\right)+\var{n*b}\\&=&\var{lowersum2}\end{eqnarray}\]

Upper sum.

\[\begin{eqnarray}\mbox{US}=\sum_{j=\var{2*c}}^{\var{2*d-1}}f(x_{\simplify{j+{side}}})\times \frac{1}{2}=\sum_{j=\var{2*c}}^{\var{2*d-1}}\left(\simplify[all,!collectNumbers,!noLeadingMinus]{{a}*((j+{side})/2)+{b}}\right)\times \frac{1}{2}&=&\frac{\var{a}}{4}\times\sum_{j=\var{2*c}}^{\var{2*d-1}}\left(\simplify[all,!collectNumbers,!noLeadingMinus]{(j+{side}) }\right)+\var{n*b}\\&=&\var{uppersum2}\end{eqnarray}\]

True value of the integral is:

\[\int_{\var{c}}^{\var{d}}\simplify{{a}*x+{b}}\;dx=\left[\simplify{{a}/2*x^2+{b}*x}\right]_{\var{c}}^{\var{d}}=\var{theint}\]

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Compute Riemann sums of a linear function,

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

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