Compute Riemann sums on an interval

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

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Name	Status	Author	Last Modified
Compute Riemann sums on an interval	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:51
Luis's copy of Compute Riemann sums on an interval	draft	Luis Hernandez	30/11/2018 19:09

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Question statement

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be the function $f(x)=\simplify{x^2+{a}*x+{b}}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	7
tuppersum	number	161.71875
c	integer	5
b	integer	5
uppersum2	number	157.930
theint	number	154.167
n	integer	20
lowersum	number	146.719
lowersum2	number	150.430
tlowersum	number	146.71875
tol	number	0.001
uppersum	number	161.719

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Generated value: `integer`

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lowersum2
theint
tlowersum
tuppersum
uppersum2

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Ask the student a question, and give any hints about how they should answer this part.

Find $a$ such that $f$ is an increasing function for $x \ge a$ and a decreasing function for $x \le a$.

$a=\;$Unnamed gap (input as a decimal).

Advice

$f(x)$ is a quadratic and its stationary point is given by $\frac{df}{dx}=0$ which after solving gives the minimum point $x=-\simplify[all]{{a}/2}$.

{plotf(a,b,c)}

We see that the function is increasing in the range $[0,\var{c}]$ and it follows that the lower sums are given by taking the value of $f$ on the left hand side of each interval, multiplying by the interval length and summing up over all intervals. The upper sums by taking the right hand side of each interval etc.

The following diagrams illustrate this. Move the slider to see the effect on the upper and lower sums by varying the number $n$ of subintervals.

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

Finding the Upper and Lower sums Directly.

First you have to know the following sums of sequences:

\[\begin{eqnarray*}\sum_{r=1}^m r &=& \frac{m(m+1)}{2}\\ \sum_{r=1}^{m} r^2&=&\frac{m(m+1)(2m+1)}{6}\end{eqnarray*}\]

Also the interval points between $0$ and $\var{c}$ are given by $x_j=\simplify[all,fractionNumbers]{({c}/{n})j},\;j=0,\dots,\var{n}$

For the lower sums, each rectangle with base the interval $[x_r,x_{r+1}]$ contributes an area $f(x_r)(x_{r+1}-x_r)=\simplify[all,fractionNumbers]{f({c}/{n}*r)*({c}/{n})},\;r=0,\ldots,\var{n-1}$.

We sum from $r=0$ to $r=\var{n-1}$ as we have to take the left value of each interval.

So the lower sum is:

\[\begin{eqnarray}\mbox{LS}=\sum_{r=0}^{\var{n-1}}\simplify[all,fractionNumbers]{f(({c}/{n})*r)*({c}/{n})}&=&\sum_{r=0}^{\var{n-1}}\simplify[all,fractionNumbers,!collectNumbers]{((({c}/{n})*r)^2+{a}*(({c}/{n})*r)+{b})*({c}/{n})}\\
&=&\sum_{r=0}^{\var{n-1}}\simplify[all,fractionNumbers,!collectNumbers]{({c}/{n})^3*r^2}+\var{a}\sum_{r=0}^{\var{n-1}}\simplify[all,fractionNumbers,!collectNumbers]{({c}/{n})^2*r}+\var{b}\sum_{r=0}^{\var{n-1}}\simplify[all,fractionNumbers,!collectNumbers]{({c}/{n})}\\
&=&\simplify[all,fractionNumbers]{({c}/{n})^3}\sum_{r=0}^{\var{n-1}}r^2+\var{a}\times\simplify[all,fractionNumbers]{({c}/{n})^2}\sum_{r=0}^{\var{n-1}}r+\var{b}\times\var{c}\end{eqnarray}\]

Using \[\sum_{r=0}^{\var{n-1}}r=\frac{\var{n-1}\times \var{n}}{2}=\var{(n-1)*n/2},\;\;\;\sum_{r=0}^{\var{n-1}}r^2=\frac{\var{n-1}\times \var{n}\times \var{2*n-1}}{6}=\var{(n-1)*n*(2*n-1)/6}\]

and substituting into the above equation we obtain:

LS = $\var{tlowersum}=\var{lowersum}$ to 3 decimal places.

As for the upper sum we have $\mbox{US}=\mbox{LS}+(f(\var{c})-f(0))\times\simplify[all,fractionNumbers]{{c}/{n}}=\mbox{LS}+(\var{c^2+a*c+b}-\var{b})\times \simplify[all,fractionNumbers]{{c}/{n}}=\var{tlowersum}+\var{tuppersum-tlowersum}=\var{uppersum}$ to 3 decimal places .

Or if we take the more long-winded approach using the same method as for LS we have:

\[\begin{eqnarray}\mbox{US}=\sum_{r=1}^{\var{n}}\simplify[all,fractionNumbers]{f(({c}/{n})*r)*({c}/{n})}&=&\sum_{r=1}^{\var{n}}\simplify[all,fractionNumbers,!collectNumbers]{((({c}/{n})*r)^2+{a}*(({c}/{n})*r)+{b})*({c}/{n})}\\&=&\sum_{r=1}^{\var{n}}\simplify[all,fractionNumbers,!collectNumbers]{({c}/{n})^3*r^2}+\var{a}\sum_{r=1}^{\var{n}}\simplify[all,fractionNumbers,!collectNumbers]{({c}/{n})^2*r}+\var{b}\sum_{r=1}^{\var{n}}\simplify[all,fractionNumbers,!collectNumbers]{({c}/{n})}\\&=&\simplify[all,fractionNumbers]{({c}/{n})^3}\sum_{r=1}^{\var{n}}r^2+\var{a}\times\simplify[all,fractionNumbers]{({c}/{n})^2}\sum_{r=1}^{\var{n}}r+\var{b}\times\var{c}\\&=&\var{tuppersum}=\var{uppersum}\end{eqnarray}\]

to 3 decimal places.

The true value is given by the integral

\[\int_{0}^{\var{c}} \simplify[all]{x^2+{a}*x+{b}}\;dx=\left[\simplify[all]{x^3/3+{a}/2*x^2+{b}*x}\right]_{0}^{\var{c}}=\var{theint}\] to 3 decimal places.

For $\var{2*n}$ intervals we obtain:

$\mbox{LS}=\var{lowersum2}$.

$\mbox{US}=\var{uppersum2}$.

Both to 3 decimal places.

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Compute Riemann sums on an interval

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Taxonomy: mathcentre

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`