For the function $2$

$1)($

x

xf $=,$ find the US and LS in the interval $[1,2]$ for

n $=100.$

Let$$s first get the function defined, since it is not a function in the Matlab library.

$>>$ f$=$inline$(1./$x$.^2)$

The dot before the operation symbols tells Matlab to generate a function that can be

evaluated on arrays, in particular vectors.

Next, use the graphing capabilities of Matlab to graph this function on this interval $([1,$

$2]).$ As you can see when you do so$,$ because our function is decreasing in any

subinterval the maximum is always on the left end and the minimum in the right end.

Let$$s first define the $\backslash$Delta x $.$

$>>$ deltaX $=100/)12($ ;

Now generate the vector with the left ends of the intervals. This will be used to give us

the US$.$

$>>$ xu $=$ linspace $$ deltaX $)$;$100,2,1($

Now the vector with the right ends of the intervals fro the LS

$>>$ xl $=$ linspace $+$ deltaX $)$;$100,2,1($

Finally a simple command gives us the Riemann sums.

$>>=$ deltaXUS xufsum $))((*$

MATH $2414$ Lab $1$ Matlab Version Page $3$of $4$

$>>=$ deltaXLS xlfsum $))((*$

Remember that we know now that

$<=<=$ USALS

If we define our approximation as the middle point $2$

LSUS Aest

$+=,$ then the largest error

we could possibly be making is given by $2$

LSUS

error $=,$ what means that

est $||<=$ errorAA

Now we can compute the estimation of the area and the error

$>>$ Estimate $+=$ LSUS $2/)($

$>>=$ LSUSerror $2/)($

Now find an n that is a power of $10$ such that the error in the estimation of the area is

less than or equal to $310001\mathrm{.}0=.$