$L_n={\displaystyle \underset{i=0}{\overset{n-1}{}}}f(x_i)x$

or the right end$-$point approximation

$R_n={\displaystyle \underset{i=1}{\overset{n}{}}}f(x_i)x$

There are functions which have no anti$-$derivatives and thus one can't use the Fundamental Theorem of Calculus and instead must be estimated.

Hopefully you're all aware that increasing $n$ which is the number of rectangles in the estimate usually increases the accuracy.

The idea is to increase $$ until the difference between the two approximations is less than some tolerance which we call $lon.$ For example, imagine that

the estimate for $n_1$ and $n_2$ rectangles are respectively $R_{n_1}=5\mathrm{.}001$ and $R_{n_2}=5\mathrm{.}002$ the difference between the two errors is

$|R_{n_2}-R_{n_1}|=0\mathrm{.}001$

if the difference of the errors is less than the tolerance $lon$ then we stop increasing the number of rectangles and use the last estimate as the estimating

value.

Program

You will create a program which estimates the definite integral of $f(x)$ between $x=$a and $x=b$ using either the left$-$end point or right$-$end point

approximation. The program will increase $n$ until the difference of the errors is less than the given tolerance.

Use the header for some of the functions.

$f(x)=x^2+3x+6$ between $x=1$ and $x=5$ with $lon=0\mathrm{.}2.$ What was the value of $n?$ Compare to the actual value.

$f(x)=cos(3x)$ between $x=0$ and $x=0\mathrm{.}5$ with $lon=0\mathrm{.}1.$ What was the value of $n?$ Compare to the actual value.

$f(x)=3^{2x+1}$ between $x=4$ and $x=5$ with $lon=0\mathrm{.}02.$ What was the value of $n?$ Compare to the actual value. cpp$!$