# RAM (Rectangle Approximation Method/Riemann Sum)

Simple Riemann approximation using rectangles
6:45 — by Khan Academy

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• Not every function (and not every interesting and important function) has an antiderivative that is finitely expressible using the algebraic operations: addition, subtraction, multiplication, division and extraction of roots.

Two examples:

Natural Logarithm
${\int}_{1}^{b} \frac{1}{x} \mathrm{dx}$

The natural log must be approximated using some approximation technique -- by approximating the integral or by some series approximation.

Probablity and Statistics
The standard Normal (or Bell or Gaussian) curve

graph{e^(-1/2 x^2)/sqrt(2pi) [-2.398, 2.469, -0.55, 1.883]}

$\frac{1}{\sqrt{2 \pi}} {\int}_{0}^{z} {e}^{- \frac{1}{2} {x}^{2}} \mathrm{dx}$ gives the probability of a random variable having a standard normal value between $0$ and $z$

This integral cannot be expressed finitely using algebraic operations and must be approximated numerically. (As must ${e}^{x}$ itself.)

• This key question hasn't been answered yet.
• The interval $\left[1 , 2\right]$ is divided into 5 equal subintervals

$\left[1 , 1.2\right] , \left[1.2 , 1.4\right] , \left[1.4 , 1.6\right] , \left[1.6 , 1.8\right] , \mathmr{and} \left[1.8 , 2\right]$.

Each interval are of length $\Delta x = \frac{b - a}{n} = \frac{2 - 1}{5} = 0.2$.

The midpoints of the above subintervals are

$1.1 , 1.3 , 1.5 , 1.7 , \mathmr{and} 1.9$.

Using the above midpoints to determine the heights of the approximating rectangles, we have

${M}_{5} = \left[f \left(1.1\right) + f \left(1.3\right) + f \left(1.5\right) + f \left(1.7\right) + f \left(1.9\right)\right] \Delta x$

$= \left(\frac{1}{1.1} + \frac{1}{1.3} + \frac{1}{1.5} + \frac{1}{1.7} + \frac{1}{1.9}\right) \cdot 0.2 \approx 0.692$

By Midpoint Rule,

${\int}_{1}^{2} \frac{1}{x} \mathrm{dx} \approx 0.692$.

I hope that this was helpful.

• The area A of the region under the graph of $f$ above the $x$-axis from $x = a$ to $b$ can be found by

$A = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$,

where ${x}_{i} = a + i \Delta x$ and $\Delta x = \frac{b - a}{n}$.

Let us find the area of the region under the graph of $y = 2 x + 1$ from $x = 1$ to $3$.

By definition,

$A = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \left[2 \left(1 + \frac{2}{n} i\right) + 1\right] \frac{2}{n}$

by simplifying the expression inside the summation,

$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \left(\frac{8}{n} ^ 2 i + \frac{6}{n}\right)$

by splitting the summation and pulling out constants,

$= {\lim}_{n \to \infty} \left(\frac{8}{n} ^ 2 {\sum}_{i = 1}^{n} i + \frac{6}{n} {\sum}_{i = 1}^{n} 1\right)$

by the summation formulas ${\sum}_{i = 1}^{n} i = \frac{n \left(n + 1\right)}{2}$ and ${\sum}_{i = 1}^{n} 1 = n$,

$= {\lim}_{n \to \infty} \left(\frac{8}{n} ^ 2 \cdot \frac{n \left(n + 1\right)}{2} + \frac{6}{n} \cdot n\right)$

by cancelling out $n$'s,

$= {\lim}_{n \to \infty} \left[4 \left(1 + \frac{1}{n}\right) + 6\right] = 4 \left(1 + 0\right) + 6 = 10$

I hope that this was helpful.

• ${\int}_{1}^{3} {e}^{x} \mathrm{dx} \approx {R}_{n} = {\sum}_{i = 1}^{n} {e}^{1 + i \Delta x} \Delta x$,

where $\Delta x = \frac{3 - 1}{n} = \frac{2}{n}$

$= {\sum}_{i = 1}^{n} {e}^{1 + \frac{2}{n} i} \frac{2}{n}$

$= \frac{2 e}{n} {\sum}_{i = 1}^{n} {\left({e}^{\frac{2}{n}}\right)}^{i}$

$= \frac{2 e}{n} \frac{{e}^{\frac{2}{n}} \left[1 - {\left({e}^{\frac{2}{n}}\right)}^{n}\right]}{1 - {e}^{\frac{2}{n}}}$

$= \frac{2 {e}^{1 + \frac{2}{n}} \left(1 - {e}^{2}\right)}{n \left(1 - {e}^{\frac{2}{n}}\right)}$

you can continue from here.

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