Dale's questions via Facebook about Riemann Sums

In summary, we have found that the area under the curve $\displaystyle \begin{align*} g(x) = x^3 \end{align*}$ between x = 0 and x = b can be approximated by dividing the interval into n rectangles and taking the sum of their areas. This can be written as $\displaystyle \begin{align*} A_i = g \left( x_i \right) \Delta x \end{align*}$ where $\displaystyle \begin{align*} x_i = \left( 2\,i - 1 \right) \frac{b}{2\,n} \end{align*}$ and $\displaystyle \begin{align*
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(a) Since a = 2, that means $\displaystyle \begin{align*} \Delta x = \frac{2}{5} \end{align*}$

(b)
$\displaystyle \begin{align*} f \left( \frac{7\,\Delta x}{2} \right) &= f \left( \frac{7}{5} \right) \\ &= \left( \frac{7}{5} \right) ^2 \\ &= \frac{49}{25} \end{align*}$

(c)
$\displaystyle \begin{align*} A_4 &= f \left( \frac{7}{5} \right) \cdot \Delta x \\ &= \frac{49}{25} \cdot \frac{2}{5} \\ &= \frac{98}{125} \,\textrm{units}^2 \end{align*}$

(d)
$\displaystyle \begin{align*} A_i &= f \left( x_i \right) \cdot \Delta x \end{align*}$

(e)
$\displaystyle \begin{align*} S_5 &= \sum_{i = 1}^5{ \left[ f \left( x_i \right) \cdot \Delta x \right] } \\ &= f \left( x_1 \right) \cdot \Delta x + f \left( x_2 \right) \cdot \Delta x + f \left( x_3 \right) \cdot \Delta x + f \left( x_4 \right) \cdot \Delta x + f \left( x_5 \right) \cdot \Delta x \\ &= \left( \frac{1}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{3}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{5}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{7}{5} \right) ^2 \cdot \frac{2}{5} + \left( \frac{9}{5} \right) ^2 \cdot \frac{2}{5} \\ &= \frac{1}{25} \cdot \frac{2}{5} + \frac{9}{25} \cdot \frac{2}{5} + \frac{25}{25} \cdot \frac{2}{5} + \frac{49}{25} \cdot \frac{2}{5} + \frac{81}{25} \cdot \frac{2}{5} \\ &= \frac{2}{125} + \frac{18}{125} + \frac{50}{125} + \frac{98}{125} + \frac{162}{125} \\ &= \frac{330}{125} \\ &= \frac{66}{25} \,\textrm{units}^2 \end{align*}$

(f)
$\displaystyle \begin{align*} A &= \int_0^a{f\left( x \right) \,\mathrm{d}x} \\ &= \int_0^2{ x^2\,\mathrm{d}x } \\ &= \left[ \frac{x^3}{3} \right] _0^2 \\ &= \frac{2^3}{3} - \frac{0^3}{3} \\ &= \frac{8}{3} \,\textrm{units}^2 \end{align*}$

(g)
$\displaystyle \begin{align*} \textrm{Error} &= \frac{66}{25} - \frac{8}{3} \\ &= \frac{198}{75} - \frac{200}{75} \\ &= -\frac{2}{75} \end{align*}$

So the percentage error is

$\displaystyle \begin{align*} \frac{-\frac{2}{75}}{\frac{8}{3}} \cdot 100\% &= -\frac{2}{75} \cdot \frac{3}{8} \cdot 100\% \\ &= -\frac{600\%}{600} \\ &= -1\% \end{align*}$
 

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(a) Since the distance along the x-axis is "b" and it is being divided into "n" rectangles, that means $\displaystyle \begin{align*} \Delta x = \frac{b}{n} \end{align*}$.

(b) Each interval can be written as $\displaystyle \begin{align*} \left[ \left( i - 1 \right) \Delta x , i\,\Delta x \right] \end{align*}$, so that means the midpoint of each interval is

$\displaystyle \begin{align*} x_i &= \frac{\left( i - 1 \right) \Delta x + i\,\Delta x}{2} \\ &= \frac{\left( i - 1 + i \right) \Delta x}{2} \\ &= \frac{\left( 2\,i - 1 \right) \Delta x}{2} \end{align*}$

View attachment 7144

(c) Since we have found $\displaystyle \begin{align*} \Delta x = \frac{b}{n} \end{align*}$ that means

$\displaystyle \begin{align*} x_i &= \frac{\left( 2\,i - 1 \right) \Delta x}{2} \\ &= \frac{\left( 2\,i - 1 \right) \frac{b}{n}}{2} \\ &= \left( 2\,i - 1 \right) \frac{b}{2\,n} \end{align*}$

(d) The length of each rectangle is $\displaystyle \begin{align*} \Delta x \end{align*}$ and the width is $\displaystyle \begin{align*} g \left( x_i \right) \end{align*}$ so the area of each rectangle is $\displaystyle \begin{align*} A_i = g \left( x_i \right) \Delta x \end{align*}$

(e) Since $\displaystyle \begin{align*} x_i = \left( 2\,i - 1 \right) \frac{b}{2\,n} \end{align*}$ that means $\displaystyle \begin{align*} g \left( x_i \right) = g \left( \left( 2\,i - 1 \right) \frac{b}{2\,n} \right) \end{align*}$ and $\displaystyle \begin{align*} \Delta x = \frac{b}{n} \end{align*}$ that means $\displaystyle \begin{align*} g \left( \left( 2\,i - 1 \right) \frac{b}{2\,n} \right) \frac{b}{n} = g \left( x_i \right) \Delta x \end{align*}$, so

$\displaystyle \begin{align*} \sum_{i = 1}^n{ \left[ g\left( \left( 2\,i - 1 \right) \frac{b}{2\,n}\right) \Delta x \right] } &= \sum_{i = 1}^n{ \left[ g \left( x_i \right) \Delta x \right] } \\ &= \sum_{i = 1}^n{A_i} \end{align*}$

(f) (i) If $\displaystyle \begin{align*} g \left( x \right) = x^3 \end{align*}$, that means

$\displaystyle \begin{align*} \sum_{i = 1}^n{ A_i } &= \sum_{i = 1}^n{ \left[ g \left( \left( 2\,i - 1 \right) \frac{b}{2\,n} \right) \Delta x \right] } \\ &= \sum_{i = 1}^n{ \left\{ \left[ \left( 2\,i - 1 \right) \frac{b}{2\,n} \right] ^3 \,\frac{b}{n} \right\} } \end{align*}$

so when $\displaystyle \begin{align*} n = 100 \end{align*}$ we have

$\displaystyle \begin{align*} \sum_{i = 1}^{100}{ \left\{ \left[ \left( 2\,i - 1 \right) \frac{b}{100}\right] ^3\,\frac{b}{100} \right\} } &= \frac{19\,999\,b^4}{80\,000} \end{align*}$

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(ii)
$\displaystyle \begin{align*} A &= \lim_{n \to \infty}{ \left\{ \left[ \left( 2\,i - 1 \right) \frac{b}{n} \right] ^3 \,\frac{b}{n} \right\} } \\ &= \frac{b^4}{4} \end{align*}$

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View attachment 7145(g) The exact area under the curve $\displaystyle \begin{align*} g(x) = x^3 \end{align*}$ between x = 0 and x = b is

$\displaystyle \begin{align*} A &= \int_0^b{x^3\,\mathrm{d}x} &= \left[ \frac{x^4}{4} \right] _0^b \\ &= \frac{b^4}{4} - \frac{0^4}{4} \\ &= \frac{b^4}{4} \,\textrm{units}^2 \end{align*}$

(i) so the error in using 100 rectangles for the approximation is

$\displaystyle \begin{align*} \frac{b^4}{4} - \frac{19\,999\,b^4}{80\,000} &= \frac{20\,000\,b^4}{80\,000} - \frac{19\,999\,b^4}{80\,000} \\ &= \frac{b^4}{80\,000} \end{align*}$

and so the percentage error is

$\displaystyle \begin{align*} \frac{\frac{b^4}{80\,000}}{\frac{b^4}{4}} \cdot 100\% &= \frac{b^4}{80\,000} \cdot \frac{4}{b^4} \cdot 100\% \\ &= \frac{100\%}{20\,000} \\ &= 0.005\% \end{align*}$

(ii) and as the limiting value is exactly equal to the integral value, the percentage error is 0%.
 

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FAQ: Dale's questions via Facebook about Riemann Sums

What is a Riemann Sum?

A Riemann Sum is a method used in calculus to approximate the area under a curve. It involves dividing the curve into smaller rectangles and finding the sum of their areas.

How do you calculate a Riemann Sum?

To calculate a Riemann Sum, you need to first divide the curve into smaller rectangles with equal widths. Then, find the height of each rectangle by plugging in a point on the rectangle's base into the function. Finally, multiply the width and height of each rectangle and add them together to find the total area.

What is the difference between a left, right, and midpoint Riemann Sum?

In a left Riemann Sum, the height of each rectangle is determined by using the left endpoint of the base. In a right Riemann Sum, the height is determined by using the right endpoint of the base. In a midpoint Riemann Sum, the height is determined by using the midpoint of the base. The choice of which endpoint or midpoint to use can affect the accuracy of the approximation.

What is the purpose of using a Riemann Sum?

A Riemann Sum is used to approximate the area under a curve when it is not possible to find the exact area using traditional methods. It can also be used to find the area under a curve when the function is not continuous or when the curve is not in a standard shape (i.e. a circle or triangle).

How can Riemann Sums be applied in real life?

Riemann Sums have many real-life applications, such as in engineering, physics, and economics. For example, they can be used to approximate the volume of irregularly shaped objects or to estimate the total profit of a business over a period of time. They are also used in computer algorithms to solve optimization problems.

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