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Riemann sums in summation notation

AP.CALC:
LIM‑5 (EU)
,
LIM‑5.B (LO)
,
LIM‑5.B.2 (EK)
Summation notation can be used to write Riemann sums in a compact way. This is a challenging, yet important step towards a formal definition of the definite integral.
Summation notation (or sigma notation) allows us to write a long sum in a single expression. While summation notation has many uses throughout math (and specifically calculus), we want to focus on how we can use it to write Riemann sums.

Example of writing a Riemann sum in summation notation

Imagine we are approximating the area under the graph of f, left parenthesis, x, right parenthesis, equals, square root of, x, end square root between x, equals, 0, point, 5 and x, equals, 3, point, 5.
Function y = the square root of x is graphed. The x-axis goes from 0 to 4. The graph is a curve. The curve starts at (0, 0), moves upward concave down, and ends at (4, 2). The region between the curve and the x-axis, between x = 0.5 and x = 3.5, is shaded.
And say we decide to do that by writing the expression for a right Riemann sum with four equal subdivisions, using summation notation.
The graph of function y has the shaded region divided into 4 rectangles of width 0.75. Each rectangle touches the curve at the top right corner.
Let A, left parenthesis, i, right parenthesis denote the area of the i, start superscript, start text, t, h, end text, end superscript rectangle in our approximation.
The area of the rectangles are A of 1, A of 2, A of 3, and A of 4.
The entire Riemann sum can be written as follows:
A, left parenthesis, 1, right parenthesis, plus, A, left parenthesis, 2, right parenthesis, plus, A, left parenthesis, 3, right parenthesis, plus, A, left parenthesis, 4, right parenthesis, equals, sum, start subscript, i, equals, 1, end subscript, start superscript, 4, end superscript, A, left parenthesis, i, right parenthesis
What we need to do now is find the expression for A, left parenthesis, i, right parenthesis.
The width of the entire interval open bracket, 0, point, 5, comma, 3, point, 5, close bracket is 3 units and we want 4 equal subdivisions, so the start color #1fab54, start text, w, i, d, t, h, end text, end color #1fab54 of each rectangle is 3, divided by, 4, equals, start color #1fab54, 0, point, 75, end color #1fab54 units.
The start color #e07d10, start text, h, e, i, g, h, t, end text, end color #e07d10 of each rectangle is the value of f at the right endpoint of the rectangle (because this is a right Riemann sum).
Let start color #11accd, x, start subscript, i, end subscript, end color #11accd denote the right endpoint of the i, start superscript, start text, t, h, end text, end superscript rectangle. To find x, start subscript, i, end subscript for any value of i, we start at x, equals, 0, point, 5 (the left endpoint of the interval) and add the common width start color #1fab54, 0, point, 75, end color #1fab54 repeatedly.
The left side of the first rectangle is at x = 0.5. Add 0.75 4 times to get the sides of the rectangles, at x sub 1 to x sub 4.
Therefore, the formula of start color #11accd, x, start subscript, i, end subscript, end color #11accd is start color #11accd, 0, point, 5, plus, 0, point, 75, i, end color #11accd. Now, the start color #e07d10, start text, h, e, i, g, h, t, end text, end color #e07d10 of each rectangle is the value of f at its right endpoint:
start color #e07d10, f, left parenthesis, end color #e07d10, start color #11accd, x, start subscript, i, end subscript, end color #11accd, start color #e07d10, right parenthesis, end color #e07d10, equals, start color #e07d10, square root of, start color #11accd, x, start subscript, i, end subscript, end color #11accd, end square root, end color #e07d10, equals, start color #e07d10, square root of, start color #11accd, 0, point, 5, plus, 0, point, 75, i, end color #11accd, end square root, end color #e07d10
And so we've arrived at a general expression for the area of the i, start superscript, start text, t, h, end text, end superscript rectangle:
A(i)=widthheight=0.750.5+0.75i\begin{aligned} A(i)&=\greenD{\text{width}}\cdot\goldD{\text{height}} \\\\ &=\greenD{0.75}\cdot\goldD{\sqrt{\blueD{0.5+0.75i}}} \end{aligned}
Now all we have left is to sum this expression for values of i from 1 to 4:
=A(1)+A(2)+A(3)+A(4)=i=14A(i)=i=140.750.5+0.75i\begin{aligned} &\phantom{=}A(1)+A(2)+A(3)+A(4) \\\\ &=\displaystyle\sum_{i=1}^4 A(i) \\\\ &=\displaystyle\sum_{i=1}^4 0.75\cdot\sqrt{0.5+0.75i} \end{aligned}
And we're done!

Summarizing the process of writing a Riemann sum in summation notation

Imagine we want to approximate the area under the graph of f over the interval open bracket, a, comma, b, close bracket with n equal subdivisions.
Define delta, x: Let start color #1fab54, delta, x, end color #1fab54 denote the start color #1fab54, start text, w, i, d, t, h, end text, end color #1fab54 of each rectangle, then start color #1fab54, delta, x, equals, start fraction, b, minus, a, divided by, n, end fraction, end color #1fab54.
Define x, start subscript, i, end subscript: Let start color #11accd, x, start subscript, i, end subscript, end color #11accd denote the right endpoint of each rectangle, then start color #11accd, x, start subscript, i, end subscript, equals, a, plus, delta, x, dot, i, end color #11accd.
Define area of i, start superscript, start text, t, h, end text, end superscript rectangle: The start color #e07d10, start text, h, e, i, g, h, t, end text, end color #e07d10 of each rectangle is then start color #e07d10, f, left parenthesis, start color #11accd, x, start subscript, i, end subscript, end color #11accd, right parenthesis, end color #e07d10, and the area of each rectangle is start color #1fab54, delta, x, end color #1fab54, dot, start color #e07d10, f, left parenthesis, start color #11accd, x, start subscript, i, end subscript, end color #11accd, right parenthesis, end color #e07d10.
Sum the rectangles: Now we use summation notation to add all the areas. The values we use for i are different for left and right Riemann sums:
  • When we are writing a right Riemann sum, we will take values of i from 1 to n.
  • However, when we are writing a left Riemann sum, we will take values of i from 0 to n, minus, 1 (these will give us the value of f at the left endpoint of each rectangle).
Left Riemann sumRight Riemann sum
sum, start subscript, i, equals, 0, end subscript, start superscript, n, minus, 1, end superscript, start color #1fab54, delta, x, end color #1fab54, dot, start color #e07d10, f, left parenthesis, start color #11accd, x, start subscript, i, end subscript, end color #11accd, right parenthesis, end color #e07d10sum, start subscript, i, equals, 1, end subscript, start superscript, n, end superscript, start color #1fab54, delta, x, end color #1fab54, dot, start color #e07d10, f, left parenthesis, start color #11accd, x, start subscript, i, end subscript, end color #11accd, right parenthesis, end color #e07d10
Problem 1.A
  • Current
Problem set 1 will walk you through the process of approximating the area between f, left parenthesis, x, right parenthesis, equals, 0, point, 1, x, squared, plus, 1 and the x-axis on the interval open bracket, 2, comma, 7, close bracket using a left Riemann sum with 10 equal subdivisions.
Function f is graphed. The x-axis goes from negative 1 to 9. The graph is a curve. The curve starts in quadrant 2, moves downward to a relative minimum at (0, 1), moves upward and ends in quadrant 1. The region between the curve and the x-axis, between x = 2 and x = 7, is shaded.
What is the length of each rectangle, start color #1fab54, delta, x, end color #1fab54?
start color #1fab54, delta, x, end color #1fab54, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Problem 2
We want to approximate the area between g, left parenthesis, x, right parenthesis, equals, start fraction, 5, divided by, x, end fraction, plus, 2 and the x-axis on the interval open bracket, 1, comma, 7, close bracket using a right Riemann sum with 9 equal subdivisions:
Function g is graphed. The x-axis goes from negative 1 to 7. The graph is a curve. The curve starts in quadrant 1, moves downward concave up, and ends in quadrant 1. The region between the curve and the x-axis, between x = 1 and x = 7, is shaded. The shaded region is divided into 9 rectangles of equal width. Each rectangle touches the curve at the top right corner.
Which expression represents our approximation?
Choose 1 answer:

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