Riemann Sums - Revision history

MathAdmin: /* Approximating Area Under a Curve */

2015-09-28T05:35:00Z

Approximating Area Under a Curve

MathAdmin: /* Using the Definition to Evaluate a Definite Integral */

2015-09-28T05:34:07Z

Using the Definition to Evaluate a Definite Integral

MathAdmin: /* Approximating Area Under a Curve */

2015-09-28T05:32:34Z

Approximating Area Under a Curve

MathAdmin: /* Approximating Area Under a Curve */

2015-09-27T18:40:41Z

Approximating Area Under a Curve

MathAdmin: /* Approximating Area Under a Curve */

2015-09-27T18:38:37Z

Approximating Area Under a Curve

MathAdmin: /* Defining the Integral as a Limit */

2015-09-21T04:55:27Z

Defining the Integral as a Limit

MathAdmin: /* Defining the Integral as a Limit */

2015-09-21T04:54:59Z

Defining the Integral as a Limit

MathAdmin: /* Approximating Area Under a Curve */

2015-09-21T01:25:13Z

Approximating Area Under a Curve

MathAdmin: /* Approximating Area Under a Curve */

2015-09-21T00:45:43Z

Approximating Area Under a Curve

MathAdmin: /* A Left Point of View */

2015-09-21T00:43:41Z

A Left Point of View

@@ Line 41: / Line 41: @@
 :: <math style="vertical-align: 0px">{\displaystyle f\left(\frac{3}{4}\right)\cdot\Delta x\,=\,\frac{9}{16}\cdot\frac{1}{4}\,=\,\frac{9}{64}.} </math>
-Adding these four rectangles up with sigma <math style="vertical-align: -5px">(\Sigma) </math> notation, we
+Adding these four rectangles up with sigma <math style="vertical-align: -5px">(\Sigma) </math>&thinsp; notation, we
 can approximate the area under the curve as

@@ Line 161: / Line 161: @@
 :: <math style="vertical-align: 0px">\Delta x\,=\,{\displaystyle \frac{b-a}{n}\,=\,\frac{3-0}{n}\,=\,\frac{3}{n}}. </math>
-For a given <math style="vertical-align: -4px">n </math>, our leftmost interval would start at <math style="vertical-align: -5px">0, </math> and be
+For a given <math style="vertical-align: 0px">n </math>, our leftmost interval would start at <math style="vertical-align: -5px">0, </math> and be
 of length <math style="vertical-align: -5px">\Delta x=3/n. </math> This describes the interval <math style="vertical-align: -5px">[0,3/n]. </math>
 On the other hand, our next interval would start where the leftmost

@@ Line 1: / Line 1: @@
 == Approximating Area Under a Curve ==
-[[File:TallRiem.png|right|325px]]
+[[File:TallRiem.png|right|300px]]
 Graphically, we can consider a definite integral, such as

@@ Line 1: / Line 1: @@
 == Approximating Area Under a Curve ==
-[[File:Riemann.gif|right|500px]]
+[[File:TallRiem.png|right|325px]]
 Graphically, we can consider a definite integral, such as

@@ Line 1: / Line 1: @@
 == Approximating Area Under a Curve ==
+[[File:Riemann.gif|right|500px]]
 Graphically, we can consider a definite integral, such as

@@ Line 125: / Line 125: @@
 == Defining the Integral as a Limit ==
-[[File:Riemann.gif|right|450px]]
+[[File:Riemann.gif|right|500px]]
 Although associating the area under the curve with four rectangles

← Older revision		Revision as of 04:54, 21 September 2015
Line 124:		Line 124:

	== Defining the Integral as a Limit ==		== Defining the Integral as a Limit ==
		+
		+	[[File:Riemann.gif\|right\|450px]]

	Although associating the area under the curve with four rectangles		Although associating the area under the curve with four rectangles

@@ Line 7: / Line 7: @@
 to be the area "under the curve", which might be better said as
-the area that lies between the lines <math style="vertical-align: 0px">x=0</math>&thinsp; and <math style="vertical-align: -4px">x=1, </math>&thinsp; the <math style="vertical-align: 0px">x </math>-axis <math style="vertical-align: -5px">(y=0) </math>&thinsp;
+the area that lies between the line <math style="vertical-align: -4px">x=1, </math>&thinsp; the <math style="vertical-align: 0px">x </math>-axis <math style="vertical-align: -5px">(y=0) </math>&thinsp;
 and the curve <math style="vertical-align: -4px">y=x^{2}. </math>&thinsp; In order to find this area, we can begin
 with a familiar geometric object: the rectangle. In this case, we
@@ Line 55: / Line 55: @@
 Of course, we could also use right endpoints. In this case, we would
-use the endpoints <math style="vertical-align: -5px">1/4,\,1/2,\,3/4 </math> and <math style="vertical-align: -1px">1</math> for each interval from
+use the endpoints <math style="vertical-align: -5px">1/4,\,1/2,\,3/4 </math> and <math style="vertical-align: -1px">1</math> for the height above each interval from
 left to right to find

@@ Line 217: / Line 217: @@
 What would change if we approached the above integral through left
-endpoints, instead of right? We would only be changing our value for <math style="vertical-align: 0px">f(x_{i}). </math> For example, the leftmost interval is <math style="vertical-align: 0px">{\displaystyle \left[0\cdot\frac{3}{n},1\cdot\frac{3}{n}\right],} </math>
+endpoints, instead of right? We would only be changing our value for <math style="vertical-align: -5px">f(x_{i}). </math> For example, the leftmost interval is&thinsp; <math style="vertical-align: -15px">{\displaystyle \left[0\cdot\frac{3}{n},1\cdot\frac{3}{n}\right],} </math>
 so our left endpoint is <math style="vertical-align: 0px">0. </math> On the other hand, our second interval
-is <math style="vertical-align: 0px">{\displaystyle \left[1\cdot\frac{3}{n},2\cdot\frac{3}{n}\right],} </math>
+is&thinsp; <math style="vertical-align: --15px">{\displaystyle \left[1\cdot\frac{3}{n},2\cdot\frac{3}{n}\right],} </math>
-so our left endpoint is <math style="vertical-align: 0px">3/n. </math> For the interval <math style="vertical-align: 0px">I_{1}={\displaystyle \left[(i-1)\cdot\frac{3}{n},i\cdot\frac{3}{n}\right],} </math>
+so our left endpoint is <math style="vertical-align: -5px">3/n. </math> For the interval <math style="vertical-align: -15px">I_{1}={\displaystyle \left[(i-1)\cdot\frac{3}{n},i\cdot\frac{3}{n}\right],} </math>
-our left endpoint is <math style="vertical-align: 0px">3(i-1)/n. </math> Let's apply the same process to
+our left endpoint is <math style="vertical-align: -5px">3(i-1)/n. </math> Let's apply the same process as the last section to
-this value. We have
+this value. for height. We then have
 :: <math style="vertical-align: 0px">\begin{array}{rcl}
@@ Line 243: / Line 243: @@
   & = & {\displaystyle \lim_{n\rightarrow\infty}\,\left(\frac{9n(n+1)(2n+1)}{2n^{3}}+\frac{27n(n+1)}{2n^{3}}+\frac{27}{n^{2}}\right)}\\
 \\
-  & = & {\displaystyle \lim_{n\rightarrow\infty}\,\left(\frac{18n^{3}}{2n^{3}}+\frac{27n^{2}}{2n^{3}}+\frac{27}{n^{2}}\right)\qquad\qquad(\textrm{for large }n)}\\
+  & = & {\displaystyle \lim_{n\rightarrow\infty}\,\left(\frac{18n^{3}}{2n^{3}}+\frac{27n^{2}}{2n^{3}}+\frac{27}{n^{2}}\right)\qquad\qquad(\textrm{for~large~}n)}\\
 \\
   & = & 9+0+0\\
@@ Line 251: / Line 251: @@
 Thus our choice of endpoints makes no difference in the resulting
 value.
 == A More Advanced Example ==