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Unformatted text preview: Accumulation Authors: Bill Davis, Horacio Porta and Jerry Uhl 1996-2007 Publisher: Math Everywhere, Inc. Version 6.0 2.01 Integrals for Measuring Area BASICS B.1) a b f @ x D x measures the signed area between the plot of f @ x D and the x-axis for a x b . Calculus&amp; Mathematica is pleased to acknowledge the heavy influence of Emil Artin's book Calculus with Analytic Geometry (notes by G. B. Seligman), Committee on the Undergraduate Program, Mathematical Asociation of America, 1957. Although dated in spots, this short book remains a jewel. Here is the idea of the integral: Take a function y = f @ x D and plot it on an interval @ a, b D . It might look something like this: If any picture is missing, then execute all the blank cells below. a b y = f @ x D x y Look at the area between the graph of y = f @ x D and the x-axis: a b y = f @ x D x y If a part of this area lies above the x-axis, then call it positive. If a part lies below the x-axis, then call it negative. This gives the signs as shown below: a b y = f @ x D x y Now define a number that all folks call the integral of f @ x D from a to b. You signify this number by writing a b f @ x D x and you calculate its value by summing up the measurements of these areas taken with the corresponding signs. Accordingly, if x 1 &lt; x 2 &lt; x 3 are the points between a and b at which the curve crosses the x-axis as shown below: x1 x2 x3 a b y = f @ x D x y Then you calculate a b f @ x D x by taking the sum of four measurements: a x1 f @ x D x + x1 x2 f @ x D x + x2 x3 f @ x D x + x3 b f @ x D x where a x1 f @ x D x and x2 x3 f @ x D x are positive and x1 x2 f @ x D x and x3 b f @ x D x are negative. B.1.a) Given a specific function f @ x D and given specific numbers a and b with a b, when you calculate a b f @ x D x, do you get a number or do you get another function? Answer: As a signed measurement of area, a b f @ x D x is a number. B.1.b) Make the indicated area measurements: B.1.b.i) 1 4 I x 2 + 1 M x Answer: 1 4 I x 2 + 1 M x measures the area under the the plot of f @ x D = I x 2 + 1 M and above the segment @ 1, 4 D on the x-axis. Here's a picture of the area measured by 1 4 I x 2 + 1 M x: Clear @ f, x D ; f @ x _ D = x 2 + 1; Plot @ f @ x D , 8 x, 1, 4 &lt; , AxesLabel 8 &quot;x&quot; , &quot;y&quot; &lt; , Filling-&gt; Axis, AxesOrigin-&gt; 8 0, 0 &lt;D This integral calculates the area of a trapezoid with shortHeight = f @ 1 D tallHeight = f @ 4 D width = 3 3 2 3 3 Consequently the area measurement 1 4 f @ x D x = 1 4 I x 2 + 1 M x in square units is given by: 1 2 width H tallHeight + shortHeight L 27 4 Mathematica can also handle this measurement directly: Integrate BJ x 2 + 1 N , 8 x, 1, 4 &lt;F 27 4 Good going....
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This note was uploaded on 10/11/2011 for the course MATH 231 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
- Spring '08