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Antidifferentiation as Area under a Curve

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ImageAntidifferentiation as Area under a Curve 

Copyright Maplesoft, a division of Waterloo Maple Inc., 2007 

Introduction 

This application is one of a collection of examples teaching Calculus with Maple. These applications use Clickable Calculus? methods to solve problems interactively. Steps are given at every stage of the solution, and many are illustrated using short video clips.  Click on the Image buttons to watch the videos. 

 

This application is reusable. Modify the problem, then click the !!! button on the toolbar to re-execute the document to solve the new problem. 

 

Problem Statement 

Use a Riemann sum to find the area bounded by the Typesetting:-mrow(Typesetting:-mi(-axis, the lines Typesetting:-mrow(Typesetting:-mi(, Typesetting:-mrow(Typesetting:-mi(, Typesetting:-mrow(Typesetting:-mi(, and the graph of the curve corresponding to Typesetting:-mrow(Typesetting:-mi( 

 

Solution 

Enter the expression for f. 

 

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Typesetting:-mrow(Typesetting:-mn( 

`+`(1, `*`(`^`(x, 2))) (1)
 

Use the Riemann Sums task template to solve the problem. 

Select Tools > Tasks > Browse > Calculus > Integration > Riemann Sums and choose one the options (Left, Right, or Midpoint).  Click Insert Default Content to insert it into the worksheet. 

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Calculate the Midpoint Riemann Sum 

Evaluate Typesetting:-mrow(Typesetting:-munderover(Typesetting:-mo(the midpoint Riemann Sum of Typesetting:-mrow(Typesetting:-mi(, Typesetting:-mrow(Typesetting:-mi(. 

 

Enter the requisite information into the task template.  Choose the interval to be Typesetting:-mrow(Typesetting:-mfenced(Typesetting:-mrow(Typesetting:-mi( and the value of Typesetting:-mrow(Typesetting:-mi( to Typesetting:-mrow(Typesetting:-mi(.  Execute the rest of the task template 

To enter the function, use the label (1).  Enter in the interval and value of n.  Execute each of the five commands in the template. 

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The Midpoint Riemann Sum 

Enter Typesetting:-mrow(Typesetting:-mi(: 

> Typesetting:-mrow(Typesetting:-mi(
 

`+`(1, `*`(`^`(x, 2))) (2)
 

Enter the interval Typesetting:-mrow(Typesetting:-mfenced(Typesetting:-mrow(Typesetting:-mi(: 

> Typesetting:-mrow(Typesetting:-mfenced(Typesetting:-mrow(Typesetting:-mi(
 

[a, b] (3)
 

Enter the value of Typesetting:-mrow(Typesetting:-mi(: 

> Typesetting:-mrow(Typesetting:-mi(
 

n (4)
 

The midpoint Riemann sum: 

> Typesetting:-mrow(Typesetting:-mi(
Typesetting:-mrow(Typesetting:-mi(
 

`/`(`*`(`+`(b, `-`(a)), `*`(Sum(`+`(1, `*`(`^`(`+`(a, `/`(`*`(`+`(i, `/`(1, 2)), `*`(`+`(b, `-`(a)))), `*`(n))), 2))), i = 0 .. `+`(n, `-`(1))))), `*`(n)) (5)
 

Value of the Riemann sum: 

> Typesetting:-mrow(Typesetting:-mi(
 

`/`(`*`(`+`(b, `-`(a)), `*`(`+`(n, `-`(`/`(`*`(`/`(1, 12), `*`(`^`(b, 2))), `*`(n))), `-`(`/`(`*`(`/`(1, 12), `*`(`^`(a, 2))), `*`(n))), `*`(`/`(1, 3), `*`(`^`(a, 2), `*`(n))), `/`(`*`(`/`(1, 6), `*`(... (6)
 

 

Find the limit of the Riemann sum as the value of Typesetting:-mrow(Typesetting:-mi( goes to Typesetting:-mrow(Typesetting:-mo( 

Use limit option in the Expression Palette to construct a limit using label (6).  Right click on the value of the limit and choose Expand. 

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Typesetting:-mrow(Typesetting:-mi( 

`+`(`-`(`*`(`/`(1, 3), `*`(`+`(a, `-`(b)), `*`(`+`(3, `*`(`^`(a, 2)), `*`(a, `*`(b)), `*`(`^`(b, 2)))))))) (7)
 

Typesetting:-mover(Typesetting:-mo( 

`+`(`-`(a), `-`(`*`(`/`(1, 3), `*`(`^`(a, 3)))), b, `*`(`/`(1, 3), `*`(`^`(b, 3)))) (8)
 

 

Notice that if Typesetting:-mrow(Typesetting:-mi(is the antiderivative of Typesetting:-mrow(Typesetting:-mi(, then the area computed is Typesetting:-mrow(Typesetting:-mi(, which is the content of the Fundamental Theorem of Calculus.   

 

 

Legal Notice: The copyright for this application is owned by Maplesoft. The application is intended to demonstrate the use of Maple to solve a particular problem. It has been made available for product evaluation purposes only and may not be used in any other context without the express permission of Maplesoft.   

 

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