Sketch Region Enclosed by f(x) & g(x): Integrate w/ Respect to X or Y

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Homework Help Overview

The discussion revolves around sketching the region enclosed by the curves defined by the functions f(x) = 6x - x² and g(x) = x². Participants are considering whether to integrate with respect to x or y to find the area between these curves.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the integral and the limits of integration, with one attempting to calculate the area using the integral of the difference between the two functions. Others question the algebraic manipulations involved in the calculations.

Discussion Status

The discussion includes attempts to clarify algebraic errors in the calculations presented. Some participants provide corrections and alternative interpretations of the results, indicating an ongoing exploration of the problem without reaching a consensus.

Contextual Notes

There are indications of algebraic mistakes in the calculations, and participants express uncertainty about their own solutions. The original poster's approach to integrating the functions is being scrutinized, and there is a mention of using a calculator to avoid errors in future calculations.

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Homework Statement


Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y.

symimage.gif



Homework Equations



[tex]\int_a^b [f(x) - g(x)]dx[/tex]


The Attempt at a Solution



[tex]f(x) = 6x-x^2[/tex]
[tex]g(x) = x^2[/tex]

[tex]a = 0, b = 3[/tex]

[tex]\int_a^b [f(x) - g(x)]dx[/tex]

[tex]\int_0^3 [(6x-x^2)-(x^2)]dx[/tex]

[tex]((3x^2-\frac{x^3}{3})-(\frac{x^3}{3})\vert_0^3[/tex]

[tex]((3(3)^2-\frac{(3)^3}{3})-(\frac{(3)^3}{3}))[/tex]

[tex]((24)-(3))[/tex]

[tex]21[/tex]
 
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nope. you messed up. check your algebra. to make it easier, turn the 2nd line into 6x-2x^2. I am typically always wrong whenever i post on this forum, BUT I got 9
 
Yes, you did mess up on the algebra at the last part.

3 * 3^3 = 27 not 24.

And 3^3/3 = 9, not 3! You have 2 of them so 9 * 2 = 18.

27 - 18 = 9
 
GAH, thanks! I should probably just stick with a calculator to solve the last parts to eliminate errors like that.
 

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