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

In summary, the problem involves sketching a region enclosed by given curves and deciding whether to integrate with respect to x or y. The equations given are f(x) = 6x-x^2 and g(x) = x^2, with a = 0 and b = 3. The attempted solution involves finding the integral of (f(x) - g(x)) with respect to x, which simplifies to (6x-2x^2). However, there was an error in the last part of the solution where 3^3 was incorrectly calculated as 24 instead of 27, resulting in a final answer of 9.
  • #1
tangibleLime
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Homework Statement


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

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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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  • #2
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
 
  • #3
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
 
  • #4
GAH, thanks! I should probably just stick with a calculator to solve the last parts to eliminate errors like that.
 

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

What is the purpose of integrating with respect to X or Y?

The purpose of integrating with respect to X or Y is to find the area of the region enclosed by two curves, represented by the functions f(x) and g(x). This method is also known as finding the definite integral of a function.

How do you determine the limits of integration?

The limits of integration are determined by finding the points of intersection between the two curves f(x) and g(x). These points serve as the boundaries for the region enclosed by the curves.

Can the region enclosed by the curves be negative?

No, the region enclosed by the curves cannot be negative. Integrating with respect to X or Y only yields positive values, as it represents the area under the curve.

Are there any specific rules for integrating a region enclosed by two curves?

Yes, the rules for integrating a region enclosed by two curves include finding the limits of integration, setting up the integral with the correct orientation (with respect to X or Y), and applying the appropriate integration techniques (such as u-substitution or integration by parts).

How is integrating a region enclosed by two curves useful in real-world applications?

Integrating a region enclosed by two curves is useful in real-world applications such as calculating the area under a graph, finding the volume of irregular shapes, and determining the net change in a quantity over a specified interval.

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