The area of a definite integral

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Homework Help Overview

The problem involves interpreting the area under the curve of the function y = √(6 - 2x² + 4x) from x = -1 to x = 3, with a focus on understanding the geometric representation rather than performing integration.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the shape of the function, questioning whether it represents a semicircle or extends beyond that. There are attempts to complete the square to identify the center and radius of the shape represented by the function. Some participants discuss the implications of this manipulation on understanding the area.

Discussion Status

The discussion is ongoing, with participants providing insights into the geometric interpretation of the function. Some guidance has been offered regarding completing the square to find the center and radius, but there remains uncertainty about how to apply this to find the area accurately.

Contextual Notes

Participants note that integration is not permitted, and the problem must be approached through geometric interpretations of the area under the curve, leading to discussions about semicircles and ellipses.

Jack_B
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Homework Statement



The problem requires the computation of an indefinite integral but says to "interpret as areas" which basically changes the question to the following;

What is the area under the curve of the function y = root( 6 - 2x2 + 4x) from x = -1 to x = 3

Homework Equations



Relevant equations would include the area of a circle, A=pi*r2

The Attempt at a Solution



To begin with, I am not sure if the function is just a semi circle or also extends beyond just half of a circle. That being said, with the assumption that the function is a semi circle, the diameter is 4 (3 - (-1)), thus the radius is 2, making the total area (1/2)pi*22. However, with a graphing calculator, I can see that the area is 8.89.

Therefore I am led to believe that the area under the curve is a semicircle + a rectangular area. However, I am unsure as to how I can find the radius of the semicircle and the height of the rectangle.
 
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try completing the square
[tex]6+4x-2x^{2}=2(3+2x-x^{2})=2(4-(x-1)^{2})[/tex]
 
Unfortunately, integration is not allowed. The question must be completed by interpreting the area under the function as a sum of areas (a semicircle and perhaps a rectangle) but I have no idea how to, using the given function, find the semi circle's radius to be able to find its area. Unless completing the square puts the equation in a form where the semicircle's radius is more apparent, I'm not sure how it would help.
 
But it does; it gives you the location of the center and from there you can figure out the radius.
 
zcd said:
But it does; it gives you the location of the center and from there you can figure out the radius.



Really?

This is the first time in my life I've seen a semicircle function. With that manipulation, what is the location of the centre?

Thank you so much, I've been stumped on this one for about an hour.
 
[tex]y=\sqrt{2(4-(x-1)^{2})}\implies y^{2}=2(4-(x-1)^{2})[/tex] with some minor caveats in the range. [tex]y^{2}+2(x-1)^{2}=8[/tex] is an equation of?
 
It seems like the equation of a circle, the only equation of a circle I've ever seen is y^2 + x^2 = 9 which has a radius of 3 and center of (0, 0). But I'm still not quite sure how to obtain the location of the center of the circle from the newly manipulated one you gave me. It seems as the y coordinate of the center would still be 0, but I don't know about the x coordinate of the origin.

After some googling, I think that the y coordinate is 0 and the x coordinate is 1, but I'm not sure how the factor of 2 affects it.
 
Ah, apparently it's an equation of a semi-ellipse. Any idea as to how I can find its area? Either if a formula for it exists or if I must split it up into common geometric shapes and take their sum?
 
Last edited:
[tex]A=ab\pi[/tex] where a and b are the lengths of semi axes
 

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