Solving Iterated Integrals with x, y Boundaries

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Homework Help Overview

The discussion revolves around solving an iterated integral involving the function sin(y^2) with specified boundaries for x and y. Participants are exploring the implications of changing the order of integration and the corresponding limits of integration based on the geometric interpretation of the region defined by the original integral.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to determine the new limits of integration after changing the order of integration. Some participants question the boundaries for y and whether they are correctly identified based on the original integral's limits.

Discussion Status

Participants are actively discussing the boundaries of the integration region and the implications of changing the order of integration. Some guidance has been provided regarding the limits for the new order, but there remains some uncertainty about the correct interpretation of the triangular region involved.

Contextual Notes

There is mention of an attached figure that illustrates the integration region, which is central to the discussion. The original integral's limits and the geometric representation are being scrutinized to clarify assumptions about the boundaries.

stunner5000pt
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i need to solve this
[tex]\int_{0}^{\sqrt{\pi}} ( \int_{x}^{\sqrt{\pi}} sin y^2 dy) dx[/tex]

now i know i have to change the order of this
the integrand is bounded by the triangle from x = 0 to [itex]x= \sqrt{\pi}[/itex] here's where i am stuck
what is the boundary of the y?? is y bounded below by x=0 and above by x =1??

so what would the limits of integration change to?? (for the inside one from 0 to root pi?) and the outside one stays the same??

pelase help!
 
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stunner5000pt said:
i need to solve this
[tex]\int_{0}^{\sqrt{\pi}} ( \int_{x}^{\sqrt{\pi}} sin y^2 dy) dx[/tex]

now i know i have to change the order of this ...


Look at the attached figure. You have to integrate for the yellow triangle, according to the boundaries of your original integral. If you change the order of integration, that is you integrate by x first, it would go from x=0 to x=y; and then by y which goes from y=0 to y= sqrt (pi).

ehild
 
Last edited:
ehild said:
Look at the attached figure. You have to integrate for the yellow triangle, according to the boundaries of your original integral. If you change the order of integration, that is you integrate by x first, it would go from x=0 to x=y; and then by y which goes from y=0 to y= sqrt (pi).

ehild

how do you know that it is the upper triangle and not the lower triangle??
 
Because in the original integral:

[tex]\int_{0}^{\sqrt{\pi}} ( \int_{x}^{\sqrt{\pi}} sin y^2 dy) dx[/tex]

Look at the limits for the dy integral, y goes from y=x to y=root pi. If it were the white triangle in the image, then y would be going from 0 to x.
 
ehild is completely correct

marlon
 
You may want to search mathworld for Fresnel' S(x) antiderivative...

Daniel.
 
Last edited:

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