Help with bounds for integration

In summary, the student is attempting to change the bounds for an integral involving sin(x^2) with x going from 1 to 2y and y going from 0 to 1. They converted 2y=x to 1/2x=y and graphed the bounds, but their parameters of 1 for x and 1/2x for y are incorrect. They are getting a close but incorrect answer of -.0200198, while the correct answer is -.0116111. The student is also not providing their actual computations or formulas used.
  • #1
Quatros
17
1

Homework Statement



I'm trying to change the bounds for this integral
Sin(x^2)dxdy

With x going from 1 to 2y, y going from 0 to 1
(I already know the integration for sin(x^2)

The Attempt at a Solution



I converted 2y=x to 1/2x=y and graphed all the bounds.

2H53WHoyRtm4W0h7LfqX3Q.png

haj14p

I went with 1,2 for my xbounds, and the 1/2x = y to to 1, I'm getting answer close to the correct answer, but no dice.
 

Attachments

  • 2H53WHoyRtm4W0h7LfqX3Q.png
    2H53WHoyRtm4W0h7LfqX3Q.png
    2 KB · Views: 782
Physics news on Phys.org
  • #2
Quatros said:

Homework Statement



I'm trying to change the bounds for this integral
Sin(x^2)dxdy

With x going from 1 to 2y, y going from 0 to 1
(I already know the integration for sin(x^2)

The Attempt at a Solution



I converted 2y=x to 1/2x=y and graphed all the bounds.

View attachment 215034
haj14p

I went with 1,2 for my xbounds, and the 1/2x = y to to 1, I'm getting answer close to the correct answer, but no dice.

Show your actual computations, and tell us your answer.
 
  • #3
(forgot to add a +3 in sign) With wolfram (we were allowed to use tech for this one) I'm getting -.0116111 ( which is the correct answer in the book).
But with my parameters, -.0200198 with those parameters. y=.5x to 1, and x = 1 to 2. )
 
  • #4
I really need help on this.
 
  • #5
Quatros said:

Homework Statement



I'm trying to change the bounds for this integral
Sin(x^2)dxdy

With x going from 1 to 2y, y going from 0 to 1
(I already know the integration for sin(x^2)
Your problem statement isn't very clear. Is the region of integration the triangle bounded by the x-axis, the line x = 1, and the line y = (1/2)x?
And are you supposed to change the order of integration?

Based on what you wrote above, the region would be described as ##\{(x, y) | 2y \le x \le 1, 0 \le y \le 1\}##. Note that in the triangle, for a given y value, the x value on the sloping line is less than the x-value on the vertical line.
Quatros said:

The Attempt at a Solution



I converted 2y=x to 1/2x=y and graphed all the bounds.

View attachment 215034
haj14p

I went with 1,2 for my xbounds, and the 1/2x = y to to 1, I'm getting answer close to the correct answer, but no dice.
It's not as simple as just switching letters. Think about the range of x values in the triangle.
 
  • #6
Quatros said:
(forgot to add a +3 in sign) With wolfram (we were allowed to use tech for this one) I'm getting -.0116111 ( which is the correct answer in the book).
But with my parameters, -.0200198 with those parameters. y=.5x to 1, and x = 1 to 2. )

This is not a helpful answer to my suggestion that you show your work. Write down the actual formulas you used.
 

1. How do I determine the bounds for integration?

The bounds for integration can be determined by analyzing the given function and the limits of integration. The lower bound is usually the starting point of the integration and the upper bound is the ending point. You can also graph the function to visually determine the bounds.

2. What is the purpose of bounds in integration?

Bounds in integration define the limits of the area or volume that is being calculated. They help to specify the range of values over which the integration is performed, ensuring a more accurate result.

3. How do I choose the correct bounds for multiple integrals?

Choosing the correct bounds for multiple integrals can be done by converting the integral into an iterated integral, and determining the appropriate bounds for each variable. This can be done by analyzing the given function and the region of integration.

4. Can I use negative bounds in integration?

Yes, negative bounds can be used in integration. Negative bounds can represent the lower limit of the integration, while positive bounds represent the upper limit. This is commonly seen in definite integrals where the area under the x-axis is being calculated.

5. Are there any techniques for finding bounds in integration?

Yes, there are various techniques for finding bounds in integration, such as the substitution method, the u-substitution method, and the method of trigonometric substitution. These methods can help simplify the integral and make it easier to determine the bounds.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
497
  • Calculus and Beyond Homework Help
Replies
3
Views
285
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
259
  • Calculus and Beyond Homework Help
Replies
2
Views
549
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
255
  • Calculus and Beyond Homework Help
Replies
5
Views
363
  • Calculus and Beyond Homework Help
Replies
15
Views
790
  • Calculus and Beyond Homework Help
Replies
23
Views
955
Back
Top