Changing bounds for integral of sin(x²) over triangular region

[Quatros]

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.

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.

 

[Ray Vickson]

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

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.

 

[Quatros]

(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. )

 

[Quatros]

I really need help on this.

 

[Mark44]

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.

The Attempt at a Solution

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

View attachment 215034

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.

 

[Ray Vickson]

(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.