) ∫_(-1)^2 ∫_0^6▒〖x² sin(x-y)dxdy= 〗 ∫_0^6 ∫_(-1)^2▒〖x^2 sin(x-y)dydx〗
determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
It is illegible. I see backwards parentheses, unexplained shaded areas and cryptic symbols.
I suggest you make the changes
(-1)^2 = 1 and
0^6 = 1
Then it looks like you will comparing two indefinite double integrals with the order of integration reversed.
If that really is a 0^6 on both sides of your equation, that number can be replaced by zero. (NOT 1)
That would make both sides of the equation zero, and the equation valid.
I am sorry about that error, and ashamed of myself.
The given statement is true.
To determine whether the statement is true or false, we need to check if the order of integration can be switched without changing the value of the double integral. This property is called Fubini's theorem.
In this case, the integral ∫_(-1)^2 ∫_0^6 (x²sin(x-y)) dxdy is a double integral where we integrate first with respect to x and then with respect to y.
To check if we can switch the order of integration, we need to rewrite this integral in terms of y first and then x. Let's do that:
∫_(-1)^2 ∫_0^6 (x²sin(x-y)) dxdy
= ∫_(-1)^2 ∫_0^6 (x²sin(x-y)) dydx
Next, we switch the order of integration:
= ∫_0^6 ∫_(-1)^2 (x²sin(x-y)) dydx
Now, we have the integral ∫_0^6 ∫_(-1)^2 (x²sin(x-y)) dydx, which is the same as the original integral. Therefore, the statement is true, and we can switch the order of integration without changing the value of the double integral.
In general, Fubini's theorem applies when the integrand is integrable (finite), and the region of integration is a rectangle or a bounded region in the Cartesian plane.