StupidMathGuy
New member
- Joined
- Apr 2, 2020
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Hello all! This is obviously my first post. I don't ask questions at all generally, but this time I am in a great trouble since this is due in 3 days. Anyone who can help me, please do 
I am starting by writing down the problem as it was given:
Exact Wording:
Prove using only the Mean Value Theorem the following inequality:
[MATH]0 \leq \int \int_{R} sin(\pi x) cos(\pi y) \leq 1/32 [/MATH], where [MATH]R = [0,\frac{1}{4}] \times [\frac{1}{4}, \frac{1}{2}][/MATH]
My attempt at an answer:
The Mean Value Theorem for Double Integrals states that, given certain conditions which the above function clearly satisfies, we can have that :
[MATH] \iint_D f(x, y) \: dA = A f(x_0, y_0) [/MATH]
So, let [MATH]f(x,y) = sin(\pi x) cos(\pi y)[/MATH] and [MATH]g(x,y) = 1[/MATH]. Using the theorem, we can get
[MATH]\int_{0}^{\frac{1}{4}} \int_{\frac{1}{4}}^{\frac{1}{2}} sin(\pi x) cos(\pi y) dydx = f(x_0,y_0)\int_{0}^{\frac{1}{4}} \int_{\frac{1}{4}}^{\frac{1}{2}} 1dydx[/MATH]
Calculating the above integrals, we get that:
[MATH]\dfrac{3-2 \sqrt{2}}{2 \pi^2} = f(x_0,y_0) \dfrac{1}{16}[/MATH]
And thus that
[MATH]f(x_0,y_0) = \dfrac{24 - 16 \sqrt{2}}{\pi^2}[/MATH]
Since that [MATH]f[/MATH] is defined in a closed interval and it's also continuous in it, there will exist
[MATH]\inf f(x,y) \leq f(x_0,y_0) \leq \sup f(x,y)[/MATH]
We know that [MATH]\inf f(x,y) = 0[/MATH] and [MATH]\sup f(x,y) = \dfrac{1}{2}[/MATH]And now I am stuck. I don't even know if that is the correct path to the answer. Of course, I know how to make the inequality happen, I just cannot for the life of me find out how to apply the MVT
Please help me guys.
-A fellow mathematics student
I am starting by writing down the problem as it was given:Exact Wording:
Prove using only the Mean Value Theorem the following inequality:
[MATH]0 \leq \int \int_{R} sin(\pi x) cos(\pi y) \leq 1/32 [/MATH], where [MATH]R = [0,\frac{1}{4}] \times [\frac{1}{4}, \frac{1}{2}][/MATH]
My attempt at an answer:
The Mean Value Theorem for Double Integrals states that, given certain conditions which the above function clearly satisfies, we can have that :
[MATH] \iint_D f(x, y) \: dA = A f(x_0, y_0) [/MATH]
So, let [MATH]f(x,y) = sin(\pi x) cos(\pi y)[/MATH] and [MATH]g(x,y) = 1[/MATH]. Using the theorem, we can get
[MATH]\int_{0}^{\frac{1}{4}} \int_{\frac{1}{4}}^{\frac{1}{2}} sin(\pi x) cos(\pi y) dydx = f(x_0,y_0)\int_{0}^{\frac{1}{4}} \int_{\frac{1}{4}}^{\frac{1}{2}} 1dydx[/MATH]
Calculating the above integrals, we get that:
[MATH]\dfrac{3-2 \sqrt{2}}{2 \pi^2} = f(x_0,y_0) \dfrac{1}{16}[/MATH]
And thus that
[MATH]f(x_0,y_0) = \dfrac{24 - 16 \sqrt{2}}{\pi^2}[/MATH]
Since that [MATH]f[/MATH] is defined in a closed interval and it's also continuous in it, there will exist
[MATH]\inf f(x,y) \leq f(x_0,y_0) \leq \sup f(x,y)[/MATH]
We know that [MATH]\inf f(x,y) = 0[/MATH] and [MATH]\sup f(x,y) = \dfrac{1}{2}[/MATH]And now I am stuck. I don't even know if that is the correct path to the answer. Of course, I know how to make the inequality happen, I just cannot for the life of me find out how to apply the MVT
Please help me guys.
-A fellow mathematics student