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Using the Gauss-Legendre form. Estimate the value of$$\displaystyle\int_{1}^{2}\displaystyle\int_{3}^{4}{f(x,y)dydx}$$ where, $f(x,y)=x^3y$.

My approach: We can approximate the integral $\int_{-1}^{1}{f(x)dx}=\sum_{i=1}^{n}{A_{i}f(x_{i})}$ with the form. Gauss-Legendre, by orthogonals polynomials $$p_{n}(x)=\dfrac{(-1)^{n}}{2^{n}n!}\dfrac{d^{n}}{dx^{n}}[(1-x^2)^n]$$ $$A_{i}=\dfrac{2}{(1-x_{i}^2)[p'_{n}(x_{i})]^{2}}$$ If n=2, then $x_{1}=0.577350=-x_{2}$, and $A_{i}=1$. And ,for instance $[1,2]\to [-1,1]$, this implies that $y=2x-3$ as I continue?

GNUSupporter 8964民主女神 地下教會
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Ahna Akbar
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2 Answers2

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Note that $$ \int_{1}^{2}\int_{3}^{4}f(x,y)dydx = \int_{1}^{2}x^3dx\int_{3}^{4}ydy = \int_{0}^{1}(1 + u)^3du\int_{0}^{1}(3 + v)dv $$ where $u = x - 1$ and $v = y - 3$. Let, $u = j/m$ and $v = k/n$, $m,n \in \mathbb{N}^{\ast}$. Thus, $$ \int_{1}^{2}\int_{3}^{4}f(x,y)dydx \simeq \sum_{j=1}^{m}\biggl(1 + \dfrac{j}{m}\biggr)^3\dfrac{1}{m}\cdot \sum_{j=1}^{n}\biggl(3 + \dfrac{k}{n}\biggr)\dfrac{1}{n} \quad \Rightarrow $$ $$ \int_{1}^{2}\int_{3}^{4}f(x,y)dydx \simeq \dfrac{1}{mn}\sum_{j=1}^{m}\sum_{k=1}^{n}\biggl(1 + \dfrac{j}{m}\biggr)^3\biggl(3 + \dfrac{k}{n}\biggr) $$ Exact value: $105/8 = 13.125$ and aproximate value for $m = n = 20: \quad$ $13.8422$

Mathsource
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Some pretty good progress. Since the integrand is at most cubic in $x$ and $y$ the result should be exact for the Gauss $2$-point formula. In the $x$-dimension, we want $u=-1$ when $x=1$ and $u=1$ when $x=2$. So $$\frac{x-1}{u+1}=\frac{2-1}{1+1}=\frac12$$ So $x=\frac12u+\frac32$. In the $y$-dimension we want $v=-1$ when $y=3$ and $v=1$ when $y=4$: $$\frac{y-3}{v+1}=\frac{4-3}{1+1}=\frac12$$ And we have $y=\frac12v+\frac72$. Then $$\begin{align}\int_1^2\int_3^4x^3y\,dy\,dx&=\int_{-1}^1\int_{-1}^1\left(\frac{u+3}2\right)^3\left(\frac{v+2}{2}\right)\frac{dv}2\frac{du}2\\ &=\sum_{i=1}^2\sum_{j=1}^2A_iA_j\frac1{64}(x_i+3)^3(x_j+7)\\ &=\frac1{64}(0.577350+3)^3(0.577350+7)+\frac1{64}(0.577350+3)^3(-0.577350+7)\\ &\quad+\frac1{64}(-0.577350+3)^3(0.577350+7)+\frac1{64}(-0.577350+3)^3(-0.577350+7)\\ &=5.420279376+4.594291848+1.683485587+1.426941966\\ &=13.12499878\end{align}$$ This compares well with the exact value of $\frac{105}8$.

user5713492
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