Trapezoidal and rectangle rule for double integral

Question

Tell me please, how to use trapezoidal and middle rectangles rules to calculate double integral without dividing the integration interval ($n=1$).

$$ \int_1^4\int_1^4 (x^3 + y^2)~{\rm d}x{\rm d}y $$

I tried to use the trapezoidal rule in this way:

$$ \int_1^4\int_1^4 (x^3 + y^2)~{\rm d}x{\rm d}y = \frac{f(1,1) + f(1,4) + f(4,1) + f(4,4)}{2}(4-1)(4-1) = 729 $$

Answer 1

Just apply the method to each integral, for example, for the trapezoidal rule

\begin{eqnarray} \int_1^4 \color{blue}{\left(\int_1^4f(x,y){\rm d}x\right)}{\rm d}y &=& \int_1^4 \color{blue}{\left(\frac{4-1}{2}[f(1,y) + f(4,y)]\right)}{\rm d}y \\ &=& \frac{3}{2}\left\{ \color{red}{\int_1^4f(1,y){\rm d}y} + \color{orange}{\int_1^4f(4,y){\rm d}y}\right\} \\ &=& \frac{3}{2}\left\{\color{red}{\frac{4-1}{2}[f(1,1) + f(1,4)]} + \color{orange}{\frac{4-1}{2}[f(4,1) + f(4,4)]} \right\} \\ &=& \frac{9}{4}\left[ f(1,1) + f(1,4) + f(4,1) + f(4,4)\right] \end{eqnarray}

Answer 2

You need to divide by $4$ to get the average height of the points, instead of dividing by $2$.

Everything else you did was correct.