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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

I wanted to 'challange' myself with solve a problem with midpoint and rule and the double integral f over the rectangle R.

This is a problem from midpoint.

"Use the Midpoint Rule m=n=2 to estimate the value of the integrab \(\displaystyle \int\int_r(x-3y^2)dA\), where \(\displaystyle R= {(x,y)| 0\leq x \leq 2, 1 \leq y \leq 2}.\)

Let's start with mid point. I Always start with draw it

and the center of those four subrectangles is \(\displaystyle x_1=\frac{1}{2}\), \(\displaystyle x_2= \frac{3}{2}\), \(\displaystyle y_1=\frac{5}{4}\) and \(\displaystyle y_2=\frac{7}{4}\) and the area of the sub rectangle is \(\displaystyle A= \frac{1}{2}\) to make it more clear we got \(\displaystyle \frac{1}{2}(f(\frac{1}{2},\frac{5}{4})+f(\frac{1}{2},\frac{7}{4})+f(\frac{3}{2},\frac{5}{4})+f(\frac{3}{2},\frac{7}{4})=-11.875\)

this question is aproxomite same as

\(\displaystyle \int_1^2\int_0^2 x-3y^2 dxdy\) and that will give result \(\displaystyle -13\)

Now for the most chalange one that I can't solve

I can only solve it with \(\displaystyle m=2\)\(\displaystyle n=1\)(we got two subs in x and 1 in y. cause then we get I start as usually draw it then we got the area of sub square \(\displaystyle A=1\) so we got these point \(\displaystyle f(1,2)+f(2,2)\)

and we get result \(\displaystyle -9\). Now afterwards I type I notice we cant use the rule right? \(\displaystyle x-3y^2=z\) and for any of our point we will get \(\displaystyle z<0\) and then we cant use defination. but are my two above correct?

Regards,