- #1
DryRun
Gold Member
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Homework Statement
Homework Equations
For example, for f(x,y)=x+y-2
The Attempt at a Solution
I've figured out part (a) which is quite simple. I simply used the relevant equations above for ##f(x,y)= 3(x^2+y^2)##
I know i should use the given hint to figure out the value of I, which is ##L_f(P)\leq I\leq U_f(P)##
But I'm not sure how to solve for the value of I, which I've calculated to be:
$$\frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n \left[ (x^2_i+x^2_{i-1}) + (y^2_j+y^2_{j-1})\right]\Delta x_i \Delta y_j (wrong)$$
$$(Fixed) \frac{3}{4}\sum_{i=1}^m \sum_{j=1}^n \left[ (x_i+x_{i-1})^2 + (y_j+y_{j-1})^2\right]\Delta x_i \Delta y_j\\ $$
I've tried the old-fashioned way of expanding the terms to find the sum, but it gets messy after expanding all the terms in i and j. But in this case, since a hint is given, i guess i should be using it to simplify the inequality. But how?
Here's my next attempt:
Using the given hint: ##0\leq s \leq t## and applying it to the given limits for x and y:
##0\leq x \leq b## gives the corresponding inequality: ##3x^2\leq b^2+bx+x^2 \leq 3b^2##
##0\leq y \leq d## gives the corresponding inequality: ##3y^2\leq d^2+dy+y^2 \leq 3d^2##
I tried adding both of these inequalities but the answer for the middle term, I, doesn't make sense, as it retains x and y, along with b and d.
##3x^2 + 3y^2\leq b^2+d^2+bx+dy+x^2+y^2 \leq 3b^2+3d^2##
I'm stuck. The final answer is not given in my notes, but based on previous calculations of the same type, the answer should be only in terms of b and d, meaning no x and y terms are involved.
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