Riemann Sum Optimization At the moment, I am taking a calculus course at my high school (we are starting to learn about integrals) and thought of an interesting problem while learning about Riemann Sums. Once I develop a method on how to solve this problem for any general curve, I hope to write a script in python.
Problem
Consider the following parameters:


*

*$y = f(x) = (x - 1.5)^{1/3} + 2$

*Domain: $[-2, 6]$

*$n = 4$

*The width of each subdivision, $\Delta w_k$, does not have to be equal.


Using left-hand endpoints, right-hand endpoints, midpoints, or a combination of all three, how could you orient the rectangles to cover the greatest area underneath the curve? Ideally, the intended python script I want to write would involve an algorithm that can be applied to any particular curve, any specific domain interval, and any number of subdivisions.
How would you even approach this problem? What steps would you outline?

Graph of $y = f(x) = (x - 1.5)^{1/3} + 2$

Same Graph with Varying Widths
This is an example that involves the midpoint approximation with varying widths ($\Delta w_k$).


OK, we know that the Riemann sum is an approximation to the integral $I = \int_a^b f(x) \, dx$.  In this case, the value of the integral is $I = 16 + (3/4) ((9/2)^{4/3} - (7/2)^{4/3}) \approx 17.5865$.   The Riemann sum using the midpoint rule as you specify is
$$S(x_1,x_2,x_3) = \frac{1}{4} \left [ f \left (\frac{a+x_1}{2} \right )(x_1-a) + f \left (\frac{x_1 + x_2}{2} \right )(x_2-x_1)  + f \left (\frac{x_2+x_3}{2} \right )(x_3-x_2)  + f \left (\frac{x_3+b}{2} \right )(b-x_3) \right ]$$
We want to minimize the difference between $S$ and $I$.  Actually, that's not very accurate: we want to minimize the absolute value of the difference between S and I.  And if we have access to a multivariable minimization routine, such as simplex, we could do that.  However, if we want to analyze things more analytically for your Python code, we would then choose to minimize the square of the difference between $S$ and I:
$$\Delta(x_1,x_2,x_3) = (I - S(x_1,x_2,x_3))^2$$
Now, if this is your first course in Calculus, then what we do here is going to be new; you typically don't lean these concepts until Calculus III.  What we do next is to find the derivatives of $\Delta$ with respect to each of the $x_i$, holding the other $x_j$ constant; this is called a partial derivative with respect to $x_i$.  For example:
$$ \frac{\partial \Delta}{\partial x_1} = -2 (I - S(x_1,x_2,x_3)) \frac{\partial S}{\partial x_1} $$
$$ \frac{\partial S}{\partial x_1} = \frac{1}{4} \left [\frac{1}{2} f' \left (\frac{a+x_1}{2} \right )(x_1-a) + f \left (\frac{a+x_1}{2} \right ) + \frac{1}{2} f' \left (\frac{x_1 + x_2}{2} \right )(x_2-x_1) - f \left (\frac{x_1 + x_2}{2} \right ) \right ]  $$
Do the same for the other points $x_2$ and $x_3$.  Then set each of the $\frac{\partial \Delta}{\partial x_i} = 0$, which is equivalent to setting $\frac{\partial S}{\partial x_i} = 0$.  You'll have 3 equations and 3 unknowns in a nonlinear system of equations.  You can try to solve for the $x_i$ analytically, or you can use a numerical routine to find them.
I hope that helps.  Good luck!
