How to plot graph regarding the definite integral by using Midpoint rule for the function 𝑦 =x^1/3?

This is what I have done and I'm not sure is it correct.

I don't know how to continue to plot the graph of it.

y = @(x) nthroot(x,3); %Function to integrate

a = 0; b = 2; %Interval where a is the lower boundary and b is the upper boundary.

n = 10; %Number of subintervals

dx = (b-a/n); %Width of each rectangle

mpr = 0;

for i = a:dx:b

mpr = mpr + f(i+dx/2);

end

I = dx*mpr;

[VBBV]

Compare the exact & approximated ones using Midpoint rule

f = @(x) nthroot(x,3);    %Function to integrate ...assign to correct variable 
a = 0; b = 2;   %Interval where a is the lower boundary and b is the upper boundary.
n = 10;     %Number of subintervals
dx = (b-a)/n;   % width 
mpr = 0;
x = linspace(a,b,100);
y = f(x);
plot(x,y,'linewidth',2)
hold on
L = a:dx:b; 
for i = 2:numel(L)-1
    mpr = mpr + f((L(i)+L(i-1))/2);  
    plot((L(1) + (i-0.5)*dx), f((L(i)+L(i-1))/2),'r+','linewidth',2)    
end
I = dx*mpr

I = 1.6485

legend('Exact: y = x^{1/3}','Approx (Midpoint)','location','best')
grid

[Torsten]

@VBBV

Should be

plot((L(1) + (i-1-0.5)*dx), f((L(i)+L(i-1))/2),'r+','linewidth',2)

instead of

plot((L(1) + (i-0.5)*dx), f((L(i)+L(i-1))/2),'r+','linewidth',2)

These are the evaluation points for the Midpoint rule. I don't know what you mean by Approx(Midpoint) in the legend. These are not approximations for an integral.

[VBBV]

Yes, you are right.

[L]

How to code for the straight line for the midpoint?

Answer 1

[Hassaan]

% Define the function to integrate using a function handle
f = @(x) nthroot(x,3); 

% Define the interval and the number of subintervals
a = 0; b = 2; % Interval [a, b]
n = 10; % Number of subintervals

% Calculate the width of each subinterval
dx = (b - a) / n; 

% Initialize the midpoint sum
mpr = 0;

% Calculate the sum for the Midpoint Rule
for i = 1:n     
    xi = a + (i-0.5)*dx; % Midpoint of the i-th subinterval
    mpr = mpr + f(xi); % Add the value at the midpoint
end

% Calculate the integral approximation
I = dx * mpr;

% Display the result
fprintf('The approximate value of the integral is: %f\n', I);

The approximate value of the integral is: 1.896224

% Plotting the function and the rectangles
x_vals = linspace(a, b, 1000); % Generate 1000 points between a and b
y_vals = arrayfun(f, x_vals); % Evaluate the function at each x value

% Plot the function
plot(x_vals, y_vals, 'b-', 'LineWidth', 1.5);
hold on; % Hold on to the current plot

% Plot each rectangle
for i = 1:n
    xi = a + (i-0.5)*dx; % Midpoint of the i-th subinterval
    % Plot the rectangle
    rect_x = [xi - dx/2, xi - dx/2, xi + dx/2, xi + dx/2];
    rect_y = [0, f(xi), f(xi), 0];
    patch(rect_x, rect_y, 'r', 'EdgeColor', 'r', 'FaceAlpha', 0.1);
end

% Formatting the plot
xlabel('x');
ylabel('y');
title('Midpoint Rule Approximation');
legend('y = nthroot(x, 3)', 'Midpoint rectangles');
hold off; % Release the plot hold

MATLAB Learning Material:

MATLAB Onramp | Self-Paced Online Courses - MATLAB & Simulink (mathworks.com)

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[L]

how do i add straight dotted line at the midpoint?