Can someone check whether my script is correct or not? I did this using the same method as the one last time.

The last work from which I took reference: https://www.mathworks.com/matlabcentral/answers/2049237-what-is-the-function-to-use-to-find-the-intersection-of-two-graphs?s_tid=prof_contriblnk

My new work:

%%Plotting the graphs
k = @(x,y) y-3.*sin((x)^2);
fimplicit(k), grid on
hold on
m = @(x,y) y- exp(x/2) + exp(-2.*x);
fimplicit(m), grid on
hold off
xlabel('x'), ylabel('y')
legend ('y = 3 + sin(x^2)','y = e^(x/2) + e^(-2*x)','location','northeast')

When I ran the script, it showed this:

Answer 1

[Dyuman Joshi]

There was a missing element-wise operator in the sin() in the definition of 'k'.

Also, I have modified the legend to show the expressions correctly.

%                        vv
k = @(x,y) y - 3.*sin((x).^2);
fimplicit(k), grid on
hold on
m = @(x,y) y - exp(x/2) + exp(-2.*x);
fimplicit(m), grid on
hold off
xlabel('x'); ylabel('y');
%             v                v       v  v
legend ('y = 3*sin(x^2)','y = e\^(x/2) - e\^(-2*x)','location','northeast')

[Image Analyst]

Yes, this works with no warning messages. 👍🏽

[Lê]

I want to find the intersection points of the graphs and use them to calculate the volume of the region bounded by the graphs, but the intersection results are only 0.7722, missing the number 1.524. Any idea why?

clear;
%% Plotting the graphs
k = @(x,y) y - 3.*sin((x).^2);
fimplicit(k), grid on
hold on
m = @(x,y) y - exp(x./2) - exp((-2).*x);
fimplicit(m), grid on
hold off
xlabel('x'); ylabel('y');
legend ('y = 3*sin(x^2)','y = e\^(x/2) + e\^(-2*x)','location','northeast')
%% Finding intersection points
x0  = [-1, 1];  % initial guesses
for j = 1:2
    fun1 = @(x) exp(x./2) + exp((-2).*x) - 3 .* sin((x).^2);    % f(x) = g(x)
    x(j) = fsolve(fun1, x0(j))
end
%% Calculating the volume
fun2 = @(x) (3.*sin(x.^2)).^2 - (exp(x./2) - exp(-2.*x)).^2;
V = pi*integral (fun2,x(1),x(2))

[Walter Roberson]

clear;
%% Plotting the graphs
k = @(x,y) y - 3.*sin((x).^2);
fimplicit(k), grid on
hold on
m = @(x,y) y - exp(x./2) - exp((-2).*x);
fimplicit(m), grid on
hold off
xlabel('x'); ylabel('y');
legend ('y = 3*sin(x^2)','y = e\^(x/2) + e\^(-2*x)','location','northeast')

%% Finding intersection points
fun1 = @(x) exp(x./2) + exp((-2).*x) - 3 .* sin((x).^2);    % f(x) = g(x)
x0  = [0 2]  % initial guesses

x0 = 1×2

0 2

for j = 1:length(x0)
    x(j) = fsolve(fun1, x0(j));
end

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.


Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

x

x = 1×2

0.7722 1.5242

uniquetol(x)

ans = 1×2

0.7722 1.5242

%% Calculating the volume
fun2 = @(x) (3.*sin(x.^2)).^2 - (exp(x./2) - exp(-2.*x)).^2;
V = pi*integral (fun2,x(1),x(2))

V = 9.2979

Answer 2

[Walter Roberson]

Are you sure you want and not ? You currently square x before taking sin() -- which is certainly a valid operation, but is much less common than taking the sin of x and squaring the result.

%%Plotting the graphs
k = @(x,y) y-3.*sin((x).^2);
fimplicit(k), grid on
hold on
m = @(x,y) y- exp(x/2) + exp(-2.*x);
fimplicit(m), grid on
hold off
xlabel('x'), ylabel('y')
legend ('y = 3 + sin(x^2)','y = e^(x/2) + e^(-2*x)','location','northeast')

[Lê]

It is indeed sin((x)^2).