*Find the area between these two curves, accurate to 6 decimal places:
y=10*ln(x+1)/exp(x+1)
x=10*ln(y+1)/exp(y+1)
Use whatever computer tools you want. Show your work.
I think I solved it…
I got 0.536594(7|8)
My strategy was to use the fact that it was symmetric about x=y.
I took the initial y=… function and rotated it by pi/4 (45 degrees) using the transformation matrix:
[y'] [cos(pi/4) sin(pi/4)] [y]
] = ] ]
[x'] [cos(pi/4) -sin(pi/4)] [x]
That allows me to essentially take the integral above the new y’ and simply multiply it by 2. I used MATLAB to get it.
clear
xmin = eval(evalin(symengine, 'numeric::solve(0==y-10*log(y+1)/exp(y+1), y=-0.5..0.5)'));
xmax = eval(evalin(symengine, 'numeric::solve(0==y-10*log(y+1)/exp(y+1), y=0.5..2)'));
error = 1;
stepsize = 3000000;
prevsum = 0;
while (abs(error > 0.000000001))
stepsize = stepsize + 10000;
th = pi/4
x0min = xmin;
x0max = xmax;
y0min = xmin;
y0max = xmax;
x0 = linspace(x0min,x0max,stepsize);
y0 = linspace(y0min,y0max,stepsize);
y = 10*log(x0+1)./exp(x0+1);
x = 10*log(y0+1)./exp(y0+1);
xp = x*cos(th) + y0*sin(th);
yp = y*cos(th) - x0*sin(th);
xp0 = x/cos(th);
plot(x0,y,x,y0,xp,yp);
axis equal
axis square
grid on
sum = 0;
for i=1:stepsize-1
sum = sum + (xp0(i+1)-xp0(i))*yp(i);
end
sum = sum*2;
error = sum - prevsum;
disp('step size');
disp(stepsize);
disp('sum');
disp(sum);
disp('error');
disp(error);
end
Rather than going through the trouble of transforming it, I just subtracted the integral of y=x, as that’s the symmetry line.
I got 0.536595 for the area – found the second intercept at .95012, then just took the integral from 0 to .95012 of [10*ln(x)/exp(x)-x] and multiplied by 2.
*Nice job guys. The key was using the y=x symmetry line.
In the attached graph, the red and green lines are the two curves. The cyan line y=x is the axis of symmetry. If you subtract the cyan line from the red curve, you get the black curve.
I used Maxima to find the X-axis intercept of the black curve, and then numerically integrate the black curve from zero to that value and double it.
Of course I go the complicated route. This transcends to my ideas for robots as well. I need to learn to be more elegant. 