MAT 142 Day 1

1336 days ago by kcrisman

Welcome to MAT 142!  Let's think about $$\int_0^1\sqrt{1+4x^2}dx$$ and what it might mean for us.

First, can we numerically approximate it?  Here is a left-hand sum with $m=20$ subintervals.

m=20 approx=sum([sqrt(1+4*((i-1)^2)/m^2)*1/m for i in [1..m]]) show(approx); n(approx) 
       

1.44841384355230
1.44841384355230

What exactly does this represent?

P = plot(sqrt(1+4*x^2),(x,0,1),fill=True,figsize=[3,3]) for i in [1..m]: P += line([((i-1)/m,0),((i-1)/m,sqrt(1+4*((i-1)^2)/m^2)),((i)/m,sqrt(1+4*((i-1)^2)/m^2)),(i/m,0)],color='red') show(P) 
       

There are tools to have arbitrary approximation of such an integral, of course.

numerical_approx(integral(sqrt(1+4*x^2),x,0,1)) 
       
1.47894285754460
1.47894285754460

Or we can try for a symbolic integration, and check what that approximates to.

show(integral(sqrt(1+4*x^2),x,0,1)); n(integral(sqrt(1+4*x^2),x,0,1)) 
       

1.47894285754460
1.47894285754460

Well, it turns out this is actually the length of a curve!

plot(x^2,(x,0,1))+plot(x,(x,0,1),linestyle='--',color='red') 
       

We have access to more sophistication - for example, known error.

numerical_integral(sqrt(1+4*x^2),0,1) 
       
(1.478942857544597, 1.6419564125509345e-14)
(1.478942857544597, 1.6419564125509345e-14)

But what if we use the table of integrals in the back of the book and the FTC?

f(x)=1/2*(x*sqrt(1+4*x^2)+ln(2*x+sqrt(1+4*x^2))/2) show(f(x));show(f(1)-f(0));n(f(1)-f(0)) 
       


1.47894285754460
1.47894285754460