Question

Ley y= f(x) be a smooth curve on the closed interval [a,b]. Prove that if m and M are nonnegative numbbers such that m<= abs(f'(x)) <= M for all x in[a,b], then the arc length L of y= f(x) over the interval [a,b]satisfies the inequalities:
(b-a)*sqrt(1+m^2) <= L<= (b-a)*sqrt(1+M^2)

Answer 1

Ok, so I think I have this.
The formula for L is:\[L=\int_a^b \sqrt{1+f'(x)^2}~\mathrm{d}x\]The *mean value theorem for integrals* states that if F (capital F) is continuous on [a,b] then there exists c (a<c<b) such that:\[\int_a^b F(x)~\mathrm{d}x=(b-a)F(c).\]Now let sqrt(1+f'(x)^2)=F(x) (capital F). Since 1+f'(x)^2 is positive and continuous, F(x) is also continuous, so we may apply the theorem.

Answer 2

So it's obvious that F(c)=sqrt(1+f'(c)^2) lies between sqrt(1+m^2) and sqrt(1+M^2) because, by definition m^2<f'(c)^2<M^2 for all c in [a,b]. All we need to complete the inequality is to know that inequality is preserved through adding 1, taking the square root, and multiplying by (b-a). Obviously the first one is true. The second one is true if we are dealing with positive numbers (which we are), and the last one is true if b>a (which presumably it is). So the proof is complete.

Answer 3

I hope that is understandable.

Answer 4

thank you so much!

Answer 5

yaw