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## Homework Statement

Let f be a function such that |f(u)-f(v)|[tex]\le|u-v|[/tex] for all u and v in an interval [a,b]. Assume that f is integrable on [a,b]. Prove that

|[tex]\int_a^bf(x)dx[/tex] - (b-a)f(a)|[tex]\leq\frac{(b-a)^{2}}{2}[/tex]

That's the absolute value of the difference of the integral of f(x) from a to b and (b-a)f(a). The subtraction and absolute value signs aren't quite coming out right, but it should be good enough to interpret.

Once I prove this, I have to prove the more general statement that for any c in [a,b],

|[tex]\int_a^bf(x)dx[/tex] - (b-a)f(c)|[tex]\leq\frac{(b-a)^{2}}{2}[/tex]

## Homework Equations

Standard integration formulas. I might be able to throw the triangle equality somewhere to break up those absolute value differences.

## The Attempt at a Solution

I've realized that the given information implies that |m|[tex]\leq1[/tex] because

m = [tex]\frac{|f(u)-f(v)|}{|u-v|}[/tex] and |u-v| is less than or equal to |f(u)-f(v)|.

[tex]\frac{(b-a)^{2}}{2}[/tex] also looks like the formula for a triangle that has a height equal to its base length. So I think that if the slope were 1, then the integral would be equal to the area of the triangle. If the slope is less than 1, then I think the integral is less than the area of the triangle, because it won't reach the same height at point b that the triangle will reach. Using this, I started working with f(x) = (1)x + c, where c is the y intercept (I used c instead of b to avoid confusion). This hasn't gotten me very far, though, and I'm having trouble putting the logic into precise enough mathematical language to call it a proof. Any advice?