Vtyjkyuk

I'm trying to prove a question of my homework, but I'm stuck. The question is

Question: let $f:[0,1]rightarrow mathbbR$, such that $|f'(x)| < 1, forall x$ and $f(0) = 2$, prove that $int_0^1f(x)dx geq frac32$

Ok, so my efforts until now were these:

1. I know that if $|f'(x)| < 1$, can be proved that $f(x)-f(y) < |x-y|$, (I don't prove that inequality is strictly), so I can use the condition on hypothesis $Rightarrow$ $|f(x)-2| leq |f(x)| -2 < |x|$ $Rightarrow$ $int_0^1f(x)dx < int_0^1x-2dx$, so I only can conclude that $int_0^1f(x)dx < 5/2$.


2. My other try is to using the mean value theorem so I can write down $f'(c) = fracf(1) - f(0)1-0 <1$, for some $c in (a,b)$, then $f(1) < 3$.

I'm not sure about my first effort, but I'm confident about the second one.

Thank you in advance!

You are on the right way. We have that

$$|f(x)-f(0)|le x.$$

Edit

Because of the mean value theorem for $xin (0,1]$ we have that there exist $cin (0,x)$ such that $f(x)-f(0)=f'(c)(x-0).$ Thus we have

$$|f(x)-2|=|f'(c)||x|<|x|.$$

End

Thus, it is

$$2-xle f(x)le 2+x.$$ (Note that the useful inequality here is $2-xle f(x)$ and not $f(x)le 2+x$.)

Integrantig over $[0,1]$ we have

$$2-dfrac12le int_0^1 f(x)dx$$ and we are done.

fou can also prove it in this way
We know that $f^'(x)>-1$ and $f(0) = 2$

$f(x) = f(0)+int_0^xf^'(t),dt$

so

$f(x)>f(0) + int_0^x-1,dt$

$f(x)>2-x$

$int_0^1f(x)dx >int_0^12-x,dx=1.5$

This image is also an explanation: