# Integral question

• Jul 11th 2011, 04:15 AM
durrrrrrrr
Integral question
Hi,
I'm having some trouble with this question. I've made a start on it but I'm not sure how to finish it.

Let f be differentiable in [a,b] and let f' be continuous in [a,b]. f(a)=0. Prove the inequality in the attached image.

What I did was deal with the left side:
LHS = (b-a)(|f(b) - f(a)|) = (b-a)|f(b)|

But I don't understand why this is necessarily bigger than the RHS.

Thanks
• Jul 11th 2011, 05:30 AM
girdav
Re: Integral question
We have $\displaystyle f(t)-f(a) =\int_a^t f'(s)ds$ hence $\displaystyle |f(t)|\leq \int_a^t|f'(s)|ds\leq \int_a^b |f'(s)|ds$. Now take the integral on both sides.
• Jul 11th 2011, 11:49 AM
durrrrrrrr
Re: Integral question
I don't understand what you've done or how it answers the question.
• Jul 11th 2011, 11:52 AM
girdav
Re: Integral question
I only used the fundamental theorem of calculus. Now if you follow my hint the result is shown.
• Jul 11th 2011, 12:22 PM
Plato
Re: Integral question
Quote:

Originally Posted by durrrrrrrr
I don't understand what you've done or how it answers the question.

Do you agree that $\displaystyle f(t) = \int_a^t {f'(s)ds}~?$

If so, then using basic theorems we get
$\displaystyle \left| {f(t)} \right| = \left| {\int_a^t {f'(s)ds} } \right| \leqslant \int_a^t {\left| {f'(s)} \right|ds} \leqslant \int_a^b {\left| {f'(s)} \right|ds}$

Here it last trick: $\displaystyle B=\int_a^b {\left| {f'(s)} \right|ds}$ is a number.

Thus $\displaystyle \int_a^b {\left| {f(x)} \right|dx} \leqslant \int_a^b {Bdx = \left( {b - a} \right)B}$