(The exercise shown below is from a past admission exam taken in order to get accepted at the École Polytechnique, France, as an international student.)
In that exercise, I was able to solve the first two items (1.1 and 1.2), but I'm having some problems in getting the right answer in the last one, 1.3.
To solve item 1.3, I'm using the fact that
$$ \left| \int_{a}^{b} f(t) dt \right| \leq \int_{a}^{b} \left| f(t) \right| dt \qquad (*)$$
Furthermore, I'm using the first result presented in the item 1.1. By integrating the inequality shown in that result from $a$ to $b$, I was able to get that
$$ \int_{a}^{b} \left| f(t) \right| dt \leq K\frac{(b-a)^2}{2}$$
Then, it follows from (*) that
$$ \left| \int_{a}^{b} f(t) dt \right| \leq K\frac{(b-a)^2}{2} $$
However, the answer expected is $ \left| \int_{a}^{b} f(t) dt \right| \leq K\frac{(b-a)^2}{4} $
It seems to me that I should also use the result obtained at item 1.2, but it isn't really clear to me how it would help me solve the last item.
Could someone help me identify where my (possible) mistake is, or suggest a way to consider the result from item 1.2?
Thanks.
(P.S. My english isn't that good — Sorry)
Best Answer
Hint: You haven't really made a mistake as such. As you've already realised, you just haven't use all the pieces of information you've been asked to gather in parts $1.1$ and $1.2$. Besides not having used $1.2$, you also haven't used the second inequality of $(1)$. What was your answer to $(1.2)$? What is $\ \displaystyle\int_{\frac{a+b}{2}}^b(b-t)dt\ $? What is the sum of those two quantities?