Let $ f(x)= ( \log_e x) ^2 $ and (Integration by parts. Comparing integrals of different limits ) Let $ f(x)=( \log_e x) ^2 $ and let 
$ I_1= \int_{2}^{12} f(x) dx $  ,  $ I_2= \int_{5}^{15} f(x) dx $ and $ \int_{8}^{18} f(x) dx$
Then which of the following is true?
(A)$I_3 <I_1 < I_2 $
(B)$I_2 <I_3 < I_1 $
(C)$I_3 <I_2 < I_1 $
(D)$I_1 <I_2 < I_3 $
What I've done: 
Have computed the integration (by integration by parts) and computed $ I_1, I_2 $ and $ I_3 $ with the help of a calculator. So option D is correct. Also intuitively $f(x)$ is increasing function. Option D is obvious. 
But I want to know how to do without calculating the limits? Calculator won't be allowed in the test. What's the intuition?

You can see that each interval of integration is of the same length.
$I_2-I_1 = \int_2^5 f(x) dx - \int_{12}^{15} f(x) dx$ 
But f is increasing, so $\forall x \in [2,5] \forall y \in [12,15], f(x) < f(y)$ 
This imply that $\int_2^5 f(x) dx < \int_{12}^{15} f(x)$
Same idea to compare $I_2$ and $I_3$
Edit : You can, also write $\int_2^5 f(x) dx - \int_{12}^{15} f(x) dx = \int_2^5 ( f(x)-f(x+10) )dx $, and as f is increasing...
