Let $ f(x)= ( \log_e x) ^2 $ and (Integration by parts. Comparing integrals of different limits )

Question

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?

Answer 1

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...

Answer 2

Note that each of the integrals are over an interval of width $10$, and so to see which integral has a larger value, we want to find the interval in which $f$ has the largest average value. Because $f$ is increasing and $2<5<8$, we must have $I_1<I_2<I_3$.

Answer 3

Let $$ g(y)=\int_{y}^{y+10}\log^2 x\,dx.\tag{1}$$ We have to compare $g(2),g(5)$ and $g(8)$. By the fundamental theorem of Calculus: $$ g'(y) = \log^2(y+10)-\log^2(y) > 0, \tag{2}$$ hence $g$ is an increasing function and $g(2)<g(5)<g(8)$ (i.e. option $\color{red}{D}$) follows.