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Find the value of the integral $\int\limits_1^4 {\mathop {\log }\nolimits_e } [x]dx$  from the options given below:A. $\mathop {\log }\nolimits_e 2$ B. $\mathop {\log }\nolimits_e 3$ C. $\mathop {\log }\nolimits_e 6$ D. None of the above

Last updated date: 13th Jul 2024
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Hint-We will make use of the formula of integration by parts and solve it.
We have the integral I= $\int\limits_1^4 {\mathop {\log }\nolimits_e } [x]dx$
Since we have the limit of integral from 1 to 4,we will split the limits and write the integral
So, we get
I= $\;\;\int\limits_1^2 {\mathop {\log }\nolimits_e } [x]dx$ +$\;\;\int\limits_2^3 {\mathop {\log }\nolimits_e } [x]dx$ +$\;\;\int\limits_3^4 {\mathop {\log }\nolimits_e } [x]dx$
Greatest integer function is discontinuous at all integers. So we should write the definition of
the function in each of the smaller limits.So,we can write the integral I as
I=$\;\;\int\limits_1^2 {\mathop {\log }\nolimits_e } 1dx + \;\;\int\limits_2^3 {\mathop {\log }\nolimits_e } 2dx + \;\;\int\limits_3^4 {\mathop {\log }\nolimits_e } 3dx$
Let us solve this integral by putting the value of ${\log _e}1 = 0$ ,since ${\log _e}2$ and ${\log _e}3$ are constants take it out of the integral and solve
I=$\int\limits_1^2 0 dx$ +${\log _e}2\int\limits_2^3 {dx}$ +${\log _e}3\int\limits_3^4 {dx}$

We know that integral 1 dx is equal to x
So, on solving the integral further ,we get
I= 0+${\log _e}2[x]_1^2 + {\log _e}3[x]_3^4$
On applying limits, we get
I=${\log _e}2$$[2 - 1]$ +${\log _e}3[4 - 3]$
So , we get I= ${\log _e}2 + {\log _e}3$
But we know the formula which says
${\log _e}a + {\log _e}b = {\log _e}ab$
Therefore, we can write
${\log _e}2 + {\log _e}3$=${\log _e}6$
So, therefore the value of the integral I=${\log _e}6$=ln6
So, option C is the correct answer

Note: It is possible to integrate to the greatest integer functions only when the limits are
given, if the limits are not given, we cannot solve the problem.