 Hello and welcome to the session the given question says evaluate the following definite integral. So it is integral 1 upon x dx and the lower limit of integration is 2 and the upper limit as 3. So first let us learn the second fundamental theorem at the help of which we shall be evaluating the given definite integral according to this theorem integral fx dx from a to b is equal to fp minus fa where f is the empty derivative of the given function which is defined on the closed interval a to b and is continuous. So with the help of this idea we shall be evaluating the given integral. So this is our key idea. Let us now start with the solution. We have to find the value of integral 1 upon x dx from 2 to 3. Now integral 1 upon x dx is equal to lot more x therefore my second fundamental theorem we get integral 1 upon x dx from 2 to 3 equal to log mod x from 2 to 3. So first putting the upper limit that is log mod 3 minus log mod 2 this is equal to log 3 minus log 2 or we have log 3 by 2 since log a minus log b is equal to log a upon b. That is on evaluating the given definite integral our answer is log 3 upon 2. This completes the session hope you have understood it well take care and bye for now.