 Hello and welcome to the session the given question says evaluate the following definite integral as limit of sums 7 is integral x is per plus x into dx and the integral goes from 1 to 2 that is the lower limit of integration is 1 and upper limit as 2. Now by the definition of limit of sum and integral fx dx from a to b is defined by b minus a limit as n goes to infinity 1 upon n value of the function at the point a plus h plus a plus 2h the value of the function plus 1 up to f at the point a plus n h well h is equal to b minus a upon n and as n goes to infinity h goes to 0. So this is the key idea which I will be using in this problem to evaluate the given definite integral. Let us now start with the solution. Now on comparing integral fx dx from a to b with integral x plus x square dx from 1 to 2 we find here that fx is equal to x square plus x a which is the lower limit of integration is 1 and b the upper limit of integration is 2. So h is equal to b minus a upon n is equal to 2 minus 1 upon n which is equal to 1 by n. So by the formula of limit of a sum integral 1 to 2 x square plus x into dx is equal to b minus a that is 2 minus 1 into limit as n approaches to infinity 1 upon n first we have value of the function at the point a plus h and this is equal to a plus h whole square plus a plus h. So this is the value of the function at the point a plus h then the value of the function at the point a plus 2 h is given by a plus 2 h whole square plus a plus 2 h and similarly the value of the function at the point a plus n h is a plus n h whole square plus a plus n h. Now this is equal to 2 minus 1 is 1 so we have limit as n approaches to infinity 1 by n. Now opening the brackets first we have a square plus h square plus 2 h plus a plus h plus from here we have a square plus 2 square h square plus 4 h which we get on expanding a plus 2 h whole square plus a plus 2 h plus so on up to on expanding a plus n h whole square we get a square plus n square h square plus 2 n a h plus a plus n h. Now this is further equal to limit as n approaches to infinity 1 by n now a square is n times plus a is also n times plus now taking 2 h common we have inside the bracket 1 plus 2 plus so on up to n plus then taking h common inside the bracket we have 1 plus 2 plus so on up to n plus lastly taking h square common we have 1 square plus 2 square plus so on and from the last term we have n square equals to limit as n approaches to infinity 1 by n n times of a square plus n times of a plus 2 a h inside the bracket we have 1 plus 2 plus so on up to n plus h into 1 plus 2 plus so on up to n plus h square into 1 square plus 2 square plus so on up to n square now summation n that is 1 plus 2 plus up to n is equal to n into n plus 1 upon 2 and summation n square that is 1 square plus 2 square plus so on up to n square is equal to n into n plus 1 into 2 n plus 1 upon 6. Now using this here we get this is further equal to limit as n approaches to infinity 1 upon n n e square plus n a plus 2 a h now sum of 1 plus 2 up to n is given by n into n plus 1 upon 2 plus h again the sum of 1 plus 2 up to n is given by n into n plus 1 upon 2 and we have plus h square 1 square plus 2 square plus up to n square and sum is given by n into n plus 1 sorry here we have n plus 1 into 2 n plus 1 upon 6. Now taking 1 by n inside the bracket we get limit as n approaches to infinity a square plus a plus upon n into n into n plus 1 upon 2 plus here also we have h upon n into n into n plus 1 upon 2 plus h square upon n into n into n plus 1 into 2 n plus 1 upon 6 so this one cancels out with this 2 with 2 here also n with n and here also n with n so we further have limit as n approaches to infinity a square plus a plus a h into n plus 1 plus h into n plus 1 upon 2 plus h square n plus 1 into 2 n plus 1 upon 6. We know that the lower limit of integration is 1 upper limit of integration is 2 in this problem and h is 1 upon n so this limit can further be written as limit n approaches to infinity a square is 1 square that is 1 plus a a is 1 plus a is 1 and h is 1 upon n so we get 1 by n into n plus 1 plus from here we have 1 upon 2 n into n plus 1 plus this can be written as 1 upon n square into n plus 1 into 2 n plus 1 upon 6 equals to limit as n approaches to infinity from here we have 2 plus 1 plus 1 by n plus 1 upon 2 plus 1 upon 2 n plus 1 upon 6 n square and here we have 2 n square plus n plus 2 n plus 1 and this is further equal to limit as n approaches to infinity 3 plus 1 by n plus 1 upon 2 plus 1 upon 2 n now simplifying this post we get 1 upon 3 n plus 2 n is 3 n and 3 n upon 6 n square is let us write it like that and here we have 1 upon 6 n square this is further equal to limit as n approaches to infinity 3 plus 1 by n plus 1 by 2 plus 1 upon 2 n plus 1 upon 3 plus n cancels out to then 3 2s are 6 so we have 1 upon 2 n plus 1 upon 6 n square now limit as n tends to infinity this term is 0 this is also 0 this is also 0 and this is also 0 so we have 3 plus half plus 1 upon 3 and lcm is 6 and the numerator we have 23 which we get on simplifying thus on evaluating the given definite integral we get our answer as 23 upon 6 so this completes the session hope you have learned it bye and take care