 Hello and welcome to the session. The given question says evaluate the following definite integral as limit of sums. First is integral x dx lower limit of integration as A and the upper limit of integration is P. Here x is called the integrand and this x is the variable of integration and this whole is called the integral of x from A to P. Now let us learn the limit of a sum. By definition integral fx dx from A to B is equal to B minus A into limit as n approaches to infinity 1 upon n bracket value of the function f at the point A and f A plus H plus so on up to A plus n minus 1 into H where H is equal to B minus A upon n and as n approaches to infinity H approaches to 0. So with the help of this we shall be evaluating the given definite integral as limit of sums. So this is our key idea. Now let us start with the solution. Here we have to evaluate integral x dx from A to B. So I am comparing here we have fx is equal to x. Therefore f A is equal to A f A plus H will be A plus H and the last term f A plus n minus 1 into H is equal to A plus n minus 1 into H. Therefore by using this limit of sum this integral x dx from A to B can be written as B minus A into limit as n approaches to infinity 1 upon n. Inside we have first f A and its value is A plus f A plus H its value is A plus H plus then we have A plus 2 H and so on up to the last is A plus n minus 1 into H where H is equal to B minus A upon n. Now this can further be written as minus A into limit as n approaches to infinity 1 by n and here we have A n times so this can be written as n into A since we have n terms here plus we have H plus 2 H plus so on up to n minus 1 into H. This can further be written as B minus A limit as n approaches to infinity 1 by n n A plus taking H common we have 1 plus 2 plus so on up to n minus 1. The sum of n terms of an AP is given by n into n plus 1 upon 2 and here we have n minus 1 terms so the sum of n minus 1 terms is given by n minus 1 into n upon 2. So this can further be written as B minus A into limit as n approaches to infinity 1 upon n n A plus H into n minus 1 into n upon 2. This is further equal to B minus A limit as n approaches to infinity taking 1 upon n inside this bracket we have A plus H is B minus A upon n into 1 upon n this 1 by n into n minus 1 into n upon 2. So this can further be written as B minus A limit as n approaches to infinity A plus B minus A upon 2 into n minus 1 upon n this n cancels up to this n this is further equal to B minus A limit as n approaches to infinity A plus B minus A upon 2 and this can be written as 1 minus 1 by n this is further equal to B minus A limit as n approaches to infinity first finding the limit of A plus B minus A upon 2 since it is a constant and independent of n so taking it outside the limit we have limit as n approaches to 0 1 minus 1 upon n this is further equal to B minus A now this is also independent of n so A plus B minus A upon 2 and limit of this is 1 minus 1 upon infinity which is equal to 1 minus 0. So we have into 1 this is equal to B minus A into B plus A upon 2 solving this bracket and this is further equal to B square minus A square upon 2 thus on evaluating the given definite integral as the limit of sums our answer is B square minus A square upon 2 so this completes the session hope you have understood it bye and take care