 Hello and welcome to the session. Let us discuss the following question which says, evaluate the following definite integral as limit of sums. Second is integral x plus 1 into dx, the lower limit of integration is 0 and the upper limit is 5. So first let us learn what does limit of a sum by definition integral fx into dx where the lower limit of integration is a upper limit as b is equal to b minus a into limit as n approaches to infinity 1 upon n bracket, the value of the function at the point a plus f a plus h plus so on up to f a plus n minus 1 into h where h is equal to b minus a upon n such that when n approaches to infinity h approaches to 0. So this is the idea we are going to use in this problem to evaluate the given definite integral. So this is our key idea. Let us now start with the solution. Here the integrand is x plus 1 therefore f a is a plus 1, f a plus h is a plus h plus 1 and so on up to f a plus n minus 1 into h is equal to a plus n minus 1 into h plus 1. Therefore integral 0 to 5 x plus 1 into dx as the limit of a sum can be written as first we have b minus a where b is the upper limit and a is the lower limit. So here we have 5 minus 0 which is equal to 5 and limit 1 upon n where n approaches to infinity then we have f a and that is a plus 1 then we have f a plus h and its value is a plus h plus 1 and so on the last term is a plus n minus 1 into h plus 1 where h is equal to b minus a upon n. Now this can further be written as 5 limit and then approaches to infinity 1 by n. Now a we have n times since these are n terms so we have n into a plus 1 is also n times so plus n and taking h common we have 1 plus 2 plus so on up to n minus 1. This is further equal to 5 limit as n approaches to infinity 1 by n n a plus n plus h and the sum of n minus 1 terms is given by n minus 1 into n upon 2 since sum of n terms which are in AP is given by n into n plus 1 upon 2 therefore sum of n minus 1 terms is given by n minus 1 into n upon 2. First let us find this limit then we will substitute the value of b n a and h so this is further equal to 5 limit as n approaches to infinity 1 upon n n a plus n plus h is b minus a upon n into n minus 1 into n upon 2. Now n cancels out with n and this is further equal to 5 limit as n approaches to infinity 1 by n taking inside the bracket here we have a plus 1 plus b minus a upon 2 into n minus 1 upon n this is further equal to 5 limit as n approaches to infinity a plus 1 plus limit as n approaches to infinity b minus a upon 2 into 1 minus 1 upon n and this is further equal to 5 now a plus 1 is independent of n therefore limit as n approaches to infinity a plus 1 is a plus 1 then we have plus this is again independent of n so we have b minus a upon 2 and limit 1 minus 1 upon n which is further equal to 5 into a plus 1 plus b minus a upon 2 and limit of 1 minus 1 upon n is 1 since here we get 1 upon infinity which is equal to 0 so here we have only 1 and this is further equal to 5 a plus 1 plus b minus a upon 2 here we are given that the upper limit of integration that is b is equal to 5 and the lower limit of integration a is 0 so this can further be written as 5 a is 0 plus 1 plus 5 minus 0 upon 2 and this is equal to 5 into adding 1 with 5 upon 2 we have 7 upon 2 and this is equal to 35 upon 2 thus on evaluating the given definite integral as a limit of sums our answer is 35 upon 2 so this complete explanation hope you have understood it well bye and take care.