 Hello and welcome to the section. The given question says evaluate the following definite integral as the limit of sum. So by the definition of limit of a sum, if we have an integral fx dx from a to b, then it is equal to b minus a into the limit. As n approaches to infinity 1 upon n, the value of the given function at the point a plus n, the value of the given function at the point a plus h and sum on up to f a plus n minus 1 into h, where h is equal to b minus a upon n. This is the key idea we shall be using in this problem to evaluate the given definite integral. Let us now start with the solution. Here the function fx is x square minus x, therefore at a is a square minus a, also the lower limit of integration here is 1 and the upper limit of integration is 4. So this can be written as 1 square minus 1. Then we have to find f a plus h, so this is equal to a plus h whole square minus a plus h and here a is 1, so we have 1 plus h whole square minus 1 plus h. And so on, the last term is f a plus n minus 1 into h. So this can be written as a plus n minus 1 into h whole square minus a plus n minus 1 into h. Or this is further equal to a is 1, so 1 plus n minus 1 into h whole square minus 1 plus n minus 1 into h. Also h is equal to b minus a upon n, b is 4 a is 1, so we have 3 upon n. And thus the integral x square minus x dx from 1 to 4 is equal to, first we have b minus a, so 3 limit as n approaches to infinity 1 by n. Then first we have f a, its value is 1 square minus 1 plus then we have 1 plus h whole square minus 1 plus h plus so on. The last term is 1 plus n minus 1 into h whole square minus 1 plus n minus 1 into h. This is further equal to 3 limit as n approaches to infinity 1 upon n 1 square minus 1 plus 1 square plus 2 h plus h square minus 1 minus h plus so on up to 1 square plus n minus 1 whole square into h square plus 2 n minus 1 into h minus 1 minus n minus 1 into h. This can further be written as 3 limit as n approaches to infinity 1 upon n. Now here 1 square is n times so it can be written as 1 into n square minus, minus 1 is also n times so n into minus 1 gives minus n plus then taking 2 h common from the terms we have 1 plus 2 plus so on up to last term is n minus 1 then taking minus h common first we have 1 plus 2 minus n minus 1. Plus so on up to n and then taking h square common inside the bracket we have 1 square since from here h square can be written as 1 square into h square then 2 square and then so on up to n minus 1 whole square. Now this is further equal to 3 limit as n approaches to infinity taking 1 upon n inside the bracket here at this end cancels out with this end so we have 2 h upon n and sum of n minus 1 terms of an AP series whose common difference is 1 and first term is also 1 is given by n into n minus 1 upon 2 n minus h upon n similarly here also we have n into n minus 1 upon 2 plus h square upon n we have taken 1 upon n inside the bracket and the square of the natural numbers up to n minus 1 is equal to into n into 2 n minus 1. And the denominator we have 6 this is further equal to 3 times limit as n approaches to infinity inside the bracket 2 cancels out with 2 n with n and h is b minus a upon n that is 3 upon n into n minus 1. Then here also cancelling n with n h is 3 upon n into n minus 1 upon 2 plus now cancelling this n with n h square is 9 upon n square into n minus 1 into 2 n minus 1. This can further be written as 3 limit as n approaches also here we have 6 also sorry so we have limit as n approaches to infinity 3 into 1 minus 1 upon n minus 3 upon 2 1 minus 1 upon n plus 9 upon 6 can be written as 3 upon 2 and 1 n taking here in the denominator and 1 here we have 1 minus 1 upon n and here we have 2 minus 1 upon n. This is further equal to 3, 3 times limit as n approaches to infinity 1 minus 1 upon n minus 3 upon 2 limit as n approaches to infinity 1 minus 1 upon n plus 3 upon 2 limit as n approaches to infinity 1 minus 1 upon n into 2 minus 1 upon n. So this is for the equal to 3, 3 and limit of this function is 1, minus 3 upon 2, similarly limit of this function is also 1, plus 3 upon 2, limit of this function is 1 and limit of this function is 2, since when n approaches to 0, 1 upon n approaches to 0 and 2 minus 0 is 2, this is for the equal to 3 into 3 minus 3 upon 2 and this 2 cancels out with this 2, so we have plus 3. Now we have 3 times of 6 minus 3 upon 2 which is equal to 3 into taking LCM, we have 12 minus 3 in the numerator which is equal to 9 upon and in the denominator we have 2, this is equal to 27 upon 2, thus on evaluating the given definite integral as the limit of sums our answer is 27 upon 2. So this completes the session, hope you have understood it, bye and take care.