 Hello and welcome to the session. Let's discuss the following question. Let's say evaluate the integral 0 to 2 of x square plus x plus 1 dx as limit of sum. So let's now move on to the solution. Now to express any integral a to b of fx as limit of sum, we write this as b minus a limit m approaching to infinity 1 upon n into fa plus f of a plus h so on f of a plus n minus 1 into h where a is the lower limit b is the upper limit and h is b minus a upon n. So here a is 0, b is 2 and h is 0 minus 2 upon n that is 2 upon n. So in this integral 0 to 2 x square plus x plus 1 dx would be given by b minus a that is 2 into limit n approaching to infinity of 1 upon n into f of a that is f of 0 plus f of 0 plus h so on f of 0 plus n minus 1 into h. So this is again equal to 2 into limit n approaching to infinity of 1 upon n f of 0 plus f of 0 plus h is f of h plus f of h similarly we will get f of 2 h so on plus f of n minus 1 h. So now in place of x to find f of 0 we will put x as 0 so we will have 2 into limit n approaching to infinity of 1 upon n f of 0 would be 0 square plus 0 plus 1 that is 1 plus f of h would be h square plus h plus 1 then f of 2 h would be 2 h square plus 2 h plus what so on f of n minus 1 h that is n minus 1 h square plus f minus 1 into h plus 1. Now again this is equal to 2 into limit n approaching to infinity of 1 upon n into now we have 1 plus 1 plus 1 n times so this becomes n plus now we collect the terms of h square so we have h square into 1 square plus 2 square similarly we will have 3 square so on n minus 1 square right. Now we will collect the terms having h we will have h into 1 the coefficient of h plus 2 plus so on up till n minus 1 right. Now again this is equal to 2 into limit n approaching to infinity of 1 upon n into n plus h square now we have sum of the squares of n minus 1 terms now the formula for the sum of the squares of n's terms is n into n plus 1 into 2 n plus 1 upon 6. Now here we have n minus 1 terms so we will replace n by n minus 1 in this formula so we will get n minus 1 into n into 2 n minus 1 upon 6 so we have h square into n minus 1 into n into 2 n minus 1 upon 6 plus h into sum of n minus 1 terms now the sum of n terms is given by the formula n into n plus 1 by 2 so if in this formula we will replace n by n minus 1 we will get the sum for n minus 1 terms as n minus 1 into n by 2 you do remember this formulae and write these formulae while solving the question and it is very important to write these formulae so we have h into n minus 1 into n upon 2 now h is 2 by n so we will substitute the value of h here so we have again 2 into limit n approaching to infinity of 1 upon n into n plus x square that is 2 by n square into n minus 1 into n into 2 n minus 1 upon 6 plus h that is 2 by n into n minus 1 into n upon 2 now here n gets cancelled with n and here also 1n gets cancelled so we have 2 into limit n approaching to infinity of 1 upon n into n plus 2 square is 4 upon 6n here left with 1n here into n minus 1 into 2 n minus 1 plus n minus 1 now again this is equal to 2 into limit n approaching to infinity of 1 upon n into n plus 4 by 6n into now we take n common from this term and we also take n common from this term so we will have n square into 1 minus 1 upon n if we take n common from this we will have 1 minus 1 upon n similarly here we will have 2 minus 1 upon n plus n minus 1 now 1n gets cancelled here we have 2 into limit n approaching to infinity of 1 upon n into n plus 4n by 6 into 1 minus 1 upon n into 2 minus 1 upon n plus n minus 1 now again we have 2 into limit n approaching to infinity multiplying 1 by n with this whole thing we have n upon n plus 4 by n into 4n by 6 into into 1 minus 1 upon n into 2 minus 1 upon n plus n minus 1 upon n right now here n gets cancelled with n so we have 2 into limit n approaching to infinity of 1 plus 4 by 6 can be written as 2 by 3 into 1 minus 1 upon n into 2 minus 1 upon n plus n minus 1 upon n can be written as 1 minus 1 upon n so we have 2 into limit n approaching to infinity of 1 plus 2 by 3 into 1 minus 1 upon n into 2 minus 1 upon n plus 1 minus 1 upon n now we take the limit n approaching to infinity so we have 2 into now we take the limit so we have 1 plus 2 by 3 into 1 minus if we take limit n approaching to infinity of 1 by n it becomes 0 similarly here we have 2 minus limit n approaching to infinity of 1 upon n is 0 plus 1 minus 0 so we have 2 into 1 plus 2 by 3 into 1 into 2 plus 1 so this is equal to 2 into 1 plus 1 plus 1 is 2 so it is 2 plus 4 by 3 so this is again equal to 2 into 2 into 3 is 6 plus 4 is 10 by 3 taking the L7 solving so this is equal to 20 by 3 so the value of the integral as limits of sum is 20 by 3 so this completes the question do write all the formulae you use and take care of your calculations bye for now take care have a good day