 Hi and welcome to the session. I am Poorwa and I will help you with the following question. Evaluate integral limit from 0 to 1 e raised to the power 2 minus 3x dx as the limit of a sum. Let us now begin with the solution. Let us denote this definite integral by i. So we have i is equal to integral limit from 0 to 1 e raised to the power 2 minus 3x dx. Here we can clearly see that the lower limit a is equal to 0 so we have a is equal to 0, the upper limit b is equal to 1 so we have b is equal to 1 and the function fx is equal to e raised to the power 2 minus 3x. So we have fx is equal to e raised to the power 2 minus 3x and h which is given by b minus a upon n is equal to 1 upon n. Therefore we have i is equal to b minus a that is 1 limit as h tends to 0 h that is 1 upon n into f of a plus f of a plus h plus f of a plus 2h plus so on plus f of a plus n minus 1 h. And we get this is equal to limit as h tends to 0 1 upon n into now f of a. Now a is equal to 0 so we have f of 0 plus f of again a is equal to 0 so we get 0 plus h which is equal to h so we get f of h plus f of now again here a is equal to 0 so we get f of 2h plus so on f of now again here a is equal to 0 so we get n minus 1 h. This is equal to limit as h tends to 0 1 upon n into now we know that f of x is equal to e raised to the power 2 minus 3x so putting 0 in place of x we get f of 0 that is equal to e raised to the power 2 plus now putting h in place of x in the function e raised to the power 2 minus 3x we get e raised to the power 2 minus 3h plus now putting 2h in place of x in the function fx we get e raised to the power 2 minus 6x plus so on e raised to the power now putting n minus h in place of x in the function fx we get e raised to the power 2 minus 3 into n minus 1 into h this is further equal to limit as h tends to 0 1 upon n into now taking e square common from all these terms we get e square into here taking out e square common we get 1 plus here taking out e square common we get e raised to the power minus 3h plus here taking out e square common we get e raised to the power minus 6h plus so on e raised to the power here again taking out e square common we get e raised to the power minus 3 into n minus 1 into h this is equal to limit as h tends to 0 1 upon n into e raised to the power 2 into 1 plus e raised to the power minus 3h plus e raised to the power now we can write minus 6h as minus 3 into 2h right plus so on e raised to the power minus 3 into n minus 1 into h so on e raised to the power minus 3 into n minus 1 into h now we can write this as this is equal to limit as h tends to 0 e square upon n into now this is a geometric progression with the first term as 1 and r is equal to e raised to the power minus 3h so we get this is equal to e raised to the power minus 3h into n minus 1 upon e raised to the power minus 3h minus 1 we get this by using sum to n terms of a GP where we have the first term a is equal to 1 and r is equal to e raised to the power minus 3h and this is equal to limit as h tends to 0 e square into h because we know that 1 upon n is equal to h so we get h into e raised to the power now we can write this whole thing as minus 3 into n into h minus 1 upon e raised to the power minus 3h minus 1 now we can write this as this is equal to e raised to the power 2 limit as h tends to 0 e raised to the power minus 3 because n h is equal to 1 so we get e raised to the power minus 3 here minus 1 upon e raised to the power minus 3 minus 1 now we can write h as e raised to the power minus 3 minus 1 upon h so we get upon h here and this is since n h is equal to 1 this is equal to e square limit as h tends to 0 e raised to the power minus 3 minus 1 upon e raised to the power minus 3h minus 1 upon now multiplying the numerator and denominator by minus 1 upon 3 we get here minus 3h into minus 1 upon 3 now we know that limit as h tends to 0 e raised to the power h minus 1 upon h is equal to 1 so taking here minus 3h as h we get this denominator is equal to 1 so we write here since limit h tends to 0 e raised to the power h minus 1 upon h is equal to 1 so we get this whole thing is equal to e square into e raised to the power minus 3 minus 1 into minus 1 upon 3 and this is equal to minus 1 upon 3 into e square into e raised to the power minus 3 minus e raised to the power 2 right and we can write this as this is equal to minus 1 upon 3 into now we know that when basis are same powers are added so we get here e raised to the power minus 1 because 2 minus 3 is equal to minus 1 minus e raised to the power 2 this is equal to now taking minus common from here minus n minus will become plus we get this is equal to 1 upon 3 into e square minus e raised to the power minus 1 and we can write this as this is equal to 1 upon 3 into e square minus 1 upon e so we get our answer as 1 upon 3 into e square minus 1 upon e thus we write our answer as 1 upon 3 into e raised to the power 2 minus 1 upon e hope you have understood the solution take care and bye