 Hello and welcome to the session the given question says evaluate the following definite integral as the limit of sum. Fifth is integral e raised to the power x dx and the lower limit of integration is minus 1 and upper limit is 1. Now by the definition of limit of sum and integral fx dx from a to b is equal to b minus a limit as an approach of infinity 1 upon n fa plus fa 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 and as n approaches to infinity h approaches to zero. So 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 given function fx is e raised to the power x value of the lower limit is minus 1 and value of upper limit is 1. That's the function at the point a is e raised to the power a value of the function at the point a plus h is e raised to the power a plus h at the point a plus 2h value of the function is e raised to the power a plus 2h and so on up to value of the function at the point a plus n minus 1 into h is e raised to the power a plus n minus 1 a to h. Also h which is equal to b minus a upon n is equal to 1 minus of minus 1 upon 2 sorry n and this will be equal to 2 upon n. Therefore the given integral which is minus 1 to 1 e raised to the power x dx can be written as b minus a that is 2 limit as n approaches to infinity 1 upon n first we have f a its values e raised to the power a plus e raised to the power a plus h plus e raised to the power a plus 2h and so on up to e raised to the power a plus n minus 1 into h. This is further equal to 2 times limit as n approaches to infinity 1 upon n e raised to the power a plus e raised to the power a into e raised to the power h plus e raised to the power a into e raised to the power 2h plus so on up to e raised to the power h into e raised to the power n minus 1 into h. This is further equal to 2 times limit as n approaches to to infinity 1 upon n taking e raised to the power a common first we have 1 plus e raised to the power h plus so on up to e raised to the power n minus 1 into h. Now the sum of 1 e raised to the power h plus e raised to the power 2 h the sum of this series is equal to e raised to the power n h minus 1 upon e raised to the power h minus 1 therefore sum of this series is e raised to the power n minus 1 into h minus 1 upon e raised to the power h minus 1. So this can further be written as 2 limit n approaches to infinity 1 upon n e raised to the power a and this can be written as e raised to the power n minus 1 into h minus 1 upon e raised to the power h minus 1. Now this is further equal to 2 limit as n approaches to infinity taking 1 upon n inside we have e raised to the power a upon n into e raised to the power n minus 1 and h is 2 upon n minus 1 upon e raised to the power again 2 upon n minus 1. So this is further equal to 2 times e raised to the power a since it is independent of n limit n approaches to infinity 1 by n e raised to the power those we have 2 minus and we have minus 2 upon n minus 1 upon e raised to the power 2n minus 1. Now this is equal to e raised to the power a limit as n approaches to infinity e raised to the power 2 minus 2 upon n minus 1 upon e raised to the power 2 upon n minus 1 upon 2 upon n this is further equal to e raised to the power a limit as n approaches to infinity e raised to the power 2 minus 2 upon n minus 1 upon limit as n approaches to infinity e raised to the power 2 upon n minus 1 upon 2 upon n and as we know limit as x approaches to 0 e raised to the power x minus 1 upon x is equal to 1. Therefore value of denominator is 1 and we have e raised to the power a limit n approaches to infinity e raised to the power 2 minus 2 upon n minus 1 since as n approaches to infinity 2 upon n tends to 0 and by using this we have denominator 0. This is further equal to e raised to the power a now limit as n approaches to infinity 2 upon n tends to 0. So, here we have e square minus 1 now a is minus 1 which is the lower limit of the given integral. So, we have e square minus 1 upon e which is equal to e minus 1 upon e. Thus on evaluating the given definite integral as a limit of sum or answer is e minus 1 upon e. This completes the session. Hope you have understood it. Bye and take care.