 Hello and welcome to the session. The given question says, evaluate the following definite integral as limit of sums. Ninth is integral x plus e raised to the power 2x into dx the lower limit of integration is 0 and upper limit is 4. By the definition of limit of sums integral fx dx from a to b is defined as b minus a limit as n approaches to infinity 1 by n 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 key idea with the help of which we shall be evaluating the following definite integral. Let us now start with the solution. Here when comparing the given definite integral integral x plus e raised to the power 2x dx from 0 to 4 the integral fx dx from a to b we find that fx is equal to x plus e raised to the power 2x a which is the lower limit of integration is 0 and b is equal to 4. Therefore h which is equal to b minus a upon n is 4 minus 0 upon n which is equal to 4 by n. Therefore by the formula of limit of sum we have integral x plus e raised to the power 2x dx from 0 to 4 is equal to 4 minus 0 limit as n approaches to infinity 1 by n and the value of the function at the point a a plus e raised to the power 2a plus the value of the function at the point a plus h this is given by a plus h plus e raised to the power 2 times of a plus h plus so on up to now we have to find the value of the function at the point a plus n minus 1 into h. So we have a plus n minus 1 into h plus e raised to the power 2 times of a plus n minus 1 into h this is for the equal to 4 limit as n approaches to infinity 1 upon n now here the term a is n times taking h common because we have 1 plus 2 plus so on up to n minus 1 then plus e raised to the power 2a plus e raised to the power 2a into e raised to the power 2h plus so on up to e raised to the power 2a into e raised to the power 2 times of n minus 1 into h this is for the equal to 4 limit as n approaches to infinity 1 upon n a appears n times so we have n a plus h sum of first n minus 1 natural numbers is given by n minus 1 into n upon 2 plus e raised to the power 2a taking common we have 1 plus e raised to the power 2h plus so on up to e raised to the power 2 times of n minus 1 into h. We have a geometric progression whose first term is a and common ratio that is the ratio between any two consecutive term is r then the sum of n terms is given by a into r raised to the power n minus 1 upon r minus 1. So, here we have the geometric progression 1 plus e raised to the power 2h plus so on up to e raised to the power 2 times of n minus 1 into h so here the common ratio is e raised to the power 2h which we get e raised to the power 2h divided by 1 is equal to e raised to the power 4h divided by e raised to the power 2h and so on we get e raised to the power 2h as the common ratio and the first term is 1 therefore this is equal to 1 into e raised to the power 2nh minus 1 upon e raised to the power 2h minus 1. Therefore, this can further be written as limit as n approaches to infinity 4 by n na plus h n minus 1 into n upon 2 plus e raised to the power 2h into e raised to the power 2nh minus 1 upon e raised to the power 2h minus 1. Now, the lower limit of integration a is 0 and h is equal to 4 by n. So, we have limit as n approaches to infinity taking 1 upon n inside the square bracket we have here 4 and from here we have a plus n cancels out within we have h into n minus 1 upon 2 plus e raised to the power 2a upon n into e raised to the power 2nh minus 1 upon e raised to the power 2h minus 1. Now, we have 4 limit as n approaches to infinity a is 0 plus value of h is 4 by n. So, we have 4 upon 2 1 minus 1 by n plus e raised to the power 2a upon n into e raised to the power 2 into n and h is 4 by n minus 1 upon e raised to the power 2 into 4 by n minus 1. This is further equal to 4 limit as n approaches to infinity 2 1 minus 1 by n plus e raised to the power 2a upon n into e raised to the power 8 minus 1 upon e raised to the power 8 upon n minus 1. This is equal to 4 limit as n approaches to infinity 2 times of 1 minus 1 by n plus e raised to the power 2a into e raised to the power 8 minus 1 or independent of n. So, we have limit as n approaches to infinity 4 upon n into 1 upon e raised to the power 8 upon n minus 1 and as n approaches to infinity 1 upon n approaches to 0. So, we have 4 into 2 plus a is the lower limit of integration which is 0 and this problem. So, we have e raised to the power 0 into e raised to the power 8 minus 1 limit as n approaches to infinity 4 by n upon e raised to the power 8 upon n minus 1. This is further equal to 8 plus e raised to the power 0 is 1. So, we have e raised to the power 8 minus 1 limit as n approaches to infinity multiplying and dividing by 2. Here we have 1 by 2 in the numerator and in the denominator let us take 8 upon n. So, we have e raised to the power 8 upon n minus 1 upon 8 upon n. Now, limit as x approaches to 0 e raised to the power x minus 1 upon e raised to the power x is equal to 1 minus 1 limit x approaches to a f x upon g x is equal to limit as x approaches to a f x upon limit as x approaches to 8 g x. So, this can be simplified as 8 plus e raised to the power 8 minus 1. Now, here limit n approaches to infinity half is half and as n approaches to infinity then 8 upon n approaches to 0. Therefore, applying limit as x approaches to 0 e raised to the power x minus 1 upon e raised to the power x here x is 8 upon n. So, this implies the value of the denominator is 1. So, we have 8 plus e raised to the power 8 minus 1 upon 2. This is further equal to 16 plus e raised to the power 8 minus 1 upon 2 and this is equal to 15 plus e raised to the power 8 upon 2. Thus, on evaluating the given definite integral our answer is 15 plus e raised to the power 8 upon 2. This completes the session. Hope you have understood it. Bye and take care.