 Hello and welcome to the session I am Deepika here. Let's discuss the question which says evaluate the following definite integral from 0 to 1 x into e raise to power x square dx. Now in this question we will use the second fundamental theorem of integral calculus. So let's start the solution. Let i is equal to integral from 0 to 1 x into e raise to power x square dx. Now we will first solve the indefinite integral x into e raise to power x square dx. Here put k is equal to t then dx is equal to dt and x dx is equal to 1 by 2 dt. So this is equal to 1 by 2 into integral of h to power t and this is equal to 1 by 2 into e raise to power t. Now here we will resubstitute for the reliability that is this is equal to 1 by 2 into e raise to power x square this is rfx that is the anti derivative of the function x into e raise to power x square by the second fundamental theorem we get is equal to of 1 minus f of 0. So this is equal to 1 by 2 is to power 1 square which is 1 minus e raise to power this is again equal to 1 by 2 into e minus now e raise to power 0 is 1 hence the answer for the above question is 1 by 2 into e minus 1. I hope the solution is clear to you. Bye and take care.