 Hello and welcome to the session my name is Asha and I am going to help you with the following question let's say evaluate the following definite integral and we have integral x plus 1 dx lower limit of integration is minus 1 and upper limit of integration is 1. So first let us learn the second fundamental theorem with the help of which we shall be evaluating the given integral. It says integral fx dx from a to b that is the lower limit of integration is a and upper limit is p is equal to fp minus fa where this is the anti-radiator and f is a continuous function defined on a closed interval a to b. So this is the key idea which we shall be using in this problem to evaluate it. Let us now start with the solution. So we have to evaluate integral x plus 1 into dx from minus 1 to 1. So first let us find the value of the integral x plus 1 into dx that is equal to integral x dot dx plus integral 1 dot dx this is what we equal to x square upon 2 plus x. So now by second fundamental theorem we get integral minus 1 to 1 x plus 1 into dx equals to x square upon 2 plus x from minus 1 to 1. This is further equal to first putting the value of x which is upper limit of integration we have 1 square upon 2 plus 1 minus now putting the lower limit we have minus 1 whole square upon 2 plus minus 1 this is further equal to 1 by 2 plus 1 minus 1 upon 2 minus 1 this is equal to 3 by 2 minus minus half which is equal to 3 by 2 plus 1 by 2 which is equal to 4 by 2 and further simplifying we get 2 thus the value of the even definite integral is 2. So this completes the session. Bye and take care.