 Hello and welcome to the session. In this session we discussed the following question which says if integral 0 to 1 3x square plus 2x plus k dx is equal to 0 then find the value of k. Let's move on to the solution now. We are given that the integral 0 to 1 3x square plus 2x plus k dx is equal to 0. So now integrating this we get 3 into x cube upon 3 plus 2 into x square upon 2 plus k into x and the limits are 0, 1 is equal to 0. Since we know integral of x raised to the power n dx is equal to x raised to the power n plus 1 upon n plus 1 and also integral of some constant k dx is equal to k into x. Now this 3 cancels with 3, 2 cancels with 2 and so we get x cube plus x square plus kx and the limit goes from 0 to 1 is equal to 0. Now putting the upper limit and the lower limit we get 1 cube plus 1 square plus k into 1 minus. Now here we have put 1 in place of x. Now this minus Now we put 0 in place of x, 0 cube plus 0 square plus k into 0 this is equal to 0 that is we have 1 plus 1 plus k minus 0 equal to 0 that is k plus 2 is equal to 0 which gives us k equal to minus 2. So we get the final value of k as minus 2. This completes the session hope you have understood the solution of this question.