 Hello and welcome to the session. The question says evaluate the following definite integral and we have integral pox cube minus 5x square plus 6x plus 9 dx from 1 to 2. So first let us learn the second fundamental theorem of integral calculus. According to the theorem F is a continuous function defined on closed interval a to b then integral fx dx from a to b that is the lower limit of integration is a upper limit is b and this is equal to fb minus fa where f is the entire derivative small f. So with the help of this theorem we are going to evaluate the given definite integral. So this is our key idea. Now let's start with the solution. We have to evaluate the integral from 1 to 2 4x cube minus 5x square plus 6x plus 9 into dx from 1 to 2. So first let us find the value of integral 4x cube minus 5x square plus 6x plus 9 into dx this can be written as integral 4x cube into dx minus integral 5x square into dx plus integral 6x into dx plus integral 9 into dx which is further equal to 4 integral x cube into dx taking the constant outside the integral sign 5 integral x square into dx plus 6 integral x into dx plus 9 integral 1 into dx this is further equal to 4 into integral x cube dx is x raised to the power 4 upon 4 minus 5 into x cube upon 3 plus 6x square upon 2 plus 9x which on further simplifying can be written as x raised to the power 4 minus 5x cube plus 3x square plus 9x. Sorry here we have to write 3 now we have to calculate this integral that is from 1 to 2 4x cube minus 5x square plus 6x plus 9 into dx. So this can be written as the value of the integral as x raised to the power 4 minus 5 upon 3x cube plus 3x square plus 9x from 1 to 2 this is equal to x raised to the power 4 from 1 to 2 minus 5 upon 3x cube taking the constant outside we have 1 to 2 plus 3x square from 1 to 2 plus 9 to x from 1 to 2. So this is further equal to first let us put the upper limit we have 2 raised to the power 4 minus 1 raised to the power 4 and we have minus 5 upon 3 2 cube minus 1 cube plus 3 2 square minus 1 square plus 9 minus 1 this is further equal to 2 raised to the power 4 is 16 minus 1 minus 5 by 3 8 minus 1 plus 3 4 minus 1 plus 9 into 1 this is equal to 15 minus on simplifying this we have 35 upon 3 plus on simplifying this we have 9 plus 9 and this is further equal to 33 minus 35 upon 3 which is further equal to 99 minus 35 upon 3 and this is equal to 64 upon 3 thus on evaluating the given definite integral its value is 64 upon 3. So this completes the session my and take care.