 Hi, and welcome to our session. Let us discuss the polling question. The question says, by using the properties of definite integrals, evaluate the polling. Integral of mod x minus pi from 2 to 8. Before solving this question, we should know one of the properties of definite integrals which says integral of fx from a to v is equal to integral of fx from a to c plus integral of fx from c to v, where c lies between a and v. So this property is the key idea in this question. Let i is equal to integral of mod x minus 5 from 2 to 8. Now consider mod of x minus 5. The mod of x minus 5 is equal to minus of x minus 5 if x is less than 5. And this is equal to plus x minus 5 if x is greater than 5. Let's name this as one. The idea, we know that integral of fx from a to b is equal to integral of fx from a to c plus integral of fx from c to v, where c lies between a and b. Now by using this property, we get i as integral of x minus 5 from 2 to 5. Now here a is equal to 2, c is equal to 5, and v is equal to 8. So we have integral of mod x minus 5 from 2 to 5 plus integral of mod x minus 5 from 5 to 8. Now when x is less than 5, then mod of x minus 5 is equal to minus of x minus 5. So this is equal to integral of minus x plus 5 from 2 to 5. And mod of x minus 5 is equal to x minus 5 if x is greater than 5. So this is equal to integral of x minus 5 from 5 to 8. Equal to integral of minus x from 2 to 5 plus 5 into integral of 1 from 2 to 5 plus integral of x from 5 to 8 minus 5 into integral of 1 from 5 to 8. Integral of xn with respect to x is equal to x to the power n plus 1 by n plus 1. Now using this integral of minus x is minus x square by 2 where the lower limit is 2 and upper limit is 5 plus 5 into integral of 1 with respect to x is x. So we have 5x where the lower limit is 2 and upper limit is 5 plus using this formula integral of x with respect to x is x square by 2 lower limit is 5 and upper limit is 8 minus 5 into integral of 1 with respect to x is x lower limit is 5 and upper limit is 8. Now by using second fundamental theorem of integral calculus this is equal to 5 square minus 2 square. We have first substituted the upper limit and then we have substituted the lower limit in x plus 5 into 5 minus 2 plus 1 by 2 into 8 square minus 5 square minus 5 into 8 minus 5. This is equal to minus 1 by 2 into 25 minus 4 plus 5 into 3 plus 1 by 2 into 64 minus 25 minus 5 into 3 and this is equal to minus 21 by 2 plus 15 plus 39 by 2 minus 15 and this is equal to minus 21 plus 39 by 2 and this is equal to 18 by 2 and this is equal to 9. Hence our required answer is 9. So this completes the session. Bye and take care.