 Hello and welcome to the session. Let us discuss the following question. Question says, by using the properties of definite integrals, evaluate the integral from 0 to 2 x square root of 2 minus x dx. First of all, let us understand the property of definite integral that definite integral from 0 to a fx dx is equal to definite integral from 0 to a f a minus x dx. This is the key idea to solve the given question. Let us now start with the solution. Now we are given with definite integral 0 to 2 x square root of 2 minus x dx. Now using the property given in key idea, we can write this definite integral as definite integral from 0 to 2 2 minus x square root of 2 minus 2 minus x dx. Now this is further equal to definite integral from 0 to 2 2 minus x square root of 2 minus 2 plus x dx. Here 2 and 2 will get cancelled and we get definite integral from 0 to 2 2 minus x multiplied by square root of x dx. Now this integral can be further written as definite integral from 0 to 2 2 root x dx minus definite integral from 0 to 2 x multiplied by root x dx. Now this is further equal to definite integral 0 to 2 root x dx multiplied by 2 minus definite integral from 0 to 2 x raised to the power 3 upon 2 dx. Here 2 is constant so we can write this integral as 2 multiplied by integral 0 to 2 square root of x dx and here bases are same so powers will get added and we get integral from 0 to 2 x raised to the power 3 upon 2 dx. Now integral of square root of x is equal to x raised to the power 1 upon 2 plus 1 upon 1 upon 2 plus 1 we will write limits here lower limit is 0 and upper limit is 2 minus sign is written as it is. And integral of x raised to the power 3 upon 2 is equal to x raised to the power 3 upon 2 plus 1 upon 3 upon 2 plus 1 and the given limits of the integral are 0 to 2. Here we have used the formula of integral that integral of x raised to the power n dx is equal to x raised to the power n plus 1 upon n plus 1 plus c. Clearly we can see value of n here is 1 upon 2 and here value of n is 3 upon 2. Now simplifying this expression further we get 2 multiplied by x raised to the power 3 upon 2 upon 3 upon 2 and the given limits are 0 to 2 minus x raised to the power 5 upon 2 upon 5 upon 2 and given limits are 0 to 2. Now this is further equal to 4 upon 3 multiplied by x raised to the power 3 upon 2 given limits are from 0 to 2 minus 2 upon 5 multiplied by x raised to the power 5 upon 2 and here lower limit is 0 upper limit is 2. Now substituting the limits we get 4 upon 3 multiplied by 2 raised to the power 3 upon 2 minus 0 minus 2 upon 5 multiplied by 2 raised to the power 5 upon 2 minus 0. Now this can be further written as 4 upon 3 multiplied by 2 root 2 minus 2 upon 5 multiplied by 4 root 2. This is further equal to 8 root 2 multiplied by 1 upon 3 minus 1 upon 5. Clearly we can see 8 root 2 is common in both these two terms. Now subtracting these two terms by taking their LCM we get 2 upon 15 so we can write it as 8 root 2 multiplied by 2 upon 15. Now multiplying these two terms we get 16 root 2 upon 15 so we get definite integral from 0 to 2 x multiplied by square root of 2 minus x dx is equal to 16 root 2 upon 15. This is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.