 Hi friends I am Purva and today we will work out the following question sketch the graph of y is equal to mod x plus 3 and evaluate integral limit is from minus x to 0 mod x plus 3 dx. Now by definition we know that modulus x is equal to x if x is greater than equal to 0 and it is equal to minus x if x is less than 0. So this is the key idea which we will use to solve this question. Let us now begin with the solution. Now by definition we know that modulus x plus 3 can be written as this is equal to x plus 3 if x plus 3 is greater than equal to 0 which implies x is greater than equal to minus 3 and it is also equal to minus of x plus 3 if x plus 3 is less than 0 which implies x is less than minus 3. Now since in the question we are given that y is equal to modulus x plus 3 therefore we get y is equal to x plus 3 if x is greater than equal to minus 3 and it is also equal to minus of x plus 3 if x is less than minus 3. Now let us mark this equation y is equal to x plus 3 as 1 and we mark this equation y is equal to minus of x plus 3 as 2. Now we can clearly see that equation 1 and equation 2 are equation of two lines. Now if we take x greater than equal to minus 3 then we have minus 3 comma 0 and 0 comma 3 are two points on line 1. If we put x is equal to minus 3 we get y is equal to 0 so we get the coordinate minus 3 comma 0 and if we put x is equal to 0 we get y is equal to 3 so we get the coordinate 0 comma 3 so these are the two points on line 1. And let us name it as AB. Now if x is less than minus 3 then we have minus 4 comma 1 and minus 5 comma 2 are two points on line 2 so if we put x is equal to minus 4 then we get y is equal to 1 in equation 2 and if we put x is equal to minus 5 in equation 2 then we get y is equal to 2 and let us name it as now let us plot these points on the graph and get two lines. So this is line AB and this is line AC and union of these two lines is the graph of y is equal to modulus x plus 3. Now we have to evaluate the limit from minus 6 to 0 modulus x plus 3 dx so integral minus x to 0 modulus x plus 3 dx is equal to integral limit from minus 6 to minus 3 minus of x plus 3 dx by equation 2 plus integral limit from minus 3 to 0 x plus 3 dx by equation 1 and this is equal to now integrating minus of x plus 3 we get minus of x square by 2 plus 3x and limit is from minus 6 to minus 3 plus now integrating x plus 3 we get x square by 2 plus 3x and limit is from minus 3 to 0 and this is equal to minus of now putting the limits we get 9 upon 2 minus 9 putting upper limit minus 3 in place of x we get 9 upon 2 minus 9 minus putting lower limit minus 6 in place of x we get 36 upon 2 plus 18 plus again putting the limits here we get putting upper limit 0 in place of x we get 0 plus 0 minus putting lower limit minus 3 in place of x we get 9 upon 2 minus 9 this is equal to minus 9 upon 2 minus into minus becomes plus so plus 9 now here cancelling out common factor 2 we get 18 in numerator and minus 18 and plus 18 cancels out so we get 0 here so we have plus 0 minus 9 by 2 and minus into minus becomes plus so plus 9 and this is equal to minus 9 by 2 minus 9 by 2 gives minus 9 plus 9 plus 9 gives 18 and this is equal to 9 hence the value of integral limit from minus 6 to 0 modulus x plus 3 dx is equal to 9 this is our answer hope you have understood the solution by and take care