Hello, students, let us take some examples on properties. So the first example says integration from minus 1 to 1 bracketed of x bracket of 1 plus sin pi x plus 1 dx.

Now over here in the example, students, we are going to use this property that is property number 3, that is I am going to break this integral into two parts. So let us break this integral. That is first integral is from minus 1 is equal to 0 f(x)dx plus integral from 0 to 1 f(x) dx. Now, students, note over here that in the first integral limits are given from minus 1 to 0. So 1 x is between minus 1 to 0. I am going to calculate sin pi x, so sin pi x will be from minus 1 to 0 because pi x, when x is between minus 1 to 0, pi x will be in third and fourth quadrant. And in third and fourth quadrant, sin pi x is from minus 1 to 0. And hence when I add 1 both the sides it will become 1 plus sin x will be from 0 to 1, substituting this using the bracketed value I add 1 plus sin pi x will be equal to zero. Because 1 plus sin pi x was from 0 to 1, so bracketed value was becoming equal to zero. Now substituting this value in the integral you can see over there the value becomes zero of the bracketed value. And now for the second integral limits are given from 0 to 1, so 1x between 0 to 1, sin pi x, since pi x is from 0 to pi, sin pi x will become 0 to 1. And now when I take, add 1 both the the side I will get 1 to 2, the range will become from 1 to 2. So bracketed of this value will become equal to 1. Now I am going to substitute this value 1 over there in the integral so it will become 1 over there in the integral. Now simplifying it further I will obtain I will be equal to integral minus 1 equal to 0. Now since the inner bracket was become equal to 0 so the integral of bracketed 1, bracket 1 is simple 1, so it becomes 1 dx. And for the second integral, it’s inner bracket has become equal to 1 but x is still over there so it becomes bracket of x plus 1.

Now, students, when I simplify it further, let us check how do we do it. For the first integral the value is very easy to simplify, that is integrating 1, we get x limits minus 1 to 0. For second integral, students, I hope you remember this that bracketed of x plus integer will be equal to bracketed x plus integer. Over here in the second integral bracket x plus 1 is there, 1 is the integer. So when I simplify it becomes 0 to 1 bracketed plus 1 dx. In the integral of bracketed limits are given from 0 to 1. Now since limit is from 0 to 1, bracketed x will become equal to 0. Hence the second integral becomes 0. And now when I simplify it further I get the answer as 2.

So, students, I hope you understood the question. Let’s move further.

Now the second integral, the second question which is given to us is I have to evaluate the value of this integral. You can see it is given in two parts. In both the parts limits are different. In first integral, limit is from minus 4 to minus 5, in second integral limit is from minus 2 to 1. Looking at this I can say we will have to use the substitution which we have studied, x is equal to b minus a into t plus a, that is this property which we studied. I convert both the limits from 0 to 1. So to do that let’s start with the first integral I have taken it as I1. So minus 4 to minus 5 sin of x square minus 3. Now substituting the limit, substituting x equal to b minus a into t plus a, over here b minus a that is upper limit minus lower limit plus a is the lower limit that is minus 4. So simplifying it x becomes minus t minus 4. And now when I substitute it in the given integral I get integral 0 to 1 using the property, students, x is replaced by minus t minus 4, taking the square of it, simplifying it further. When I simplify it further I expand the bracket, right, and then I am simplifying it and I1 becomes 0 to 1, sin of t square plus 8t plus 13. I have taken this integral aside. And I am now going to solve integral second that is I2, which is given as minus 2 to 1 integration of this. Again substituting x equal to b minus a into t plus a. b and a are the upper and the lower limits. So when I substitute the upper and lower limits over here x equals to t minus 2. Substituting this substitution in the integral I get 0 to 1 sin of substitution. And now simplifying it further I get I2 equal to 0 to 1 sin of the integral. And now simplifying it further 0 to 1 it is just a simplification I have done and obtained this.

You can see over here, friends, I1 and I2 are both of opposite signs. So when I add I1 and I2 it becomes equal to 0. So, students, the value of this I1 plus I2 becomes equal to 0.

I hope you understood this, thank you.