Look at the information given on the screen.
Can you tell me the value of this integral?
It will be equal to this.
We've seen this result in the previous video.
It tells us that
integration is just the reverse of differentiation.
Integral of the derivative of the function f of X
is just equal to the difference in the function "f of X"
evaluated at the limits of integration.
Now in this video,
we will understand how to apply this result
to find the integral of a funtion.
Consider this function "G of X".
Let's find the integral of this function from 'A' to 'B'
Can you think how we can apply this result
to find the integral ?
Let's see.
Suppose we find a function "capital G of X" such that
its derivative is this function "G of X".
Then we will get the integral of G of X is equal to
the integral of the derivative of "capital G of X".
Now according to this result we can see that
this integral will be equal to this.
So will see that the integral of the function G of X
is equal to the difference between
the values of the function
"capital G of X".
Do you know any such function?
Let's try to find the function "capital G of X".
So this is what we have.
We need to find the function "capital G of X".
Now we have seen earlier
that if we have a function "f 1 of X equal to X",
then it's derivative is equal to 1.
If we have a function "f 2 of x equal to x squared"
then its derivative is equal to "2 times X".
Notice that each time we take the derivative,
the power of X is reduced by 1.
Now we want the function whose derivative,
is equal to 'X square'.
So maybe the derivative of the function "f 3 of X"
equal to "X cubed" will give us this function.
By now, we are familiar
with how to find the derivative of a function.
Can you find the derivative of "f 3 of x"?
The derivative of 'f 3 of X' will be given by this limit.
This is something we've seen before.
We know the formula
for the cube of sum of two numbers
and the cube of difference of two numbers.
Using this here, we can easily find
that the derivative will be equal to "3 times x squared".
So we get the derivative of "X cubed"
to be three times 'x squared'.
But we want the derivative of this function
to be only 'x squared'.
Okay, so what if we consider the function
capital "G of X equal to X cubed over 3"?
What will be its derivative?
Now for this function this average rate
will get multiplied by 1 over 3.
So we will get the derivative to be equal to "x squared".
So we get the function "capital G of X"
to be equal to "X cubed over 3".
Now to find the value of the integral,
we just have to find this difference.
But wait..
What if we add a constant let's say 1,
to this function "capital G of X"
What will be the derivative of the function
"capital G 1 of X"?
Let's continue this in the next part.
So what will be the derivative of the function
"capital G 1 of X"?
Yes, notice that the function 'capital G 1 of X'
is just the function capital "G of X plus 1".
We know that the derivative of a function tells us
its instantaneous rate of change.
So adding a constant to a function
will not affect its rate of change.
We can find its derivative by taking the limit of
its average rate of change.
We will surely get it to be equal to
the derivative of "capital G of X".
So now we have two functions whose derivative
is equal to the function "G of X", that is "x squared".
So what if to find this integral we use the function
"capital G 1 of X" instead of "capital G of X"?
Notice that since this function is just
"capital G of X" plus a constant,
we will again get that same value for the integral.
That means we can use any of these two functions
to find the value of the integral.
We found these functions by going in the
reverse direction.
That is we started with the derivative "G of X" and found
the original functions whose derivative will be 'G of X'.
So these functions are called antiderivatives
of the function "G of X".
Now let me ask you a question.
Are there any other antiderivatives of "G of X"?
Correct! Any function which is equal to "G of X"
plus some constant
will be the antiderivative of 'G of X'.
Now we can use any of these antiderivatives
to find the integral of the function.
We will always get the value of this integral
to be equal to capital "G of B" minus capital "G of A".
So for this reason we denote all these antiderivatives
by one general antiderivative.
Capital "G of X plus a constant C" called
the indefinite integral of "G of X".
We denote it by the integral symbol without
the limits of integration.
We can see that once we know the indefinite integral
we can easily find this integral.
The indefinite integral of 'x squared' is equal to
"X cubed over 3 plus a constant C".
And evaluating it at the limits of integration we will get this.
Notice that for this integral
we will get a number as our answer.
Whereas for the indefinite integral we get a function.
So this integral with the limits of integration
is called as the definite integral.
Let's quickly review the whole process.
To find the definite integral of the function,
we first find its indefinite integral.
Now to find the indefinite integral of the function "G of X",
we find a function capital "G of X"
such that its derivative is equal to "G of X".
And now to find the definite integral
we just have to evaluate it at the limits of integration.
We can see that once we know the indefinite integral,
we can find any definite integral of the function.
So usually the indefinite integral is itself
called the integral of the function.
So we see that related to a function
we have two types of integrals,
indefinite integral and definite integral.
Now let's see if you can find the indefinite integral of
this function "H of X" equal to "x squared plus 6".
How can we do that?
The indefinite integral of "H of X",
will be given by a function capital "H of X plus C".
The function capital "H of X" should be such that
its derivative is equal to "H of X".
Can you find the function "capital H of X"?
Tell us your answers in the comment section below and
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