By now, we are well-acquainted with the idea
of differentiation and integration.
We know that the derivative of a function
tells us its instantaneous rate of change
and the integral of a function
tells us the area under its graph.
We can see that these two ideas were developed
to tackle two different kinds of problems.
But it turns out
that these ideas are related to each other.
In this video we will see
how the process of differentiation
and integration are opposite to each other.
But wait, there's more
this relation has one major advantage for us.
We have seen earlier
how to find the integral of a function.
First we divide the area into rectangles,
then we find the sum of the areas of the rectangles.
Then to get the area under the graph,
we take the limit of the 'sum of the areas '
as the width of the rectangles tends to zero.
We can see that this process is a bit lengthy
and cumbersome.
We can't do this every time.
So now we will see that
THIS relation comes to our rescue.
It gives us an easier way of finding
the integral of the function.
So let's get started.
Consider a car traveling on a road,
we know that the distance traveled by the car
and its speed are both functions of time.
Let's denote the distance traveled with variable 'Y'
and the speed with 'S' and the time elapsed with 'T'.
Now suppose the distance traveled by the car
is this function of time,
it's the square of 't'.
Can you tell me what will be the speed of the car at
any instant 'T'?
We know that the speed at an instant 'T' is the
instantaneous rate of change of the distance traveled.
So it will be equal to the derivative of the
distance function at the instant 'T'.
And we have seen earlier that the derivative of
this function is equal to '2 times T'.
So this function will give us the
speed of the car at time 'T'.
Its graph will look like this.
Now to see the relation between
differentiation and integration,
consider this integral of the speed function.
Here 'T1' is some arbitrary time instant.
Let's mark it here on this axis
Can you find the value of this integral?
For this we need to find this area under the graph.
But since this is a triangle,
we don't even have to use integration here.
Notice that at 't1' the value of this function
is equal to 'two times T1'.
So the base of this triangle will be equal to 'T1'
and its height will be equal to 'two times T1'
So the area of this triangle will be equal to 'T1 squared'
So we get that this integral from '0 to T1'
is equal to 'T1 squared.'
But notice that according to this distance function,
'T1 squared' is exactly the distance traveled in time 'T1'
So, what does this tell us?
We see that the area under this graph
tells us the distance travelled by the car
Since 'T1' is an arbitrary value of time 'T'.
this is true for any time instant 'T'.
So this integral is the function of time.
This is the connection between
integration and differentiation.
We have a function 'F of T'
and it's differentiation gives us this function 'f prime of T'.
Now integration of this derived function
gives us the function 'F of T' back.
So we see that the process of integration
and differentiation act opposite to each other.
Integration is the reverse of differentiation
and Differentiation is the reverse of integration.
If you remember, we mentioned earlier that
this relation can be used to find the integral
of the function easily.
Can you think of how we can use the
following relation for this?
Let's see that in the next part.
Now we understand that the process of integration
is actually the reverse of differentiation.
If we take the integral of the derivative of the function
'F of T', we get back the function 'F of T'.
Take a moment and understand this.
Observe this integral.
It tells us that for each value of 'T1',
this integral will be equal to the value of the function
F of T at T1.
For an arbitrary time 'T1'
it's equal to this area under the graph.
Now, let's see how this relation helps us
in finding an integral of a function.
Consider this integral of the function '2T'
between 'T equal to A and B'.
Here 'A' and 'B' are just some constant numbers.
Let's say they are these numbers on the 'T' axis.
So this integral will be equal to this area under the graph.
Now can you tell me
what will be the value of this integral?
Notice that this area is equal to the difference
in the areas between this triangle and this one.
Yes, the big triangle minus the small one!
We know that the area of this big triangle
is equal to this integral from '0 to B'.
And the area of this triangle
is equal to this integral from '0 to A'.
But now according to this we don't have to find
the value of these integrals.
Notice these integrals will be equal to this integral
for 'T1' equal to 'A' and 'B'.
So we get the value of this integral is equal to
'F of B' minus 'F of A'.
What will this be equal to now?
Yes, 'B squared' minus 'A squared'.
So we see that using the function 'F of T',
we can easily find any integral of the function
'F prime of T'.
Any integral of the derivative of the function
will be equal to the difference in the value of the
function evaluated at the limits of integration.
This result is true in general and it's called
the Fundamental theorem of Calculus.
Now we can use this result
to find the integral of a function.
Let's see how.
So we know the fundamental theorem of calculus!
Now consider this function 'G of X' equal to 'X squared'.
Let's say we want to find the integral of this function from 'A' to 'B'.
Can you tell me how we can use this result
to find this integral?
Suppose we find a function 'Capital G of X'
such that its derivative is equal to the function 'G of X'.
Since we get that the derivative of 'Capital G of X'
is equal to 'G of X',
we can see that the integral of this function
will be equal to this integral.
Now notice that we can apply this result here.
We will get the integral of 'G of X'
to be equal to 'Capital G of B'
minus 'Capital G of A'.
So we can see that we just have to find the
function 'Capital G of X'.
This function is such that it's derivative
is equal to 'G of X'.
Can you find this function?
Share your thoughts in the comment section below.
Such a function 'Capital G of X'
is called the anti-derivative
or the indefinite integral of the function 'G of X'.
In the next video, we will see what this means
and also find this function 'Capital G of X'.
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