Evaluation of Definite Integral Video Lecture | Mathematics (Maths) Class 12 - JEE

Professor Dave here, let’s look at indefinite
integrals.
We now know how to evaluate definite integrals,
by evaluating antiderivatives over a specific interval.
But sometimes we will be looking at integrals
with no limits of integration listed.
These are called indefinite integrals.
The main difference is that definite integrals
are numbers, like the area under a curve for
a specific interval.
But indefinite integrals, because there is
no specified interval over which to evaluate,
are actually functions, or families of functions.
This may seem totally different than what
we have described integration to be thus far,
but if we think about it, with definite integrals,
we were simply evaluating the antiderivative
over some arbitrary interval.
If that interval can be literally anything,
and we can compute absolutely any definite
integral from the antiderivative, then it
follows that the antiderivative, the function
itself, is the integral of the integrand in
the truest sense.
In short, we started off wanting areas under
curves, and found that we could get these
from antiderivatives, so we’re now just
considering the antiderivatives themselves.
Since indefinite integrals still involve the
same process of finding the antiderivative
that evaluating definite integrals requires,
it won’t be that much different, but there
are some important things to mention regarding
indefinite integrals, so let’s go over these now.
As we said, indefinite integrals are ones
with no limits of integration indicated.
This means that it won’t be evaluated to
get a number, but rather we are integrating
some function, f of x, to find some new function.
This new function will be any antiderivative
of f of x, which we can call upper case F
of x, such that big F prime will equal f of x.
We must make sure to understand that when
we say “any” antiderivative of f of x,
this is because there are indeed infinitely
many functions that qualify as antiderivatives,
because although the part of the antiderivative
with the variable must be a certain way, we
can also include any constant whatsoever.
One, ten, pi, a billion, any real number.
This is because when we take the derivative
of this function, any constant will simply
go away, leaving us with f of x.
So if we are integrating x squared, we will
get x cubed over three, as we expected, but
we also write plus C, meaning that any constant
may be present.
We didn’t care about this when we were evaluating
definite integrals, because whatever this
constant is, it would be present when evaluating
at both the upper limit and lower limit, which
means it will simply subtract, and therefore
not affect our result.
But if we are looking at indefinite integrals,
where we are simply writing a function, the
notion of this constant does indeed become
relevant, and we must write it.
Just to clarify one more time, when we take
the derivative of this antiderivative, x cubed
over three becomes x squared, as expected,
and the constant will go away, giving us the
original function, so truly any constant could
have been present.
The properties of indefinite integrals will
be precisely the same as those for definite
integrals that we already went over, so just
to review, the integral of a constant times
some function will be equal to that constant
times the integral of the function, so make
sure to remember that constants can be pulled
out of integrals.
Let’s also recall that integrals of sums
and differences can be split up into sums
and differences of integrals, and this property
will be very useful in certain circumstances.
The integral of a constant with respect to
some differential, like k dx, will always
be the constant times the variable, like kx
here, and then plus C, for the reasons we
mentioned earlier.
And lastly, the property you will be using
most frequently, the integral of (x to the
n) dx will be equal to x to the (n plus one)
over (n plus one).
Remember, this is just the precise opposite
process of taking a derivative, so that taking
the derivative of the integral gives us back
our original function, just as the fundamental
theorem of calculus demands.
We have already practiced finding antiderivatives
of polynomials when we used them to evaluate
definite integrals, so just one or two practice
examples should be enough.
Let’s try root x times the quantity x minus two.
To be clear, there is no product rule for
integration, so this won’t work the way
we are looking at it.
We have to express this some other way.
What if we change root x into x to the one
half, and then distribute that across this
other term?
That would give us (x to the three halves)
minus two (x to the one half).
This may look a little less appealing, but
it doesn’t matter, because now it is possible
to find the antiderivative, so this is the
way we have to go.
Three halves plus one is five halves, so we
get x to the five halves over five halves,
or two fifths x to the five halves.
One half plus one is three halves, so for
this term we get minus two times x to the
three halves over three halves, or two times
two thirds x to the three halves, which is
four thirds x to the three halves.
And lastly, as we discussed, if this were
a definite integral, we could just evaluate
this antiderivative over the limits of integration,
in which case we don’t have to worry about
a constant, because any constant would disappear
during the subtraction.
But as this is an indefinite integral, we
do indeed have to write plus C at the end
of this, because any constant can be here
and this expression would be valid.
How about another one.
One over e to the negative x, plus one over three x.
At first glance this seems complicated, but
it’s actually quite straightforward.
First, since this is the integral of a sum,
let’s break if up into a sum of integrals,
to make things easier.
Then, looking at the first one, if e is being
raised to a negative exponent and in the denominator,
let’s just bring it up to the numerator
and make the exponent positive.
That will be e to the x.
Then for this other term, this is the same
thing as one third times one over x.
From here, to integrate, we just have to remember
a couple things about the derivatives of special terms.
When we looked at the derivatives of logarithmic
and exponential functions, we learned the
incredible fact that the derivative of (e
to the x) is (e to the x).
It is the only function that does not change
upon differentiation.
But that means that if the derivative of (e
to the x) is (e to the x), then the integral
of (e to the x) is also (e to the x).
Whether we are going up or down on the ladder,
(e to the x) doesn’t change.
So when finding this antiderivative, we can
leave (e to the x) as it is.
Then, for this other one, we must recall that
the derivative of the natural log of x is
one over x.
That also means that the antiderivative of
one over x is the natural log of x.
So this integral will be one third ln x.
And lastly, we add the constant, C, at the
end.
So sometimes we will have to identify special
terms like these when integrating.
That (e to the x) remains (e to the x) is
easy to remember.
That one over x becomes natural log of x isn’t
so obvious, as x to the negative one seems
like it would follow the regular rule we use
for any other exponent, but it is indeed the
one exception to that rule.
We don’t get x to the zero over zero, as
that would be undefined, instead we get natural
log of x, so you’ll have to memorize this
one, and get used to spotting it within a
more complicated function.
So we should now understand the difference
between definite and indefinite integrals.
Both involve taking the antiderivative of
a function, but a definite integral will evaluate
that antiderivative at the upper and lower
limits of integration, subtract them, and
give a definite answer for the area under
the curve over this particular interval.
So a definite integral is a number.
An indefinite integral will not be a number,
but rather a function, the antiderivative
of the integrand, with the constant C listed
so that this function represents all the possible
antiderivatives.
We have practiced evaluating definite and
indefinite integrals for polynomials, but
there are many other types of functions we
will have to integrate, so let’s look at
some of those next, but first, let’s check
comprehension.