Video Transcript
In this video, we’re going to learn
how to find indefinite integrals of exponential and reciprocal functions. We’ll begin by recalling the first
part of the fundamental theorem of calculus before looking at how this helps us to
integrate exponential functions of the form 𝑒 to the power of 𝑥 and reciprocal
functions of the form 𝑎 over 𝑥.
We begin by stating the first part
of the fundamental theorem of calculus. We let 𝑓 be a continuous
real-valued function defined on some closed interval 𝑎 to 𝑏. We then let capital 𝐹 be the
function defined for all 𝑥 in this closed interval by capital 𝐹 of 𝑥 equals the
integral evaluated between 𝑎 and 𝑥 of 𝑓 of 𝑡 with respect to 𝑡. Then, capital 𝐹 is uniformly
continuous on the closed interval 𝑎 to 𝑏 and differentiable on the open interval
𝑎 to 𝑏 such that capital 𝐹 prime of 𝑥 is equal to 𝑓 of 𝑥 for all 𝑥 in the
open interval 𝑎 to 𝑏. In other words, capital 𝐹 is the
antiderivative of a function 𝑓, the function whose derivative is equal to the
original function.
And essentially all this tells us
is that integration is the reverse process to differentiation. So, let’s begin by looking at the
function 𝑓 of 𝑥 equals 𝑒 to the power of 𝑥. Well, we know that the derivative
of 𝑒 to the power of 𝑥 is simply 𝑒 to the power of 𝑥. So, the antiderivative of 𝑒 to the
power of 𝑥 is also 𝑒 to the power of 𝑥. And we can, therefore, say that the
indefinite integral of 𝑒 to the power of 𝑥 evaluated with respect to 𝑥 two is 𝑒
to the power of 𝑥. And of course, since we’re working
with an indefinite integral, we add that constant of integration. Let’s call that 𝑐.
But what about the integration of a
function 𝑎 to the power of 𝑥 for a real constant 𝑎? Well, once again, we think about
derivatives. And we recall the fact that the
derivative of 𝑎 to the power of 𝑥 is 𝑎 to the power of 𝑥 times the natural log
of 𝑎. We can therefore say that the
indefinite integral of 𝑎 to the power of 𝑥 times the natural log of 𝑎 is equal to
𝑎 to the power of 𝑥, since that’s its antiderivative. Once again, we’re working with an
indefinite integral, so we add a constant of integration 𝑐.
But this isn’t quite what we’re
after. We were wanting to evaluate the
integral of 𝑎 of the power of 𝑥. So, we take this constant, the
natural log of 𝑎, outside of our integral. And then, we divide both sides by
the natural log of 𝑎. And we’ve obtained the integral of
𝑎 to the power of 𝑥 to be 𝑎 to the power of 𝑥 over the natural log of 𝑎 plus
𝐶. Notice, too, that I’ve written this
constant as capital 𝐶. That’s to demonstrate that we’ve
divided our original constant by another constant, thereby changing this number. Now, it’s useful to know where
these results come from, but generally it’s fine to state them. We’re now going to look at some
examples demonstrating the integration of these exponential functions.
Determine the indefinite integral
of eight 𝑒 to the three 𝑥 minus 𝑒 to the two 𝑥 plus nine over seven 𝑒 to the
power of 𝑥 with respect to 𝑥.
Now, this question might look
really nasty, and you might be starting to consider how a substitution might
help. However, it’s important to notice
that we can simplify the integrand by simply dividing each part of the numerator by
seven 𝑒 to the power of 𝑥, remembering that we can then simply subtract the
exponents. When we do, we’re left with the
indefinite integral of eight-sevenths 𝑒 to the two 𝑥 minus one-seventh 𝑒 to the
𝑥 plus nine-sevenths 𝑒 to the negative 𝑥.
Next, we recall that we can
separate this integral. The integral of the sum of a number
of functions is equal to the sum of the integrals of each function. And we can also take any constant
factors outside of the integral and deal with them later. So, we’re looking at eight-sevenths
of the integral of 𝑒 to the two 𝑥 with respect to 𝑥 minus one-seventh of the
integral of 𝑒 to the 𝑥 with respect to 𝑥 plus nine-sevenths of the integral of 𝑒
to the power of negative 𝑥 with respect to 𝑥. Okay, So what next?
Well, we know that the indefinite
integral of 𝑒 to the power of 𝑥 is 𝑒 to the power of 𝑥 plus 𝑐. But what about the integral of 𝑒
of the power of two 𝑥? You might be wanting to make a
prediction as to what you think this will give. And there is a standard result that
we can quote. But let’s look at this derivation
using integration by substitution. We’re going to let 𝑢 be equal to
two 𝑥. So, d𝑢 by d𝑥 is equal to two. Now, we know that d𝑢 by d𝑥 is
absolutely not a fraction, but we can treat it a little like one for the purposes of
integration by substitution. And we see that a half d𝑢 is equal
to d𝑥.
And then, we see that we can simply
replace two 𝑥 with 𝑢 and d𝑥 with a half d𝑢 and then take out this factor of
one-half. And all we need to do now is
integrate 𝑒 to the power of 𝑢 with respect to 𝑢. Well, that’s simply a half times 𝑒
to the power of 𝑢. But of course, since 𝑢 is equal to
two 𝑥, we can say that the indefinite integral of 𝑒 to the power of two 𝑥 is a
half times 𝑒 to the power of two 𝑥 plus a constant of integration. Let’s call that 𝐴. And this is great because it
provides us with the general result for the integral of 𝑒 the power of 𝑎𝑥 for
real constants 𝑎. It’s one over 𝑎 times 𝑒 to the
power of 𝑎𝑥.
We can use this result. And we see that the integral of 𝑒
to the power of negative 𝑥 is one over negative one times 𝑒 to the power of
negative 𝑥 plus 𝑐. So, popping this altogether, we see
that our indefinite integral is eight-sevenths times a half 𝑒 to the two 𝑥 plus 𝐴
minus a seventh times 𝑒 to the 𝑥 plus 𝐵 plus nine-sevenths times one over
negative one times 𝑒 to the power of negative 𝑥 plus 𝐶. Distributing our parentheses and
combining all our constants, and we’re left with the solution as being four-sevenths
𝑒 to the power of two 𝑥 minus 𝑒 to power of 𝑥 over seven minus nine-sevenths 𝑒
to the power of negative 𝑥 plus 𝐷.
This example demonstrated really
nicely that whilst we can obtain the integral of 𝑒 to the power of 𝑎𝑥 from the
general result for 𝑒 of the power of 𝑥 and a clever substitution. It’s much more sensible to use the
result given. That is the integral of 𝑒 to the
power of 𝑎𝑥 with respect to 𝑥 is one over 𝑎 times 𝑒 to the power of 𝑎𝑥 plus
𝑐. We can even generalize further and
say that the integral of 𝑒 to the power of 𝑎𝑥 plus 𝑏 is 𝑒 the power of 𝑎𝑥
plus 𝑏 over 𝑎.
In our next example, we’re going to
consider how a substitution can help us obtain the result for the integral of 𝑎 to
the power of 𝑏𝑥, where both 𝑎 and 𝑏 are real constants.
Determine the integral of two to
the power of nine 𝑥 with respect to 𝑥.
Let’s begin by quoting what we do
know about the integral of 𝑎 to the power of 𝑥. It’s 𝑎 to the power of 𝑥 divided
by the natural log of 𝑎. Our integrand is slightly different
though; it’s a constant to the power of another constant times 𝑥. So, we’re going to use the process
of introducing something new, a new letter. We let 𝑢 be equal to nine 𝑥, and
of course this is known as integration by substitution. We obtain the derivative of d𝑢
with respect to 𝑥 to be equal to nine. Now, remember, d𝑢 by d𝑥 is
absolutely not a fraction, but we can treat it a little like one for the purposes of
integration by substitution. And we see that we can say that a
ninth d𝑢 equals d𝑥.
We replace 𝑢 with nine 𝑥 and d𝑥
with a ninth d𝑢. And then, we take out this constant
factor of a ninth. And we see that our integral is now
a ninth of the integral of two to the power of 𝑢 with respect to 𝑢. Well, the integral of two to the
power of 𝑢 is two to the power of 𝑢 over the natural log of two. And then, of course, we can use the
definition of our substitution and replace 𝑢 with nine 𝑥. And we’ve found the integral of two
to the power of nine 𝑥 with respect to 𝑥. It’s two to the power of nine 𝑥
over nine times the natural log of two plus this constant of integration 𝐶, which
I’ve made a capital 𝐶 to show that it’s different from the value we had before.
This example gives us a general
result for the integration of 𝑎 to the power of 𝑏𝑥 for real constants 𝑎 and
𝑏. It’s 𝑎 to the power of 𝑏𝑥 over
𝑏 times the natural log of 𝑎 plus that constant of integration 𝐶. We’re now going to consider some
reciprocals. That is, functions of the form 𝑎
over 𝑥. Now, you might initially think that
this is quite straightforward. We can rewrite this function as 𝑎
times 𝑥 to the power of negative one and go from there. But let’s see what happens when we
use the power rule for integration.
We add one to the power. That gives us 𝑥 to the power of
zero. And then, we divide by zero. We can’t do that though. It gives us a number which is
undefined. So, what else can we do? Well, once again, we’re going to
consider a derivative. We recall the derivative of the
natural log of 𝑥 as being one over 𝑥. And this leads us really nicely
into the result that the integral of the one over 𝑥 is the natural log of 𝑥.
But actually, we do need to
redefine this a little. The natural log of 𝑥 can only take
real positive values of 𝑥 is greater than zero. So, we can instead say that the
integral of one over 𝑥 for non-zero values of 𝑥 is the natural log of the absolute
value of 𝑥. And of course, we have that
constant of integration 𝑐. Generalizing for 𝑎 over 𝑥, we
find that the integral of 𝑎 over 𝑥 is 𝑎 times the natural log of the absolute
value of 𝑥 plus 𝑐, of course.
Determine the indefinite integral
of negative two over seven 𝑥 d𝑥.
We’re going to begin by removing
the constant factor from this expression. And that gives us negative
two-sevenths times the integral of one over 𝑥 d𝑥. We then quote the general result
for the integral of one over 𝑥. It’s the natural log of the
absolute value of 𝑥. And so, we obtain our integral to
be negative two-sevenths times the natural log of the absolute value of 𝑥 plus
𝑐. Finally, we distribute our
parentheses. And we find that the solution to
this question is negative two-sevenths times the natural log of the absolute value
of 𝑥 plus capital 𝐶.
And notice here I’ve written
capital 𝐶 to demonstrate that the original constant has been multiplied by negative
two-sevenths, thereby changing it. It’s useful to remember that we can
actually check our solution by performing the reverse process, by
differentiating. The derivative of the natural log
of 𝑥 is, of course, one over 𝑥. So, the derivative of negative
two-sevenths times the natural log of 𝑥 is negative two-sevenths times one over
𝑥. And the derivative of a constant 𝐶
is zero. Multiplying, and we end up with our
derivative to be negative two over seven 𝑥, as required.
Find, if possible, an
antiderivative capital 𝐹 of 𝑓 of 𝑥 equals one over two 𝑥 minus one that
satisfies the conditions capital 𝐹 of zero is one and capital 𝐹 of one is negative
one.
Another way of thinking about the
antiderivative is evaluating the indefinite integral. So, what we’re actually going to do
is integrate one over two 𝑥 minus one with respect to 𝑥 and consider the
conditions on it in a moment. To integrate one over two 𝑥 minus
one, we’re going to use a substitution. We’re going to let 𝑢 be equal to
two 𝑥 minus one, which means that d𝑢 by d𝑥 equals two. Now, we do know that d𝑢 by d𝑥 is
not a fraction, but we do treat it a little like one for the purposes of integration
by substitution and say that a half d𝑢 equals d𝑥.
Let’s perform some
substitutions. We replace two 𝑥 minus one with 𝑢
and d𝑥 with a half d𝑢. And we can, of course, take out the
constant factor of one-half, and we have half times the integral of one over 𝑢. Well, remember, the integral of one
over 𝑥 with respect to 𝑥 is the natural log of the absolute value of 𝑥. So, we can say that the integral of
one over two 𝑥 minus one d𝑥 is a half times the natural log of 𝑢 plus 𝑐. We replace 𝑢 with two 𝑥 minus
one, and we’ve obtained our antiderivative. Distributing the parentheses, we
see that the antiderivative 𝑓 is equal to a half times the natural log of two 𝑥
minus one plus capital 𝐶.
We’re now going to consider
conditions. But we also recall the fact that
two 𝑥 minus one cannot be equal to zero. So, we’re actually going to have a
piecewise function for our antiderivative. The first part of the function
we’re interested in is when two 𝑥 minus one is greater than zero. Solving for 𝑥, and we obtain 𝑥 to
be greater than one-half. The second is when one minus two 𝑥
is greater than zero. Solving for 𝑥 this time, and we
obtain 𝑥 to be less than one-half. Let’s clear some space and
formalize this.
We currently have that our
antiderivative, capital 𝐹 of 𝑥, is equal to a half times the natural log of one
minus two 𝑥 plus some constant when 𝑥 is less than one-half, and a half times the
natural log of two 𝑥 minus one plus some constant when 𝑥 is greater than one
half. We’re now going to use our
conditions on the antiderivative to evaluate 𝑐. The first is capital 𝐹 of zero
equals one. So, when 𝑥 is equal to zero, 𝐹 is
equal to one.
Since 𝑥 is less than one-half,
we’re going to use the first part of our function. We substitute 𝑥 equals zero and
capital 𝐹 equals one in. And we get one equals a half times
the natural log of one minus two times zero. Well, the natural log of one minus
two times zero is the natural log of one, which is of course zero. So, we obtain 𝑐 here to be
one.
The second condition we have is
that when 𝑥 is equal to one, capital 𝐹 equals negative one. This time 𝑥 is greater than
one-half, so we’re going to use the second part of the function. We see that negative one is equal
to a half times the natural log of two times one minus one plus 𝑐. Once again, the natural log of two
times one minus one is zero. So, we see that 𝑐 in the second
part is equal to negative one. And so, the antiderivative capital
𝐹 does indeed exist. Such that capital 𝐹 of 𝑥 is equal
to a half times the natural log of one minus two 𝑥 plus one for 𝑥 is less than
one-half and a half times the natural log of two 𝑥 minus one minus one for 𝑥 is
greater than one-half.
In this video, we’ve seen that we
can use the first part of the fundamental theorem of calculus to integrate
exponential reciprocal functions. We saw that the integral of 𝑒 the
power of 𝑎𝑥 is equal to one over 𝑎 times 𝑒 to the power of 𝑎𝑥. The integral of 𝑎 to the power of
𝑏𝑥 is 𝑎 to the power of the 𝑏𝑥 over 𝑏 times the natural log of 𝑎. And the integral of 𝑎 over 𝑥 with
respect to 𝑥 is 𝑎 times the natural log of the absolute value of 𝑥, given that 𝑥
is not equal to zero. And of course, since we’ve been
working with indefinite integrals, we need to assume that we’ve got this constant of
integration 𝑐 each time.
