This is absolutely basic, and if anything in calculus deserves an
intuitive, simple explanation, this is it:
(
1)
Yet this is too often presented merely as the antiderivative
of x
^{n},
or is derived using epsilonics, or in a few lines via integration by
parts  with no attempt to motivate it for the easily visualized case
where
n
is
an integer.
(After drawing the
illustrations for this page, and working out
the arguments given on the pyramids and hypercubes pages, I confess I
acquired a lot of sympathy for authors who just "push symbols" to show
this...)
On this page we present a simple motivation for the integral:
It describes the
volume of an n+1
dimensional hyperpyramid
 and that volume is something we can easily determine using
geometry. In short, the result is divided by n+1 because we
can
pack
n+1 pyramids into an
n
dimensional hypercube  so the volume of the cube, which is
r^{n+1},
is divided by n+1 to obtain the volume of a single pyramid.
The
facts we need about hyperpyramids in order to understand this were
developed on our
hyperpyramids
page and our
hypercubes
page.
The
rest of this page will be devoted to going into this in more detail.
We'll give a purely geometric derivation of equation (1) for all
positive integers
n,
and
then, using more traditional arguments, we'll extend the result to
cover fractional powers of
x,
rational powers of
x,
and
finally arbitrary real powers of
x.
1. A False Start: The Usual Visualization as Simple Area
Figure 1:
Integral of x^{0}

Figure 2:
Integral of x^{1}

We have
defined an
integral as the area under a curve. We can use that
definition directly to evaluate the integrals of
x^{0}
and
x^{1}. The area
of the rectangle shown in figure 1 is obviously
r
units; the area of the triangle shown in figure 2 is certainly
r^{2}/2
units
(since it's bounded by the line y=x, the height of the triangle
is
r, as is the base).
We can see at a glance that the area of a square
r
units on a side is
r^{2}
units; and a triangle is, of course, half a square, so the area in
figure 2 is, just as obviously,
r^{2}/2.
Figure
3: Integral of x^{2}

However, when we get to
x^{2}
the "obvious" nature of the area vanishes (figure 3). Since
the height is proportional to
r^{2},
and the base is proportional to
r,
it's not unreasonable that the area under the curve should be
proportional to
r^{3}
 but why is the constant of proportionality
1/3?
This is, of course, easy enough to
prove, but drawing it as the
area under a parabola does very
little to help with visualizing the result (for me, at least).
2. An More Successful Approach: Visualization
as a Pyramid
Rather than try to picture the integral as the
flat
area under a
curve, let's look at the formula itself and see if we can make
something more sensible of it.
Figure 4:
Integral of x^{0}

The expression
x^{n}
is the
nvolume
an
n
dimensional hypercube with length of side
x, as
we discussed
here.
But if we "sweep" an
n
dimensional hypercube along an axis perpendicular to itself, and
grow
it as we do
so, we obtain an
n+1
dimensional
hyperpyramid,
as we discussed
here.
And the volume of that pyramid is exactly what we obtain by
integrating
x^{n}.
Figure 5:
Integral of x^{2}

In figure 4, we show the integral of
x^{0}
as a
1
dimensional pyramid, or line segment. In figure
2,
we already showed the integral
of
x^{1}
as a
2 dimensional
pyramid, or triangle. And in figure 5, we show the integral
of
x^{2}
as an
ordinary
3
pyramid.
As we saw on the
pyramids
page, we can pack
n
ndimensional pyramids in an ndimensional hypercube.
Consequently, the volume of an
n
dimensional pyramid
r
units high, with edges of the base
r
units long, is (1/n)
r^{n}.
But since the integral of
x^{n}
from 0 to
r
is
the volume of an
n+1
dimensional
pyramid, that integral must also be
r^{n+1}/(n+1), which
is
equation (
1).
That's all there is to it, as long as
n is a positive
integer. The rest of this page is devoted to extending the result to
cover more general values for
n.
3. The
Integral of x^{1/n}
Figure
6: Integral of x^{1/n}

We can evaluate the integral of
x^{1/n} by
turning it into an integral of the inverse function,
y^{n}.
See figure 6. The integral of
y^{n},
plus the integral of
x^{1/n}, exactly fill the
rectangle in the figure, which has area
r ⋅
r^{1/n}.
Thus, we have
(3.1)
Turning that around, and plugging in the value we already know from
section
2 for the integral in y, we have:
(3.2)
which was to be shown.
4. The Integral of x^{m/n}
Lemma
1: dx^{n}/dx, n>0
We know that the derivative of an integral is just the function
being integrated, as we saw here.
For a positive integer n,
we know eqn (1)
is true; if we just differentiate eqn (1),
we get,
(L1)
or, replacing n+1
with n and rearranging,
(L2) 
At this point we need to start "pushing symbols". To extend our
results to all positive rational powers of
x, we
substitute:
which gives us,
(4.1)
From Lemma 1, we know
(4.2)
so, plugging that into (4.1) and multiplying out, we have
(4.3)
The integral on the right is a positive integer power of
u,
and we already know how to do that, so we finally have
(4.4)
which was to be shown.
Note that, when
m=1, this covers the case of a fractional
power of
x which we treated separately in
section
3; we didn't actually need to treat that case separately.
However, I like the graphic argument given
above,
even though it doesn't cover all cases; the symbolpushing we used here to
prove the general case of
x^{m/n} seems less
satisfying to me.
5. The Integral of x^{s}, for any positive real s
We've taken this out of order compared with the "usual" approach, so the
first thing we need to do here is
define x^{s}
for an irrational
s.
We can find
x^{s} for any rational value of
s:
For any real
s, we can find rational values
m/n
which are arbitrarily close to
s, and we can find
x^{m/n}
for each of them; we just define
x^{s} to be
the limit of the sequence of values,
x^{m/n}, given
a sequence of rational numbers
m/n which converges on
s.
In turn, the integral of
x^{s} is the limit
of the sequence of integrals of those values.
To make this concrete, given an arbitrary real number
s,
write out its (infinite) decimal expansion:
s =
n.
abcdefg....
We can find an example of a sequence of rational numbers which converges
to
s
by chopping all but the first decimal place for the first term, all but
the the first two places for the second term, and so forth:
n.
a
n.
ab
n.
abc
n.
abcd
....
This
sequence actually converges rather rapidly; on average each term is
1/10 as far from the "final value" as the preceding term. After
the first few terms, the "total change" in all terms for the entire
rest of the series is going to be very small.
The integral of
x^{s}, then, will be defined as the
limit of a sequence of integrals:
It should be apparent that this sequence converges to a fixed value.
If it isn't, pick a value for
s, and draw a few
curves representing the first few functions of the form
x^{n.abcd...}.
Then look at the area under these curves. It should be
clear that, as we add decimal places, the area rapidly converges to a
fixed value: the changes caused by adding more decimal places rapidly
become minuscule.
Furthermore, the values of the integrals given
above, which we can find from the results of the previous section,
converge to exactly the result given by eqn (
1):
The
arguments in this section can be made rigorous without much difficulty,
simply by taking the actual limits in each case, but I don't think it
adds a lot to the clarity of the explanation to do that, so I won't.
(We're already doing far too much "symbol pushing" on this page.)
6. The Integral of x^{s} for Arbitrary
Bounds
We've been discussion the integral from 0 to
r, where
r
is a positive real number.
For any
positive values
a and
b,
from the definition of the integral as the area under the curve, we have:
(6.1)
On the other hand, if either
a or
b is
negative
we may have a problem, in that fractional powers of negative numbers
are not generally real. Dealing with complex integrands is beyond
the scope of this page, so for negative bounds we will only consider
integral powers of
x.
To extend eqn (6.1) to negative bounds as well, all we need is to
determine the integral from 0 to
r where
r
is positive. From the definition of the integral, flipping the
sign of the integrand flips the sign of the integral, just as swapping
the bounds does; so, for integer
n, we have:
(6.2)
7. The Integral of x^{s}, for any positive real
s ≠ 1
I would love to present a good picture for this, but I don't have one.
What's worse, the only proofs I've been able to come up with, and
the only ones I've seen, either require that one know the behavior of
the exponential function, which we haven't dealt with yet, or require
the derivative of
x^{s}, which we
haven't derived yet. As the lesser of two evils, I'll use the
latter, and
assume that
(Lemma 2)
(We will prove it later, when we deal with derivatives.)
We will now proceed with a completely opaque proofbysymbol.
Given:
(7.1)
Substitute:
(7.2)
where
t is a nonzero number whose value we will specify
later. We obtain:
(7.3)
Applying Lemma 2, and collecting terms, we obtain
(7.4)
In order to apply the result of
section (6) to
this, the exponent on
u must be
positive.
Thus, we need to choose
t such that
(7.5)
We can certainly do that, as long as
s ≠
1.
And then, applying our earlier results, we obtain
(7.6)
which was to be shown.
Page
created on 10/29/2007; first posted on 11/04/2007