Path:  physics insights > basics > Calculus > The Integral of xn

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 xⁿ, 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 rⁿ⁺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 x0

Figure 2: Integral of x1

We have defined an integral as the area under a curve.  We can use that definition directly to evaluate the integrals of x⁰ and x¹.  The area of the rectangle shown in figure 1 is obviously r units; the area of the triangle shown in figure 2 is certainly r²/2 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 r² units; and a triangle is, of course, half a square, so the area in figure 2 is, just as obviously, r²/2.

Figure 3: Integral of x2

However, when we get to x² the "obvious" nature of the area vanishes (figure 3).  Since the height is proportional to r², and the base is proportional to r, it's not unreasonable that the area under the curve should be proportional to r³ -- 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 x0

The expression xⁿ is the n-volume 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 xⁿ.

Figure 5: Integral of x2

In figure 4, we show the integral of x⁰ as a 1 dimensional pyramid, or line segment.  In figure 2, we already showed the integral of x¹ as a 2 dimensional pyramid, or triangle.  And in figure 5, we show the integral of x² as an ordinary 3 pyramid.  

As we saw on the pyramids page, we can pack n n-dimensional pyramids in an n-dimensional hypercube.  Consequently, the volume of an n dimensional pyramid r units high, with edges of the base r units long, is (1/n) rⁿ.

But since the integral of xⁿ from 0 to r is the volume of an n+1 dimensional pyramid, that integral must also be rⁿ⁺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 x1/n

We can evaluate the integral of x^1/n by turning it into an integral of the inverse function, yⁿ.  See figure 6.  The integral of yⁿ, 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: dxn/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 symbol-pushing 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 proof-by-symbol.

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