1
The power rule reversed has nothing to say about 1/t
1/t is t⁻¹, and the reversal would give t⁰/0. A division by zero is not an antiderivative. No polynomial has derivative 1/t, because a polynomial's derivative never contains a term in t⁻¹.
Lesson · Accumulation and the Integral · Notation and a second integral
The Integral of 1/t
The power rule reversed divides by one more than the power, and for one over t that is a division by zero. The sums still work: left, right and trapezoid sums on the interval from 1 to 2 are exact fractions that bracket a number the lab can print only as a rounded decimal. The lesson states that the number is ln 2 and, more generally, that the integral from 1 to x of one over t is ln x, as a claim the sums support.
tⁿ → tⁿ⁺¹/(n + 1) fails for n = −1
L₄ = 319/420, R₄ = 533/840 on [1, 2]
ln 2 ≈ 0.693147 sits between R₄ and L₄
claim: ∫₁ˣ dt/t = ln x
every sum is a fraction; ln 2 is not
The one integral the power rule cannot reach
Core ideas
What this lesson establishes
Three ideas: why the usual reversal fails, what the sums do instead, and what the number is called.
2
The sums bracket a number that no fraction equals
On the interval from 1 to 2 with four pieces, the left sum is 319/420 ≈ 0.759524 and the right sum is 533/840 ≈ 0.634524. The rate falls, so the total lies between: ≈ 0.693147, rounded.
3
The number is ln x, as a claim
The integral from 1 to x of 1/t is written ln x, the natural logarithm. The lab prints it rounded and labelled. That the sums approach it is demonstrated here, not proved.
The material
A rate with no polynomial antiderivative
The failed reversal, the exact sums, what refining them shows, and a scaling that makes the logarithm's addition rule visible.
Every integral so far had an antiderivative from the reversed power rule: raise the power, divide by the new one. Try it on 1/t = t⁻¹. The new power is 0 and the division is by 0. Nothing is wrong with the rate: 1/t is a perfectly good falling rate on the interval from 1 to 2, and the sums of the earlier lessons apply to it unchanged. What fails is the shortcut.
1/t on [1, 2], n = 4, h = 1/4
left ends: 1 5/4 3/2 7/4
rates: 1 4/5 2/3 4/7
L₄ = (1/4)·(1 + 4/5 + 2/3 + 4/7) = 319/420 ≈ 0.759524
R₄ = (1/4)·(4/5 + 2/3 + 4/7 + 1/2) = 533/840 ≈ 0.634524
T₄ = (L₄ + R₄)/2 = 1171/1680 ≈ 0.697024
The rate falls, so by “Left and Right Sums” the total lies between the right sum and the left sum, and the window has width h·(f(1) − f(2)) = (1/4)·(1/2) = 1/8. The lab's exact-total tile prints ≈ 0.693147 (ln 2, rounded). The ≈ and the word “rounded” are there because that figure is a double-precision evaluation, printed to six figures; the sums are exact and the total is not.
Why no refinement ends in a fraction
Refine the pieces and the left sum drops: n = 8 gives 52279/72072 ≈ 0.725372, and n = 16 gives ≈ 0.709016. Each is a different exact fraction, with a larger denominator, and each is above the target by about half what the last one was. The target ln 2 is not a fraction. That is a known fact of mathematics and is not proved here; what the lab shows is that no sum it can print is equal to the target.
Theorem · The integral of 1/t
For x > 1 the left and right sums of 1/t on the interval from 1 to x approach a number written ln x, the natural logarithm, and ∫₁ˣ dt/t = ln x.
This is a claim, supported by the brackets above and by the presets for x = 3 and x = 4, and not proved. Here ln x is the natural logarithm of “Common and Natural Logarithms” in Algebra's Exponential and Logarithmic Functions, the power to which e is raised to give x; that accumulating 1/t produces it is the claim.
Theorem · Scaling leaves the sums of 1/t unchanged
For every n, the left sum of 1/t on the interval from 2a to 2b with n pieces equals the left sum on the interval from a to b with n pieces.
On the interval from a to b a piece has left end t and width h, contributing (1/t)·h. On the doubled interval the matching piece has left end 2t and width 2h, contributing (1/(2t))·2h = (1/t)·h.
Each piece contributes the same amount, so the two sums are equal term by term.
□
Example · The sum for 2 to 4 is the sum for 1 to 2
Select the scaled preset: 1/t with limits 2 and 4 and four pieces. The left ends are 2, 5/2, 3, 7/2, the rates 1/2, 2/5, 1/3, 2/7, the width 1/2, and the sum is (1/2)·(1/2 + 2/5 + 1/3 + 2/7) = 319/420, exactly the sum for 1 to 2.
So the total from 2 to 4 equals the total from 1 to 2, and by additivity the total from 1 to 4 is twice it. The ln4 preset agrees: with six pieces, L₆ = 223/140 ≈ 1.59286 and R₆ = 341/280 ≈ 1.21786, and between them the lab's total ≈ 1.38629, which is 2·ln 2, rounded.
That is the addition rule of logarithms, ln 4 = 2·ln 2, appearing as an exact equality of fractions for every n, with only the limit left as a claim.
Interactive
Sums of a rate
Left, right, trapezoid and midpoint sums as fractions
Method
Bracketing the integral of 1/t
The method is the one of the earlier lessons in this course. What changes is the last move, because the answer is a rounded number.
1
Check the interval avoids t = 0
1/t has no value at 0. The lab refuses a pole inside the interval, and so should you.
2
Choose the pieces and list the rates as fractions
h = (b − a)/n; the left ends; then 1/t at each, kept as a fraction.
3
Form the left and right sums
Add the rates, multiply by h. For a falling rate the left sum is the larger. Their difference is h·(f(a) − f(b)) in size, a check on both.
4
State the bracket
Right sum ≤ total ≤ left sum, as exact fractions.
5
Quote the total rounded and labelled
The total is ln(b/a): print it with ≈, say it is rounded, and never write it as a fraction or as a bare decimal.
Sums that bracket a logarithm
Score 0 / 0Choose the answer you can justify.
The integral of 1/t from 1 to 2, bracketed
Compute the left, right and trapezoid sums of 1/t on the interval from 1 to 2 with four pieces, and place ln 2 among them.
n = 4, h = 1/4
rates at left ends: 1, 4/5, 2/3, 4/7
L₄ = (1/4)·(319/105) = 319/420
R₄ = L₄ − h·(f(1) − f(2)) = 319/420 − 1/8 = 533/840
T₄ = (319/420 + 533/840)/2 = 1171/1680
rounded: L₄ ≈ 0.759524, R₄ ≈ 0.634524, T₄ ≈ 0.697024
ln 2 ≈ 0.693147, between R₄ and L₄
The window is 1/8 wide, and the target is nearer the trapezoid sum than either end. Nothing in the sums, however fine, makes the target a fraction; the lab keeps it as a labelled decimal for that reason.
For a rehearsal, take the ln3 preset, which runs from 1 to 3. The supplied first move is that the left ends are 1, 3/2, 2, 5/2. You should find L₄ = 77/60 and R₄ = 19/20, around ln 3 ≈ 1.09861.
Common mistakes
1
Applying the power rule to 1/t and writing t⁰/0The rule sends tⁿ to tⁿ⁺¹/(n + 1), and for n = −1 the denominator is 0. The expression has no value, so it names no function. The honest response is the one the lab takes: sum the rate, bracket the total, and report that it is the number written ln 2.
2
Writing the rounded total as if it were exact0.693147 is not ln 2; it is ln 2 rounded to six figures, and 693147/1000000 is not ln 2 either. The lab prints ≈ 0.693147 with the label “rounded”, and the sentence that quotes it should keep the ≈.
3
Concluding that the total is a fraction because every sum isEach of 319/420, 52279/72072 and the rest is an exact fraction, and each is a sum, not the total. A list of fractions can close in on a number that is none of them, which is what happens here; the total is the number they approach, not one of them.
Check your understanding
✓
Finish when you can bracket the integral of 1/t with exact sums and report it rounded and labelled.
You should be able to say why the reversed power rule fails for 1/t, compute the left, right and trapezoid sums of 1/t on an interval as exact fractions, bracket the total, quote it with ≈ and a rounding label, and state the claim ∫₁ˣ dt/t = ln x as a claim.
Note: That ends the calculus the Subject needs: a rate can be differentiated and totalled, and the totals are exact whenever a polynomial is involved, with 1/t bracketed and named. The next course, Differential Equations and Euler's Method, turns the question around and asks for a quantity given only a rule for its rate.