Reversing the power rule turns a polynomial rate into a function whose derivative is that rate. The total change over an interval of a quantity with that rate is then the value of that function at the right end minus its value at the left end, an exact fraction found without a single sum. The lesson states this as the claim the refining sums of the earlier lessons were supporting, shows why it is plausible, and says what the lab demonstrates and what it does not prove.
Core ideas
Three ideas: what an antiderivative is, the one test that catches a wrong one, and why its two end values give the total.
A function F is an antiderivative of the rate f when F′ = f. The power rule sends tⁿ to n·tⁿ⁻¹; reversing it sends tⁿ to tⁿ⁺¹/(n + 1), one power higher and divided by the new power. For f = t² that is F = t³/3.
Whatever F you write down, differentiate it. (t³/3)′ = 3t²/3 = t² passes. (t³)′ = 3t² does not, and it fails by exactly the factor the division was meant to remove. The check needs no lab and no sum.
If F′ = f, the total change from a to b of a quantity whose rate is f is F(b) − F(a). For t² on the interval from 0 to 1 it is 1/3 − 0 = 1/3, the number the left sums 7/32, 35/128, 155/512 were closing in on.
The material
The reversal, the check, the claim, and the reason the claim is plausible.
The earlier lessons added a rate piece by piece and compared every sum with an exact total printed beside it. That total came from a different method, and this lesson is the method. It runs the derivative backwards: given the rate f, find a function F whose rate of change is f. If you can, the total change the rate f produces over an interval is read from the two ends of F.
A function F is an antiderivative of f when F′ = f. For a polynomial it is found term by term: a term c·tⁿ becomes c·tⁿ⁺¹/(n + 1), and a constant term c becomes c·t.
The power rule of “The Derivative as a Function” is the same rule read from right to left.
Take the rate t². Raise the power to get t³, divide by the new power to get t³/3, and check: (t³/3)′ = 3t²/3 = t². Then F(1) − F(0) = 1/3 − 0 = 1/3. The sums of “Refining the Partition” were 7/32, 35/128, 155/512, each a little nearer a third, and none of them equal to it.
Cut the interval into pieces. The change of F across one piece is F(t + h) − F(t), and those changes add up to F(b) − F(a) exactly, because every interior value is added once and subtracted once. For the ramp 2t, with F = t² and four pieces of width 1/2, the piece changes are 1/4, 3/4, 5/4, 7/4, and they add to 4.
The left sum charges each piece at the rate at its left end, 0, 1, 2, 3, so the pieces contribute 0, 1/2, 1, 3/2, adding to the 3 of “Total Change from a Rate”. Each of those contributions is a flat guess at the matching change of F, and it is a good guess because F′ = f: over a short piece, F changes by about its rate times the width. Shorter pieces make each guess better. That is the whole argument, and it is an argument, not a proof.
If F′ = f on the interval from a to b, the left and right sums of f approach F(b) − F(a) as the number of pieces grows, and F(b) − F(a) is the total change, over the interval, of the quantity whose rate is f.
The identity that the piece changes of F telescope to F(b) − F(a) is algebra and holds exactly. The claim that the sums approach that number is what the tables of the earlier lessons show for particular rates and particular piece counts. Neither the tables nor the lab prove it for every rate; a first analysis course does, and this Subject states it and uses it.
A constant added to F changes nothing here. If G = F + 5 then G(b) − G(a) = (F(b) + 5) − (F(a) + 5) = F(b) − F(a), so the total does not depend on which antiderivative was chosen. Where a constant does matter is the subject of the next lesson.
The rate 3t² − 4t + 1 on the interval from 1 to 3. Term by term, F = t³ − 2t² + t, and the check F′ = 3t² − 4t + 1 passes. Then F(3) = 27 − 18 + 3 = 12 and F(1) = 1 − 2 + 1 = 0.
The total change is 12 − 0 = 12. The lab prints 12 on the mixed preset, and the same value would appear however many pieces you chose, because no pieces were used.
Interactive
F, its constant, and F(b) − F(a)
Method
Five moves, and the third is the one that catches a wrong answer before it is used.
Expand first if you must. A constant is the term with n = 0. Each term is handled alone and the results are added.
c·tⁿ becomes c·tⁿ⁺¹/(n + 1). The division is the step that is easy to drop, because for 2t it is invisible: 2t²/2 = t².
Apply the power rule to your F. If the result is not exactly f, term by term, there is an error in the second move; do not go on.
Compute F(b) and F(a) as fractions, then F(b) − F(a). Keep the order: the right end first.
It is the total change over the interval of the quantity whose rate is f. It is exact, and it is the number the sums of the earlier lessons approach, by a claim this course demonstrates and does not prove.
Find the total change the rate t² produces on the interval from 0 to 1 by an antiderivative, check the antiderivative, and compare the answer with the left sums.
One subtraction gives the number the sums were closing in on, and the gaps shrink the way “Refining the Partition” found, by a factor near two each time the pieces doubled. The antiderivative has no such gap: it is the total, not an estimate of it.
For a rehearsal, take the mixed preset. The supplied first move is that the three terms become t³, −2t² and t. Differentiate your result to see 3t² − 4t + 1 come back, then evaluate at 3 and at 1 and subtract.
You should be able to reverse the power rule on every term of a polynomial, confirm the result by differentiating it, evaluate F(b) − F(a) as an exact fraction, and say why the answer does not depend on a constant added to F and why it is a claim and not a proof that the sums approach it.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.