The symbol with limits is shorthand for the total change a rate produces over an interval, and it is read aloud as that. Two rules follow from the antiderivative and are checked here by computing both sides as exact fractions: constants and sums pass through the sign, and an interval can be cut at any interior point. A third, tempting rule is false, and one example refutes it.
Core ideas
Three ideas: how to read the symbol, the two rules that hold, and the rule that does not.
∫ₐᵇ f(t) dt is read aloud as “the integral from a to b of f of t, d t”. It is the number the sums of f approach, and for a polynomial it is F(b) − F(a). The dt recalls the width h in rate times width, and the stretched S recalls a sum.
∫ₐᵇ (c·f + d·g) dt = c·∫ₐᵇ f dt + d·∫ₐᵇ g dt, and for m between a and b, ∫ₐᵐ f dt + ∫ₘᵇ f dt = ∫ₐᵇ f dt. The lab computes both sides of each and says whether they are equal.
∫₀¹ t·t dt = 1/3, but ∫₀¹ t dt times ∫₀¹ t dt is 1/4. The integral of a product is not the product of the integrals.
The material
The notation, the two rules with proofs that use the antiderivative, and the rule that fails.
The phrase “the total change the rate f produces from a to b” has appeared in every lesson of this course, and a short form for it makes the rules below possible to write. The short form is built from the sums. A left sum is f(t)·h added over the pieces; the integral sign is a stretched S for the adding, f(t) is the rate, dt stands where the width h stood, and the limits a and b are the ends.
For a polynomial rate f with antiderivative F, ∫ₐᵇ f(t) dt = F(b) − F(a). It is the total change from a to b of a quantity whose rate is f, a number, and the number the sums of f approach.
The letter t is a placeholder: ∫₀² 3t² dt and ∫₀² 3s² ds are both 8, and neither depends on a variable t.
Read aloud, ∫₀² (3t² + 2t) dt is “the integral from 0 to 2 of 3 t squared plus 2 t, d t”, and in words it is the total change from 0 to 2 of a quantity whose rate is 3t² + 2t. With F = t³ + t², the value is F(2) − F(0) = 12.
For polynomials f and g and numbers c and d, ∫ₐᵇ (c·f + d·g) dt = c·∫ₐᵇ f dt + d·∫ₐᵇ g dt.
Let F′ = f and G′ = g. By the sum and constant rules of “The Derivative as a Function”, (c·F + d·G)′ = c·f + d·g, so c·F + d·G is an antiderivative of c·f + d·g.
The left side is therefore (c·F(b) + d·G(b)) − (c·F(a) + d·G(a)), and regrouping gives c·(F(b) − F(a)) + d·(G(b) − G(a)), which is the right side.
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For m between a and b, ∫ₐᵐ f dt + ∫ₘᵇ f dt = ∫ₐᵇ f dt.
The left side is (F(m) − F(a)) + (F(b) − F(m)). The two F(m) cancel, and what is left is F(b) − F(a).
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Both proofs lean on ∫ₐᵇ f dt = F(b) − F(a), the claim of “The Antiderivative and the Fundamental Theorem”. They are exact for the polynomials the lab accepts, and they are only as strong as that claim. On the scaled preset, 6·∫₀¹ t² dt − 2·∫₀¹ t dt = 6·(1/3) − 2·(1/2) = 2 − 1 = 1, and the lab prints both sides.
Put f = g = t on the interval from 0 to 1. The product is t·t = t², and ∫₀¹ t² dt = 1/3. Each factor has ∫₀¹ t dt = 1/2, and the product of those is 1/4.
1/3 is not 1/4. The proofs above worked because (c·F + d·G)′ = c·f + d·g. For a product, (F·G)′ is f·G + F·g by “The Product Rule”, not f·g, so F·G is not an antiderivative of f·g. The scaled preset already runs from 0 to 1 and its definite tile prints 1/3 for the rate t²; type t in its place and the tile prints 1/2.
Interactive
F, its constant, and F(b) − F(a)
Method
Read first, compute second, and use a rule only when its conditions are in front of you.
Say “the integral from a to b of f of t, d t”. Name the rate, the two limits and the placeholder. The result will be a number.
Reverse the power rule term by term and check by differentiating.
Right end first. That is the value, as an exact fraction.
A sum or a constant multiple: pull the constants out and split the sum. An interior point m: compute ∫ₐᵐ f dt and ∫ₘᵇ f dt and add. Compute at least one by the other route as a check.
If the rate is a product, multiply out and integrate the polynomial. Never multiply two integrals to get the integral of a product.
Compute ∫₀² (3t² + 2t) dt three ways and see that they agree.
Three routes, one fraction. The lab prints each of them: the linear preset shows 12 = 8 + 4 and the split preset 12 = 2 + 10. Agreement is not luck, since the cancellations of the two proofs are exactly what happens in each subtraction.
For a rehearsal, take the scaled preset. The supplied first move is that the coefficients are 6 and −2. Compute ∫₀¹ t² dt and ∫₀¹ t dt separately, combine, and then compute ∫₀¹ (6t² − 2t) dt directly.
You should be able to say ∫ₐᵇ f(t) dt in words, evaluate it as F(b) − F(a), compute both sides of linearity with coefficients and of additivity at an interior point and compare them, and refute the product rule for integrals with one example.
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