Practice: Integration and the fundamental theorem

Self-check

Six short questions. Answer each one in your head (or on paper) before opening the collapsible. Trying to retrieve the answer is where the learning sticks; rereading feels productive but does much less.

1. What is the definite integral, and how is it defined?

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∫_a^b f(x) dx is the area under the curve y = f(x) from a to b. It is defined as the limit of Riemann sums: chop [a, b] into n thin strips of width Δx, approximate each by a rectangle of area f(x_i)·Δx, add them, and let n -> ∞ (strips shrink to zero). The sum approaches the true area. It is the precise version of the slice-and-add that opened the track.

2. State the fundamental theorem of calculus.

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∫_a^b f(x) dx = F(b) - F(a), where F is any antiderivative of f (a function with F'(x) = f(x)). To accumulate f over [a, b], find a function whose rate of change is f, evaluate at the endpoints, and subtract. You do not sum rectangles; you run differentiation backward.

3. Why are antiderivatives just “derivative rules run backward”?

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Because an antiderivative of f is a function whose derivative is f, so every derivative rule reverses into an integration rule. The power rule reverses to ∫ x^n dx = x^(n+1)/(n+1) + C; e^x is its own antiderivative; ∫ sin x dx = -cos x + C. Your differentiation knowledge is your integration knowledge, read the other way.

4. What is special about the n = -1 case of the power rule for antiderivatives?

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The formula x^(n+1)/(n+1) divides by zero when n = -1. So the antiderivative of 1/x is not a power; it is ln|x| + C. This works precisely because d/dx(ln x) = 1/x (from the implicit-differentiation lesson). It is why the area under 1/x from 1 to a number is the natural log of that number.

5. What is the difference between a definite and an indefinite integral, and where does + C come from?

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A definite integral ∫_a^b f dx has limits and produces a number (the accumulated total). An indefinite integral ∫ f dx has no limits and produces a function, the general antiderivative F(x) + C. The + C is there because adding any constant leaves the derivative unchanged, so f has a whole family of antiderivatives differing by a constant. In a definite integral the C cancels in F(b) - F(a).

6. Why is the fundamental theorem “the central fact of calculus”?

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Because it welds the subject’s two halves together: it says differentiation and integration are inverse operations. The rate of change of an accumulation is the thing being accumulated, and conversely, to accumulate you reverse the rate. The first lesson glimpsed it on the circle (A(R) = πR² had derivative 2πR, the circumference being accumulated); the theorem says that is no coincidence.

Try it yourself, part 1: compute definite integrals

Pen and paper, about 6 minutes. For each, find an antiderivative and apply F(b) - F(a).

(a) ∫_0^2 x³ dx

(b) ∫_1^3 (1/x) dx

(c) ∫_0^(π/2) cos(x) dx

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(a) Antiderivative of x³ is x⁴/4 (raise the exponent, divide by 4):

∫_0^2 x³ dx = [x⁴/4]_0^2 = 2⁴/4 - 0 = 16/4 = 4

(b) This is the n = -1 case, so the antiderivative is ln|x|, not a power:

∫_1^3 (1/x) dx = [ln|x|]_1^3 = ln 3 - ln 1 = ln 3 ≈ 1.099

(c) Antiderivative of cos x is sin x:

∫_0^(π/2) cos(x) dx = [sin x]_0^(π/2) = sin(π/2) - sin 0 = 1 - 0 = 1

The area under one quarter-wave of cosine, from 0 to its first zero, is exactly 1.

Try it yourself, part 2: an indefinite integral (and check it)

About 4 minutes. Find the indefinite integral ∫ (4x³ + e^x - 1/x) dx (include the + C), then verify your answer by differentiating it back.

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Integrate term by term, reversing each derivative rule:

∫ 4x³ dx = 4·(x⁴/4) = x⁴
∫ e^x dx = e^x
∫ (1/x) dx = ln|x|        (the n = -1 case)
result:  x⁴ + e^x - ln|x| + C

Check by differentiating back: d/dx(x⁴ + e^x - ln|x| + C) = 4x³ + e^x - 1/x + 0, which is exactly the integrand. (The + C differentiates to 0, which is why it does not show up in the check, and why it must be written: a whole family of functions has this same derivative.)

Flashcards

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Q. What is the definite integral, and how is it defined?

∫_a^b f(x) dx is the area under f from a to b, defined as the limit of Riemann sums: thin rectangles f(x_i)·Δx summed as their width shrinks to zero. The precise version of slice-and-add.

Q. State the fundamental theorem of calculus.

∫_a^b f(x) dx = F(b) - F(a), where F is any antiderivative of f (F' = f). To accumulate f, find a function whose rate of change is f, evaluate at the endpoints, and subtract.

Q. What does the integral notation mean, piece by piece?

∫ is an elongated S for “sum”; dx is the width of a strip shrinking to zero. So ∫_a^b f(x) dx reads “sum f(x) times a tiny width dx, across x from a to b, in the limit as the width vanishes.”

Q. What is the antiderivative of x^n?

∫ x^n dx = x^(n+1)/(n+1) + C, for n ≠ -1 (raise the exponent, divide by the new exponent). Check by differentiating back. The n = -1 case is the exception.

Q. What is the antiderivative of 1/x, and why is it special?

∫ (1/x) dx = ln|x| + C. The power-rule formula x^(n+1)/(n+1) divides by zero at n = -1, so 1/x is the exception. It works because d/dx(ln x) = 1/x.

Q. Definite versus indefinite integral?

Definite ∫_a^b f dx: has limits, gives a number (the accumulated total). Indefinite ∫ f dx: no limits, gives a function F(x) + C (the family of antiderivatives). The + C cancels in a definite integral.

Q. Why is there a + C on an indefinite integral?

Because adding any constant leaves the derivative unchanged (a constant’s derivative is 0), so f has infinitely many antiderivatives differing by a constant. Omitting C falsely claims there is only one.

Q. Compute ∫_0^1 x² dx with the fundamental theorem.

Antiderivative of x² is x³/3, so ∫_0^1 x² dx = [x³/3]_0^1 = 1/3 - 0 = 1/3. (The Riemann sums 0.469, 0.385, 0.338 for n = 4, 10, 100 confirm they approach 1/3.)

Q. Why does integration matter in machine learning?

It is the math of continuous probability: a density integrates to 1, P(a≤X≤b) = ∫_a^b f dx, expected value is ∫ x·f(x) dx, and entropy/KL divergence (in generative-model losses) are integrals. Continuous-time models (neural ODEs, diffusion) solve integrals in the forward pass.