Formula & theorem reference

The vertical line x = a consists of exactly the points whose first coordinate is a — so the curve's intersections with that line are exactly the points (a, y) the curve assigns to input a.

If the curve is the graph of a function, input a has exactly one output, so x = a meets it in at most one point — for every a. Conversely, if some vertical line x = a met the curve twice, at (a, y_1) and (a, y_2) with y_1 \neq y_2, that would mean input a produced two different outputs, which is precisely what the definition of a function forbids. So the curve fails to be a function exactly when some vertical line crosses it more than once.

A real nth root \sqrt[n]{x} is, by definition, a real number y with y^n = x. When n is even, y^n \ge 0 for every real y — an even power is never negative — so x must be \ge 0 for such a y to exist. That's the bare-radical, n-even case. When n is odd, y^n ranges over all of \mathbb{R} as y does, since odd powers preserve sign and grow without bound in both directions, so every real x has a real nth root — the bare-radical, n-odd case.

The two denominator cases layer the ordinary "a denominator can't be zero" restriction on top of that: the boundary g(x)=0 is excluded whenever the radical sits under a fraction, regardless of n's parity, since \sqrt[n]{0}=0 either way.

The horizontal line y = k meets the graph of f exactly at the points (x, k) with f(x) = k — one point for every input that produces output k.

If f is one-to-one, then f(a) = f(b) forces a = b, so no two distinct inputs share an output: every horizontal line meets the graph at most once. Conversely, suppose some horizontal line y = k met the graph twice, at (a, k) and (b, k) with a \neq b. Then f(a) = k = f(b) with a \neq b — two different inputs sharing an output, which is exactly the negation of one-to-one. So f fails to be one-to-one exactly when some horizontal line crosses its graph more than once.

Both scales measure the same full rotation, just in different units: 360^\circ = 2\pi \text{ rad}. Divide both sides by 360 to find what one degree equals in radians: 1^\circ = \frac{2\pi}{360} \text{ rad} = \frac{\pi}{180} \text{ rad}. Multiplying by \theta_{\text{deg}} scales this up to any angle: \theta_{\text{rad}} = \theta_{\text{deg}} \cdot \frac{\pi}{180}. The reverse conversion is the reciprocal factor, \frac{180}{\pi}.

Each law reduces to the same style of argument used to make a limit claim rigorous (see this course's section on the precise definition of a limit): to bound |f(x)+g(x) - (L+M)|, use the triangle inequality to split it as |f(x)-L| + |g(x)-M|, then make each piece small enough by staying close enough to a to satisfy both limit definitions at once. The product and quotient laws need one more step — bounding f(x) near a so a factor of g(x)-M doesn't get multiplied by something unbounded — but the shape of the argument is identical. This is the precise version of the intuition "if two things are close to their targets, arithmetic combinations of them are close to the combined target."

For any tolerance around L, the definitions of the two outer limits each give a window around a that traps g(x) and h(x) within that tolerance; take the smaller of the two windows so both hold at once. Since g(x) \leq f(x) \leq h(x), f(x) is trapped in that same window — it has nowhere else to go. (See this course's section on the precise definition of a limit for this argument made fully rigorous.)

For any tolerance around L, the definitions of the two outer limits each give a window around a that traps g(x) and h(x) within that tolerance; take the smaller of the two windows so both hold at once. Since g(x) \leq f(x) \leq h(x), f(x) is trapped in that same window — it has nowhere else to go. (See this course's section on the precise definition of a limit for this argument made fully rigorous.)

Write p(x) = (x-c)^kP(x) and q(x) = (x-c)^nQ(x) with P(c) \neq 0 and Q(c) \neq 0. For x \neq c, cancel the shared factor: f(x) = \frac{(x-c)^kP(x)}{(x-c)^nQ(x)} = (x-c)^{k-n}\,\frac{P(x)}{Q(x)}.

k \geq n: the hole. The exponent k - n is \geq 0, so (x-c)^{k-n}\frac{P(x)}{Q(x)} is an ordinary polynomial-or-rational expression that's perfectly defined at x = c (recall Q(c) \neq 0). f agrees with this reduced expression at every x \neq c, and is simply undefined at the single point x = c where the original q(x) vanished — a removable gap, i.e. a hole.

k < n: the asymptote. Here f(x) = \dfrac{P(x)/Q(x)}{(x-c)^{n-k}} with n - k > 0. As x \to c, the numerator \frac{P(x)}{Q(x)} approaches the nonzero number \frac{P(c)}{Q(c)}, while the denominator (x-c)^{n-k} shrinks toward 0 — a nonzero number divided by something shrinking to 0 grows without bound, so |f(x)| \to \infty.

For the sign: (x-c)^{n-k} is positive on both sides of c when n-k is even (the sign-erasing behavior an even power always has), so f blows up toward the same infinity on both sides. When n-k is odd, (x-c)^{n-k} carries the sign of (x-c) itself — negative to the left of c, positive to the right — so f blows up toward opposite infinities on the two sides.

The idea is intuitive — a graph that starts below the x-axis and ends above it, without ever jumping, has to cross somewhere — but making "without ever jumping" precise is a statement about the real numbers themselves. A rigorous proof needs the completeness property of \mathbb R (typically via the least-upper-bound axiom, showing the crossing point is the supremum of \{x \in [a,b] : f(x) < 0\}), which belongs to a real analysis course, not this one. It's taken on faith here — a polynomial is continuous (its graph really is one unbroken curve), and continuity is exactly the property the sign-change test relies on.

Continuity on the whole closed interval is what does the work — it rules out the graph ever jumping over a height without passing through it. The polynomial-only version from College Algebra (the sign-change test, N = 0) is the special case where f(a) and f(b) have opposite signs; this is that same argument with the hypothesis weakened from "polynomial" to "continuous," and the conclusion widened from "crosses zero" to "hits any value in between." A rigorous construction (typically the least-upper-bound / completeness argument, showing c is the supremum of \{x \in [a,b] : f(x) < N\}) belongs to real analysis; here it's enough to know continuity is the ONLY property the proof actually needs.

For x\neq a near a, f(x)-f(a) = \frac{f(x)-f(a)}{x-a}\cdot(x-a). Taking the limit as x\to a, the first factor approaches f'(a) (a finite number, by hypothesis) and the second factor approaches 0, so their product approaches f'(a)\cdot 0 = 0. That means \lim_{x\to a}f(x) = f(a), which is exactly continuity at a.

The converse is false: f(x)=|x| is continuous at 0 but not differentiable there, since the left-hand and right-hand difference quotients approach -1 and 1 respectively instead of agreeing. A corner, a cusp, or a vertical tangent can each break differentiability while leaving continuity untouched — differentiability is the strictly stronger condition of the two.

Everything traces back to one limit, \displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1, plus its companion \displaystyle\lim_{x\to 0}\frac{\cos x - 1}{x}=0 (provable the same way). For f(x)=\sin x, the difference quotient is \frac{\sin(x+h)-\sin x}{h}=\frac{\sin x\cos h+\cos x\sin h-\sin x}{h} (angle-sum identity) =\sin x\cdot\dfrac{\cos h-1}{h}+\cos x\cdot\dfrac{\sin h}{h}. As h\to 0 the first term vanishes and the second approaches \cos x, giving \frac{d}{dx}[\sin x]=\cos x. \cos x's derivative follows from a cofunction identity applied to the same limit. The remaining four all reduce to the quotient rule applied to \tan x=\sin x/\cos x and its reciprocals, using only the two derivatives just found.

Differentiate both sides of f(f^{-1}(x))=x with respect to x using the chain rule: f'(f^{-1}(x))\cdot(f^{-1})'(x) = 1. Dividing by f'(f^{-1}(x)) — valid exactly because that quantity is f'(a)\neq0 by hypothesis — isolates (f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))}.

This formula never needs an explicit closed-form expression for f^{-1} itself — only f' evaluated at the ALREADY-KNOWN point a where f(a)=b. The hypothesis f'(a)\neq0 is essential: geometrically, reflecting a horizontal tangent line (f'(a)=0) across y=x produces a vertical tangent line on f^{-1}'s graph, where the derivative fails to exist.

Each follows from implicit differentiation on the inverse relation plus one Pythagorean identity. For y=\arctan x, so \tan y=x: differentiating implicitly gives \sec^2y\cdot\dfrac{dy}{dx}=1, so \dfrac{dy}{dx}=\dfrac{1}{\sec^2y}. Since \sec^2y=1+\tan^2y=1+x^2, \frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}. The same move on y=\arcsin x (\sin y=x, \cos y\cdot y'=1, and \cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2} for y in \arcsin's range) gives \dfrac{d}{dx}[\arcsin x] = \dfrac{1}{\sqrt{1-x^2}}, and \arccos x = \tfrac{\pi}{2}-\arcsin x immediately gives \dfrac{d}{dx}[\arccos x] = -\dfrac{1}{\sqrt{1-x^2}} — same magnitude, opposite sign, since the two graphs are vertical mirror images of each other.

The idea is intuitive — a graph that starts below the x-axis and ends above it, without ever jumping, has to cross somewhere — but making "without ever jumping" precise is a statement about the real numbers themselves. A rigorous proof needs the completeness property of \mathbb R (typically via the least-upper-bound axiom, showing the crossing point is the supremum of \{x \in [a,b] : f(x) < 0\}), which belongs to a real analysis course, not this one. It's taken on faith here — a polynomial is continuous (its graph really is one unbroken curve), and continuity is exactly the property the sign-change test relies on.

Both hypotheses are load-bearing, and dropping either one produces a function with no maximum, no minimum, or neither. f(x) = x on the OPEN interval (0, 1) is continuous but has no largest value -- for any candidate close to 1, a larger one is still available, since 1 itself is never reached. f(x) = 1/x on the closed interval [-1, 1], with f(0) left undefined (or defined arbitrarily), fails continuity at a single interior point and its values run off to \pm\infty on either side -- no maximum or minimum exists at all. A closed, bounded interval together with continuity is exactly the pair of hypotheses that rules out both failure modes: closedness keeps the interval's own endpoints in play as candidates, and continuity keeps the function's values from escaping to infinity or leaving a gap where a max or min would otherwise sit. A full proof (typically via the Bolzano–Weierstrass theorem, showing a maximizing sequence has a convergent subsequence whose limit is itself in [a,b]) belongs to real analysis; here it's enough to know EVT is what guarantees the "candidates test" in this chapter's next section always has a winner to find.

Say f has a local maximum at c (a local minimum runs the identical argument with every inequality flipped). Since c is interior, both one-sided difference quotients are defined near c. For h>0 small, f(c+h)\leq f(c), so \dfrac{f(c+h)-f(c)}{h}\leq 0; letting h\to0^+ gives f'(c)\leq0. For h<0 small, f(c+h)\leq f(c) still holds, but dividing by the negative h flips the inequality: \dfrac{f(c+h)-f(c)}{h}\geq 0; letting h\to0^- gives f'(c)\geq0. Both one-sided limits equal f'(c) (since f'(c) is assumed to exist), so f'(c)\leq0 and f'(c)\geq0 force f'(c)=0.

The converse is false: f(x)=x^3 has f'(0)=0 but no extremum at 0 at all (it's an inflection point). The "interior" hypothesis also matters — f(x)=x on [0,1] attains its maximum at the endpoint c=1, where f'(1)=1\neq0, because c=1 is not an interior point and the two-sided argument above never gets to run.

Both hypotheses are load-bearing, and dropping either one produces a function with no maximum, no minimum, or neither. f(x) = x on the OPEN interval (0, 1) is continuous but has no largest value -- for any candidate close to 1, a larger one is still available, since 1 itself is never reached. f(x) = 1/x on the closed interval [-1, 1], with f(0) left undefined (or defined arbitrarily), fails continuity at a single interior point and its values run off to \pm\infty on either side -- no maximum or minimum exists at all. A closed, bounded interval together with continuity is exactly the pair of hypotheses that rules out both failure modes: closedness keeps the interval's own endpoints in play as candidates, and continuity keeps the function's values from escaping to infinity or leaving a gap where a max or min would otherwise sit. A full proof (typically via the Bolzano–Weierstrass theorem, showing a maximizing sequence has a convergent subsequence whose limit is itself in [a,b]) belongs to real analysis; here it's enough to know EVT is what guarantees the "candidates test" in this chapter's next section always has a winner to find.

Say f has a local maximum at c (a local minimum runs the identical argument with every inequality flipped). Since c is interior, both one-sided difference quotients are defined near c. For h>0 small, f(c+h)\leq f(c), so \dfrac{f(c+h)-f(c)}{h}\leq 0; letting h\to0^+ gives f'(c)\leq0. For h<0 small, f(c+h)\leq f(c) still holds, but dividing by the negative h flips the inequality: \dfrac{f(c+h)-f(c)}{h}\geq 0; letting h\to0^- gives f'(c)\geq0. Both one-sided limits equal f'(c) (since f'(c) is assumed to exist), so f'(c)\leq0 and f'(c)\geq0 force f'(c)=0.

The converse is false: f(x)=x^3 has f'(0)=0 but no extremum at 0 at all (it's an inflection point). The "interior" hypothesis also matters — f(x)=x on [0,1] attains its maximum at the endpoint c=1, where f'(1)=1\neq0, because c=1 is not an interior point and the two-sided argument above never gets to run.

If f is constant on [a, b], every point is such a c and there is nothing more to show. Otherwise f takes some value different from f(a) = f(b) somewhere in the interval, so by the Extreme Value Theorem f attains its absolute maximum and absolute minimum on [a, b], and at least one of those two extreme values is different from the shared endpoint value f(a). That extreme value is therefore attained at some interior point c in (a, b) -- not at an endpoint, since both endpoints share the one value already ruled out. An extreme value attained at an INTERIOR point of an interval where f is differentiable is a local extremum, so Fermat's theorem applies: f'(c) = 0.

Every hypothesis here is necessary. f(x) = |x| on [-1, 1] has f(-1) = f(1) = 1 but is not differentiable at the one interior point, x=0, where f' would need to vanish -- and indeed f'(x) = \pm 1 everywhere it exists, never 0. Drop continuity instead and the conclusion fails just as easily: a function equal to 0 on [0, 1) and jumped to 1 at x=1 satisfies f(0)=0 \ne 1 = f(1)... but even keeping f(a)=f(b) by patching the jump elsewhere, a single discontinuity is enough to route the graph from f(a) back to f(b) without ever flattening out in between. Rolle's Theorem is the special case f(a)=f(b) of the Mean Value Theorem below, proved first because the general MVT proof is built by tilting the picture until it reduces to exactly this one.

Build a new function g measuring the vertical GAP between f's graph and its own secant line through (a, f(a)) and (b, f(b)): g(x) = f(x) - \left[f(a) + \frac{f(b)-f(a)}{b-a}(x-a)\right]. g inherits continuity on [a,b] and differentiability on (a,b) directly from f (the bracketed term is linear, hence both everywhere), and g(a) = g(b) = 0 by construction -- the secant line agrees with f at both endpoints, so subtracting it out zeroes the gap there. That is exactly Rolle's Theorem's hypothesis on g, so some c in (a,b) has g'(c) = 0. Differentiating, g'(x) = f'(x) - \frac{f(b)-f(a)}{b-a}, so g'(c)=0 says precisely f'(c) = \frac{f(b)-f(a)}{b-a} -- the claim.

MVT is Rolle's Theorem tilted: Rolle's is the special case where the secant is already horizontal (f(a)=f(b), slope 0), and the general proof above works by subtracting off whatever secant IS there to reduce the tilted picture back to a horizontal one. The same differentiability hypothesis Rolle's needed carries over unchanged, and fails the same way: f(x)=|x| on [-1,1] has secant slope \frac{f(1)-f(-1)}{1-(-1)} = \frac{1-1}{2} = 0, so MVT would demand some c with f'(c)=0 -- but f'(x)=\pm1 everywhere it's defined on (-1,1) and doesn't exist at all at x=0, so no such c exists. The corner at x=0 is the only place a horizontal tangent could have occurred, and it's exactly the one point where f isn't differentiable.

Suppose f'>0 immediately to the left of c and f'<0 immediately to the right. By the Mean Value Theorem consequence that a positive derivative on an interval forces f increasing there, f is increasing on the left piece and decreasing on the right piece. So every nearby point on either side has a smaller f-value than f(c) — exactly the definition of a local maximum. The negative-to-positive case mirrors this into a local minimum, and if f' keeps the same sign on both sides, f is monotonic straight through c with no extremum there at all — for instance f(x)=x^3 at c=0, where f'(x)=3x^2\geq0 never actually changes sign.

f''(c)=0 (or f''(c) undefined) is necessary for an inflection point but never sufficient by itself — concavity must genuinely change sign across c, not merely touch zero at c. f(x)=x^4 has f''(0)=0, but f''(x)=12x^2\geq0 never goes negative on either side, so concavity stays "up" straight through 0 and x=0 is not an inflection point despite f''(0)=0. By contrast f(x)=x^3 has f''(x)=6x, negative for x<0 and positive for x>0 — concavity genuinely flips at 0, so x=0 is a genuine inflection point there.

Say f''(c)>0. Since f'' is continuous, f''>0 on some interval around c, so f is concave up throughout that interval. A concave-up function lies entirely above each of its tangent lines, and the tangent line at c is horizontal (since f'(c)=0) sitting at height f(c) — so f(x)\geq f(c) for every nearby x, exactly a local minimum. The f''(c)<0 case mirrors this into a local maximum via concave down.

When f''(c)=0, all three outcomes are genuinely possible: f(x)=x^4 has f''(0)=0 and a local minimum at 0; f(x)=-x^4 has f''(0)=0 and a local maximum; f(x)=x^3 has f''(0)=0 and no extremum at all. No amount of examining f''(c)=0 alone can distinguish these — the First Derivative Test is the required fallback whenever this happens.

Suppose f'>0 immediately to the left of c and f'<0 immediately to the right. By the Mean Value Theorem consequence that a positive derivative on an interval forces f increasing there, f is increasing on the left piece and decreasing on the right piece. So every nearby point on either side has a smaller f-value than f(c) — exactly the definition of a local maximum. The negative-to-positive case mirrors this into a local minimum, and if f' keeps the same sign on both sides, f is monotonic straight through c with no extremum there at all — for instance f(x)=x^3 at c=0, where f'(x)=3x^2\geq0 never actually changes sign.

Say f''(c)>0. Since f'' is continuous, f''>0 on some interval around c, so f is concave up throughout that interval. A concave-up function lies entirely above each of its tangent lines, and the tangent line at c is horizontal (since f'(c)=0) sitting at height f(c) — so f(x)\geq f(c) for every nearby x, exactly a local minimum. The f''(c)<0 case mirrors this into a local maximum via concave down.

When f''(c)=0, all three outcomes are genuinely possible: f(x)=x^4 has f''(0)=0 and a local minimum at 0; f(x)=-x^4 has f''(0)=0 and a local maximum; f(x)=x^3 has f''(0)=0 and no extremum at all. No amount of examining f''(c)=0 alone can distinguish these — the First Derivative Test is the required fallback whenever this happens.

The \frac{0}{0} case follows from the Generalized (Cauchy) Mean Value Theorem: for x near a, there is a c between a and x with \dfrac{f(x)-f(a)}{g(x)-g(a)} = \dfrac{f'(c)}{g'(c)}. Since f(a)=g(a)=0 in the \frac{0}{0} case, the left side is exactly \frac{f(x)}{g(x)}; letting x \to a forces c \to a too, giving the limit of f'/g'. The \frac{\infty}{\infty} case uses the same idea with a more careful algebraic argument and is not reproduced here.

Say f''(c)>0. Since f'' is continuous, f''>0 on some interval around c, so f is concave up throughout that interval. A concave-up function lies entirely above each of its tangent lines, and the tangent line at c is horizontal (since f'(c)=0) sitting at height f(c) — so f(x)\geq f(c) for every nearby x, exactly a local minimum. The f''(c)<0 case mirrors this into a local maximum via concave down.

When f''(c)=0, all three outcomes are genuinely possible: f(x)=x^4 has f''(0)=0 and a local minimum at 0; f(x)=-x^4 has f''(0)=0 and a local maximum; f(x)=x^3 has f''(0)=0 and no extremum at all. No amount of examining f''(c)=0 alone can distinguish these — the First Derivative Test is the required fallback whenever this happens.

Both hypotheses are load-bearing, and dropping either one produces a function with no maximum, no minimum, or neither. f(x) = x on the OPEN interval (0, 1) is continuous but has no largest value -- for any candidate close to 1, a larger one is still available, since 1 itself is never reached. f(x) = 1/x on the closed interval [-1, 1], with f(0) left undefined (or defined arbitrarily), fails continuity at a single interior point and its values run off to \pm\infty on either side -- no maximum or minimum exists at all. A closed, bounded interval together with continuity is exactly the pair of hypotheses that rules out both failure modes: closedness keeps the interval's own endpoints in play as candidates, and continuity keeps the function's values from escaping to infinity or leaving a gap where a max or min would otherwise sit. A full proof (typically via the Bolzano–Weierstrass theorem, showing a maximizing sequence has a convergent subsequence whose limit is itself in [a,b]) belongs to real analysis; here it's enough to know EVT is what guarantees the "candidates test" in this chapter's next section always has a winner to find.

Both hypotheses are load-bearing, and dropping either one produces a function with no maximum, no minimum, or neither. f(x) = x on the OPEN interval (0, 1) is continuous but has no largest value -- for any candidate close to 1, a larger one is still available, since 1 itself is never reached. f(x) = 1/x on the closed interval [-1, 1], with f(0) left undefined (or defined arbitrarily), fails continuity at a single interior point and its values run off to \pm\infty on either side -- no maximum or minimum exists at all. A closed, bounded interval together with continuity is exactly the pair of hypotheses that rules out both failure modes: closedness keeps the interval's own endpoints in play as candidates, and continuity keeps the function's values from escaping to infinity or leaving a gap where a max or min would otherwise sit. A full proof (typically via the Bolzano–Weierstrass theorem, showing a maximizing sequence has a convergent subsequence whose limit is itself in [a,b]) belongs to real analysis; here it's enough to know EVT is what guarantees the "candidates test" in this chapter's next section always has a winner to find.

Fix x in (a,b) and form the difference quotient defining A'(x): \frac{A(x+h)-A(x)}{h} = \frac{1}{h}\left[\int_a^{x+h}f(t)\,dt -\int_a^x f(t)\,dt\right] = \frac{1}{h}\int_x^{x+h} f(t)\,dt, the last step by additivity of the definite integral over adjacent intervals. The remaining integral is f's net signed area over a sliver of width h -- and by the Mean Value Theorem for Integrals (f continuous on [x,x+h]), that sliver equals f(c)\cdot h for SOME c between x and x+h, so \frac{A(x+h)-A(x)}{h} = f(c) \quad\text{for some $c$ strictly between $x$ and $x+h$.} As h\to0, c is squeezed between x and x+h and is forced to x too, and f's continuity at x then forces f(c)\to f(x). Taking the limit, A'(x) = \lim_{h\to0}\frac{A(x+h)-A(x)}{h} = \lim_{h\to0} f(c) = f(x).

The same conclusion falls out of the Riemann-sum definition directly, without invoking the Mean Value Theorem for Integrals: for small h, \int_x^{x+h} f(t)\,dt is itself approximated by a single thin rectangle of width h and height approximately f(x) (since f barely changes over an interval that short, by continuity), so \frac{1}{h}\int_x^{x+h}f(t)\,dt\approx f(x), with the approximation becoming exact in the limit h\to0 -- the same "thin slice, height \approx f(x)" idea Chapter 7 built Riemann sums from in the first place, just run over a single infinitesimally short interval instead of the whole [a,b].

Continuity of f is load-bearing in both arguments: it is what supplies the Mean Value Theorem for Integrals' own hypothesis, and it is what forces f(c)\to f(x) as the squeezed point c\to x -- a discontinuous f can make A fail to be differentiable exactly at the discontinuity, even though A itself (an accumulated area) still exists there.

FTC Part 1 already exhibits ONE particular antiderivative of f: the accumulation function A(x) = \int_a^x f(t)\,dt satisfies A'(x) = f(x) on (a,b). Since F is also given to be an antiderivative of the same f, the General Antiderivative Theorem (a Mean Value Theorem corollary, Chapter 7 Section 1) says F and A differ by a constant: F(x) = A(x) + C \quad\text{for some constant } C. Evaluate this identity at x=a: A(a) = \int_a^a f(t)\,dt = 0 (a degenerate interval of zero width bounds zero signed area), so F(a) = A(a) + C = 0 + C = C -- the constant C is exactly F(a). Evaluate at x=b: F(b) = A(b) + C = A(b) + F(a), so A(b) = F(b) - F(a). But A(b) = \int_a^b f(t)\,dt by A's own definition, which is exactly \int_a^b f(x)\,dx with the dummy variable renamed -- giving \int_a^b f(x)\,dx = F(b)-F(a), the claim.

This is why FTC Part 2 makes definite integrals COMPUTABLE without ever building a Riemann sum: Chapter 7 could only ever approach \int_a^b f as a limit of finite sums, refining n forever without literally reaching it by hand. FTC Part 2 replaces that limit with a single subtraction, F(b)-F(a), PROVIDED an antiderivative F can be written down in closed form -- which every basic antiderivative rule from Chapter 7 Sections 1-4 (power, constant multiple, sum/difference, trig, exponential) now feeds directly into. The theorem does not care WHICH antiderivative F is used: any two antiderivatives of f differ by a constant C, and that same C cancels exactly in F(b)-F(a), so the computation is independent of which member of the antiderivative family gets chosen -- picking C=0 is always a valid, and always the simplest, choice.

FTC Part 1 already exhibits ONE particular antiderivative of f: the accumulation function A(x) = \int_a^x f(t)\,dt satisfies A'(x) = f(x) on (a,b). Since F is also given to be an antiderivative of the same f, the General Antiderivative Theorem (a Mean Value Theorem corollary, Chapter 7 Section 1) says F and A differ by a constant: F(x) = A(x) + C \quad\text{for some constant } C. Evaluate this identity at x=a: A(a) = \int_a^a f(t)\,dt = 0 (a degenerate interval of zero width bounds zero signed area), so F(a) = A(a) + C = 0 + C = C -- the constant C is exactly F(a). Evaluate at x=b: F(b) = A(b) + C = A(b) + F(a), so A(b) = F(b) - F(a). But A(b) = \int_a^b f(t)\,dt by A's own definition, which is exactly \int_a^b f(x)\,dx with the dummy variable renamed -- giving \int_a^b f(x)\,dx = F(b)-F(a), the claim.

This is why FTC Part 2 makes definite integrals COMPUTABLE without ever building a Riemann sum: Chapter 7 could only ever approach \int_a^b f as a limit of finite sums, refining n forever without literally reaching it by hand. FTC Part 2 replaces that limit with a single subtraction, F(b)-F(a), PROVIDED an antiderivative F can be written down in closed form -- which every basic antiderivative rule from Chapter 7 Sections 1-4 (power, constant multiple, sum/difference, trig, exponential) now feeds directly into. The theorem does not care WHICH antiderivative F is used: any two antiderivatives of f differ by a constant C, and that same C cancels exactly in F(b)-F(a), so the computation is independent of which member of the antiderivative family gets chosen -- picking C=0 is always a valid, and always the simplest, choice.

A direct Riemann-sum argument, identical in spirit to a single-curve area. Slice [a,b] into n thin strips of width \Delta x. Each strip is very nearly a rectangle of width \Delta x and height \text{top}(x_i^*)-\text{bottom}(x_i^*), the vertical gap between the two curves at that strip's location. Summing every strip's area and taking the limit as n\to\infty turns the Riemann sum into exactly the integral above. Finding which curve is top and which is bottom on the whole interval is not optional bookkeeping — integrating \text{bottom}-\text{top} instead produces a negative number for a quantity that is, by definition, a positive area.

This is the cross-sectional slicing formula V=\int_a^b A(x)\,dx applied to a specific shape: revolving the segment from the axis up to f(x) sweeps out a full circle of radius f(x) at each x, so the cross-section there is a disk with area A(x) = \pi\big[f(x)\big]^2. Substituting that area function directly into the slicing formula gives V = \pi\int_a^b [f(x)]^2\,dx; no new idea is needed beyond identifying the disk's area correctly.

Revolving the outer boundary alone would sweep out a solid disk of area \pi[f(x)]^2 at each x, by the disk method. But the inner boundary carves a circular hole of area \pi[g(x)]^2 out of that disk at the same x, leaving a washer — an annulus — of area \pi[f(x)]^2 - \pi[g(x)]^2. Substituting that washer's area into the cross-sectional slicing formula V=\int_a^b A(x)\,dx gives the stated integral.

Let A(x) = \int_a^x f(t)\,dt. By the ordinary Mean Value Theorem applied to A (differentiable by FTC Part 1, with A'=f), there is a c in (a,b) with A'(c) = \dfrac{A(b)-A(a)}{b-a} = \dfrac{1}{b-a}\int_a^b f(t)\,dt. Since A'(c) = f(c), that's exactly the claim.

FTC Part 1 already exhibits ONE particular antiderivative of f: the accumulation function A(x) = \int_a^x f(t)\,dt satisfies A'(x) = f(x) on (a,b). Since F is also given to be an antiderivative of the same f, the General Antiderivative Theorem (a Mean Value Theorem corollary, Chapter 7 Section 1) says F and A differ by a constant: F(x) = A(x) + C \quad\text{for some constant } C. Evaluate this identity at x=a: A(a) = \int_a^a f(t)\,dt = 0 (a degenerate interval of zero width bounds zero signed area), so F(a) = A(a) + C = 0 + C = C -- the constant C is exactly F(a). Evaluate at x=b: F(b) = A(b) + C = A(b) + F(a), so A(b) = F(b) - F(a). But A(b) = \int_a^b f(t)\,dt by A's own definition, which is exactly \int_a^b f(x)\,dx with the dummy variable renamed -- giving \int_a^b f(x)\,dx = F(b)-F(a), the claim.

This is why FTC Part 2 makes definite integrals COMPUTABLE without ever building a Riemann sum: Chapter 7 could only ever approach \int_a^b f as a limit of finite sums, refining n forever without literally reaching it by hand. FTC Part 2 replaces that limit with a single subtraction, F(b)-F(a), PROVIDED an antiderivative F can be written down in closed form -- which every basic antiderivative rule from Chapter 7 Sections 1-4 (power, constant multiple, sum/difference, trig, exponential) now feeds directly into. The theorem does not care WHICH antiderivative F is used: any two antiderivatives of f differ by a constant C, and that same C cancels exactly in F(b)-F(a), so the computation is independent of which member of the antiderivative family gets chosen -- picking C=0 is always a valid, and always the simplest, choice.