Formula & theorem reference

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.

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.

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.

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.

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.

Slice the solid into n thin slabs, each of thickness \Delta x. The i-th slab is nearly a cylinder of cross-sectional area A(x_i^*) and height \Delta x, so its volume is approximately A(x_i^*)\Delta x. Summing over all n slabs gives a Riemann sum \sum_{i=1}^n A(x_i^*)\Delta x that approximates the total volume, and the approximation only improves as the slabs get thinner. Taking the limit as n\to\infty turns the Riemann sum into the definite integral \int_a^b A(x)\,dx by the same limiting argument that defines any definite integral.

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.

Instead of slicing perpendicular to the axis of revolution, slice parallel to it: a thin vertical strip at position x, of width dx and height f(x), sweeps out a thin cylindrical shell when revolved about the y-axis. That shell has radius x, height f(x), and thickness dx, so unrolling it into a flat rectangular slab gives a volume (\text{circumference})\times(\text{height})\times(\text{thickness}) = (2\pi x)\,f(x)\,dx. Summing infinitely many such shells from x=a to x=b — the same limiting Riemann-sum argument behind every slicing formula — turns the sum into V = 2\pi\int_a^b x f(x)\,dx.

Approximate the curve by a polygonal path connecting points at x_0,x_1,\ldots,x_n. Between consecutive points, the path rises \Delta y_i over a run of \Delta x, so that segment's length is \sqrt{(\Delta x)^2+(\Delta y_i)^2} = \sqrt{1+(\Delta y_i/\Delta x)^2}\,\Delta x. By the Mean Value Theorem, \Delta y_i/\Delta x = f'(x_i^*) for some point x_i^* in that subinterval, so the segment's length is \sqrt{1+[f'(x_i^*)]^2}\,\Delta x. Summing over all n segments gives a Riemann sum for \sqrt{1+[f'(x)]^2}, and letting n\to\infty turns that sum into the stated integral, with the polygonal path's total length converging to the true arc length.

Approximate the curve by short straight segments, as in the arc length formula. Revolving one segment of length ds=\sqrt{1+[f'(x_i^*)]^2}\,\Delta x about the x-axis, where the segment sits at height \approx f(x_i^*), sweeps out a thin frustum (a truncated cone). A frustum's lateral surface area is 2\pi\,(\text{average radius})\times(\text{slant length}), and for a short enough segment the average radius is well approximated by f(x_i^*) itself, giving area \approx 2\pi f(x_i^*)\,ds. Summing over all segments gives a Riemann sum for 2\pi f(x)\sqrt{1+[f'(x)]^2}, and the limit as the segments shrink turns that sum into the stated integral.

Over a short displacement \Delta x, the force is nearly constant at F(x_i^*), so the work done over that piece is approximately F(x_i^*)\,\Delta x — the constant-force formula "work equals force times distance," applied on a small enough scale that the force barely changes. Summing over n such pieces from a to b gives a Riemann sum \sum_{i=1}^n F(x_i^*)\Delta x approximating the total work, and letting the pieces shrink (n\to\infty) turns that sum into \int_a^b F(x)\,dx by the definition of the definite integral.

Slice the plate into thin horizontal strips of thickness \Delta x at depth x. Pressure at depth x is w x (weight-density times depth), and pressure is uniform across a thin horizontal strip since every point on it sits at (nearly) the same depth — unlike a vertical strip, whose depth varies top to bottom. That strip has area \approx \ell(x)\,\Delta x, so the force on it is (\text{pressure})\times(\text{area}) \approx w x\,\ell(x)\,\Delta x. Summing over all strips gives a Riemann sum for w x \ell(x), and the limit as the strips thin out turns that sum into F = w\int_a^b x\,\ell(x)\,dx.

Treat the region as built from thin vertical strips of width dx at position x, each with area \big[f(x)-g(x)\big]dx and, being thin, essentially a point mass at x for the purpose of \bar x, and at its own midpoint height \frac{f(x)+g(x)}{2} for the purpose of \bar y. Moments add: the total first moment about the y-axis is \int_a^b x\big[f(x)-g(x)\big]dx, and the total first moment about the x-axis is \int_a^b \frac{f(x)+g(x)}{2}\big[f(x)-g(x)\big]dx = \int_a^b\frac12\big([f(x)]^2-[g(x)]^2\big)dx. Dividing each moment by the total area A — the same "moment over mass" rule used for a system of discrete point masses — gives \bar x and \bar y.

The two conditions are exactly the continuous analogue of a discrete probability distribution: individual probabilities can't be negative, which forces f(x)\ge 0, and the probabilities of all possible outcomes must add to 1, which becomes \int_a^b f(x)\,dx=1 once "adding over all outcomes" is replaced by integrating over the continuum from a to b. The same reasoning that turns a sum into an integral for area turns "sum the probability over every outcome in [c,d]" into \int_c^d f(x)\,dx — each infinitesimal width dx near x carries "probability density f(x) times width dx" worth of probability, and those slivers are added by the integral exactly as area slivers are.

This is the continuous version of the discrete weighted average E(X)=\sum_i x_i p_i. Partition [a,b] into thin slivers of width \Delta x; the sliver at x carries approximately f(x)\Delta x worth of probability (by the definition of a density), so it contributes x\cdot f(x)\Delta x to the weighted sum, the same way a discrete outcome x_i contributes x_i p_i. Summing every sliver's contribution gives a Riemann sum \sum x\,f(x)\Delta x, and letting the slivers shrink turns that sum into \int_a^b x f(x)\,dx.

Start from the product rule, \dfrac{d}{dx}[uv] = u\,\dfrac{dv}{dx} + v\,\dfrac{du}{dx}, and integrate both sides with respect to x: uv = \int u\,\frac{dv}{dx}\,dx + \int v\,\frac{du}{dx}\,dx = \int u\,dv + \int v\,du. Solving for \int u\,dv gives the formula directly. No information is added or lost — it's the product rule read from right to left, with the two terms it produces separated onto opposite sides of the equation.

Each row is one honest application of integration by parts: \int u\,dv = uv - \int v\,du moves one derivative onto u and one antiderivative onto dv, exactly matching the next D/I row. Repeating this on the leftover integral \int v\,du each time is what produces the alternating sign — every fresh application contributes another minus sign in front of its own leftover integral. The process terminates because u's derivative column reaches 0 after finitely many steps whenever u is a polynomial, at which point the "leftover integral" is \int 0\cdot(\text{anything})\,dx = 0 and nothing remains to integrate by parts again — the table's diagonal products are the whole answer.

Write q(x) = (x-r)^k m(x) with m(r) \neq 0 (every other factor of q). The cover-up value A_1 = \dfrac{p(r)}{m(r)} is well-defined since m(r)\neq0, and \frac{p(x)}{q(x)} - \frac{A_1}{(x-r)^k} = \frac{p(x) - A_1\,m(x)}{(x-r)^k m(x)}. The new numerator vanishes at x=r, since p(r) - A_1 m(r) = p(r) - p(r) = 0 by exactly how A_1 was chosen — so by the Factor Theorem, (x-r) divides it: p(x) - A_1 m(x) = (x-r)\,p_1(x) for some polynomial p_1. Substituting back, \frac{p(x)}{q(x)} - \frac{A_1}{(x-r)^k} = \frac{(x-r)\,p_1(x)}{(x-r)^k m(x)} = \frac{p_1(x)}{(x-r)^{k-1}m(x)}, a rational function with the SAME shape as the one just started with, except the power of (x-r) in the denominator dropped from k to k-1. Repeat the identical step on \frac{p_1(x)}{(x-r)^{k-1}m(x)} to peel off A_2 and drop to k-2, and so on. After k repetitions every power of (x-r) has been stripped out, leaving exactly \frac{A_1}{x-r}+\cdots+\frac{A_k}{(x-r)^k} pulled out and a remaining proper fraction over m(x) — the other factors of q(x), handled the same way by the rest of the Partial Fraction Decomposition Recipe.

Let F(t)=\int_a^t f(x)\,dx. Since f\ge0, F is nondecreasing in t, so F(t) either grows without bound or approaches a finite limit — there is no other option for a nondecreasing function. If \int_a^\infty g\,dx converges to some value L, then for every t\ge a, F(t)=\int_a^t f\,dx \le \int_a^t g\,dx \le L, since f\le g pointwise and g\ge 0 makes \int_a^t g\,dx itself nondecreasing in t with limit L. A nondecreasing function bounded above by L must converge to some limit at most L — it cannot grow without bound — so \int_a^\infty f\,dx converges. The second bullet is the direct contrapositive: if f's integral grew without bound, the same inequality F(t)\le\int_a^t g\,dx would force g's integral to grow without bound as well.

The construction is exactly the tangent-line approximation, applied repeatedly instead of once. At (x_n, y_n) the differential equation itself hands over the slope, f(x_n,y_n), since that's what \frac{dy}{dx} evaluates to there. Walking a horizontal distance h along the line through (x_n,y_n) with that slope lands at height y_n + h\cdot f(x_n,y_n) — the linear approximation formula y \approx y_n + f'(x_n)(x-x_n) with x - x_n = h. Treating that new point as the start of the next step and repeating is the entire method; no higher-order information is ever used, which is exactly why the approximation drifts from the true solution as the curve bends away from each step's tangent line.

Treat \frac{dy}{dx} as a genuine ratio for a moment (this is a shortcut that a full proof by the chain rule justifies rigorously): dividing both sides by h(y) gives \frac{1}{h(y)}\,dy = g(x)\,dx. Integrate the left side with respect to y and the right side with respect to x — the two sides need not share a variable of integration at all, since each is being integrated on its own. The chain rule confirms this is legitimate: if H(y) is an antiderivative of \frac{1}{h(y)}, differentiating H(y(x)) = \int g(x)\,dx + C implicitly with respect to x gives H'(y)\cdot\frac{dy}{dx} = g(x), i.e. \frac{1}{h(y)}\cdot\frac{dy}{dx} = g(x) — exactly the original equation, recovered.

Separate variables: \frac{1}{y}\,dy = k\,dx (valid wherever y \neq 0). Integrating both sides, \ln|y| = kx + C_1. Exponentiating, |y| = e^{kx+C_1} = e^{C_1}e^{kx}, and absorbing the sign together with e^{C_1} into a single constant C gives y = Ce^{kx} for any real C (including C=0, the constant solution y\equiv0, which the division step above had to set aside). Setting x=0 pins down C = y(0) = y_0 exactly, giving the stated formula.

Let u = T - T_s, the object's temperature gap above (or below) the medium. Since T_s is constant, \frac{du}{dt} = \frac{dT}{dt} = -k(T-T_s) = -ku — the same \frac{dy}{dx}=ky pattern as ordinary exponential decay, with rate constant -k. Its solution is u = u_0 e^{-kt} where u_0 = T_0 - T_s. Substituting back T = T_s + u gives the stated formula. The sign is what forces the object toward T_s regardless of whether it starts above or below it: if T>T_s the right side is negative and T falls; if T<T_s the right side is positive and T rises — both cases push T toward T_s, never away from it.

Separate variables: \dfrac{dP}{P(1-P/K)} = r\,dt. Rewrite the left side by partial fractions: \dfrac{1}{P(1-P/K)} = \dfrac{1}{P} + \dfrac{1/K}{1-P/K} (verified by combining the right side back over a common denominator). Integrating, \ln|P| - \ln|1 - P/K| = rt + C_1, i.e. \ln\left|\dfrac{P}{1-P/K}\right| = rt+C_1. Exponentiating and solving for P eventually isolates P = \dfrac{K}{1+Ae^{-rt}} for a constant A absorbing C_1 and the sign. Evaluating at t=0, P_0 = \dfrac{K}{1+A}, which rearranges to A = \dfrac{K-P_0}{P_0}.

y(x)=c solves \dfrac{dy}{dx}=f(y) exactly when its derivative, which is identically 0, matches f(c) — hence the requirement f(c)=0. The stability classification follows from the sign of \frac{dy}{dx}=f(y) alone, without solving the equation: wherever f(y)>0, any solution curve is increasing, so a solution starting just below c climbs toward it; wherever f(y)<0, a solution is decreasing, so one starting just above c falls toward it. Stable means both of those hold at once; unstable means both directions point away from c instead; semi-stable mixes the two, one side attracting and the other repelling.

By the product rule, \dfrac{d}{dx}\big[\mu(x)y\big] = \mu(x)y' + \mu'(x)y. Choosing \mu(x) = e^{\int P(x)\,dx} gives \mu'(x) = P(x)e^{\int P(x)\,dx} = P(x)\mu(x) by the chain rule, so \dfrac{d}{dx}\big[\mu(x)y\big] = \mu(x)y' + P(x)\mu(x)y = \mu(x)\big(y'+P(x)y\big) — exactly \mu(x) times the left side of the standard-form equation. Multiplying the whole equation by \mu(x) therefore turns it into \dfrac{d}{dx}\big[\mu(x)y\big] = \mu(x)Q(x), which integrates directly to \mu(x)y = \int \mu(x)Q(x)\,dx + C, and dividing by \mu(x) (never zero, since it's an exponential) isolates y.

Near a value t_0 where x'(t_0)\neq 0, the Inverse Function Theorem guarantees x=x(t) has a differentiable inverse t=t(x) on some interval around x_0=x(t_0), so y can locally be viewed as the composite function y(t(x)). The chain rule gives \dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx}, and the inverse function rule gives \dfrac{dt}{dx} = 1/\dfrac{dx}{dt}. Substituting recovers \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}.

Write g(t) = \dfrac{dy}{dx}, itself a function of t once dy/dx has been computed as y'(t)/x'(t). The curve's second derivative \dfrac{d^2y}{dx^2} means \dfrac{d}{dx}\big[g(t)\big], and by the same chain-rule argument used for the first derivative, \dfrac{d}{dx}\big[g(t)\big] = \dfrac{dg/dt}{dx/dt} — differentiate g(t) with respect to t, then divide by x'(t) again, never by x''(t).

Approximate the curve by a polygonal path through the points at t_0,t_1,\ldots,t_n. The segment from t_{i-1} to t_i runs \Delta x_i across and \Delta y_i up, so its length is \sqrt{(\Delta x_i)^2+(\Delta y_i)^2}. Factoring out \Delta t gives \sqrt{(\Delta x_i/\Delta t)^2+(\Delta y_i/\Delta t)^2}\,\Delta t, and the Mean Value Theorem replaces each difference quotient with x'(t_i^*) and y'(t_i^*) at (possibly different) points in the subinterval. Summing over all n segments produces a Riemann sum for \sqrt{[x'(t)]^2+[y'(t)]^2}, and letting n\to\infty turns that sum into the stated integral, with the polygonal path's length converging to the curve's true arc length.

The ordinary area formula A=\int_a^b y\,dx still applies once x is recognized as a function of t: substituting x=x(t) turns dx into x'(t)\,dt, and the bounds x=a,\,x=b become t=\alpha,\,t=\beta through a=x(\alpha), b=x(\beta). No new geometry is involved — this is u-substitution run on the area integral itself, with the parameter t playing the role of u.

Approximate the curve by short segments, as in the parametric arc length derivation. Revolving one segment about the x-axis, where the segment sits at height \approx y(t_i^*) and has length ds\approx\sqrt{[x'(t_i^*)]^2+[y'(t_i^*)]^2}\,\Delta t, sweeps out a thin frustum of lateral area \approx 2\pi\,y(t_i^*)\,ds. Summing over all segments gives a Riemann sum for 2\pi\,y(t)\sqrt{[x'(t)]^2+[y'(t)]^2}, and the limit as the segments shrink turns that sum into the stated integral.

A polar pair (r, \theta) names a point by facing the direction \theta from the origin and walking a signed distance r along that ray — positive r forward, negative r backward along the same line.

Adding 2\pi k. Turning through a full extra 2\pi (or several) returns to facing exactly the same direction — \theta and \theta+2\pi k are coterminal angles, sharing one terminal ray. Walking the same signed distance r along the same ray lands on the same point.

Flipping the sign and adding \pi. Turning through an extra \pi reverses the facing direction to point exactly the opposite way. Walking distance -r backward along that reversed ray retraces the same path as walking distance r forward along the original ray — same point, reached from the opposite facing.

This last claim can be checked algebraically using the reduction formulas \cos(\theta+\pi) = -\cos\theta and \sin(\theta+\pi) = -\sin\theta: rectangular coordinates built from (-r, \theta+\pi) come out to (-r)\cos(\theta+\pi) = (-r)(-\cos\theta) = r\cos\theta, \qquad (-r)\sin(\theta+\pi) = (-r)(-\sin\theta) = r\sin\theta, exactly matching the coordinates built from (r, \theta) — confirming both name the same point.

Since either move alone leaves the point unchanged, applying both in any combination still leaves it unchanged, which is why the two rules combine freely.

This is nothing but the unit-circle definitions of cosine and sine — \cos\theta and \sin\theta are the x- and y-coordinates on a circle of radius 1 — stretched from radius 1 out to radius r.

r = \sqrt{x^2+y^2} is just the distance formula from the origin to (x,y). Dividing the two conversion equations above, \frac{y}{x} = \frac{r\sin\theta}{r\cos\theta} = \tan\theta, whenever x \neq 0.

Differentiate x and y with respect to \theta using the product rule on each: \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta, \qquad \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta. Since x and y are both functions of the single parameter \theta, the parametric-curve slope formula applies directly: \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}, which is exactly the ratio above once both derivatives are substituted in.

A sector with angle \theta is one particular fraction of the whole circle — specifically, the fraction \frac{\theta}{2\pi}, since 2\pi is the angle of the complete circle. The whole circle has area \pi r^2, so the sector's area is that same fraction of the total: A = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{\theta \pi r^2}{2\pi} = \frac{1}{2}r^2\theta. The \pi in the numerator and denominator cancel, leaving a formula with no \pi in it at all — it's already baked into \theta being in radians.

Partition [\alpha,\beta] into n subintervals of width \Delta\theta. On each subinterval, approximate the thin polar region by a circular SECTOR of radius r(\theta_i^*) (for some sample angle \theta_i^* in that subinterval) and central angle \Delta\theta. By the sector area formula, that one sector has area \tfrac12\big[r(\theta_i^*)\big]^2\Delta\theta. Summing all n sectors gives \sum_{i=1}^n \frac12\big[r(\theta_i^*)\big]^2\,\Delta\theta, a Riemann sum for \tfrac12\big[r(\theta)\big]^2. As n\to\infty and \Delta\theta\to0, the sector approximation's error shrinks to zero and the sum converges to the stated integral.

Write the curve parametrically as x=r(\theta)\cos\theta, y=r(\theta)\sin\theta, and apply the parametric arc-length integral L=\int_\alpha^\beta\sqrt{(dx/d\theta)^2+(dy/d\theta)^2}\,d\theta. Using \dfrac{dx}{d\theta}=r'\cos\theta-r\sin\theta and \dfrac{dy}{d\theta}=r'\sin\theta+r\cos\theta (product rule on each, writing r' for dr/d\theta), \left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2 = (r'\cos\theta-r\sin\theta)^2 + (r'\sin\theta+r\cos\theta)^2. Expanding both squares, the cross terms -2rr'\sin\theta\cos\theta and +2rr'\sin\theta\cos\theta cancel exactly, leaving (r')^2\cos^2\theta + r^2\sin^2\theta + (r')^2\sin^2\theta + r^2\cos^2\theta = (r')^2\big(\cos^2\theta+\sin^2\theta\big) + r^2\big(\sin^2\theta+\cos^2\theta\big) = (r')^2 + r^2, by the Pythagorean identity. Substituting this simplified sum under the square root gives the stated formula.

These are the ordinary limit laws for functions, transported to sequences through one observation: a sequence \{a_n\} is a function of n whose domain happens to be the positive integers instead of the real line, and "n \to \infty" is exactly the same kind of limit-at-infinity statement already proved for functions. Every argument used there — bounding |a_n + b_n - (L+M)| by the triangle inequality, controlling a product by first bounding one factor near its limit — carries over verbatim with n restricted to integers.

Identical in spirit to the function version, with n \to \infty playing the role of x \to a: past whatever index makes both outer sequences land within a given tolerance of L, a_n is trapped between them and has nowhere else to land either. The only difference is that "near \infty" means "past some index N" instead of "inside some window around a" — the trapping argument itself doesn't change.

Take the non-decreasing, bounded-above case (the other case is the same argument flipped). The set of all terms \{a_n\} is bounded above, so by the completeness of the real numbers it has a least upper bound L. Given any tolerance around L, L minus that tolerance is no longer an upper bound (it's smaller than the least one), so some term a_N exceeds it. Because the sequence is non-decreasing, every later term a_n for n \geq N is at least a_N — and still can't exceed L, since L IS an upper bound. So every term from a_N onward is squeezed inside that tolerance of L, which is exactly what a_n \to L means.

If \sum a_n converged to a sum S, its partial sums S_n would converge to S, and so would S_{n-1} (the same sequence, shifted one index — a shift by one term can't change what a sequence converges to). But a_n = S_n - S_{n-1} for n \geq 2, so by the sequence limit laws, a_n \to S - S = 0. Contrapositive: a_n \not\to 0 forces \sum a_n to diverge.

Each partial sum of the combined series splits as \sum_{k=1}^n (a_k \pm b_k) = \sum_{k=1}^n a_k \pm \sum_{k=1}^n b_k, a finite sum being rearranged, which is always legal. Taking n \to \infty on both sides and applying the sequence limit laws to the two partial-sum sequences (which converge to A and B by hypothesis) gives the result directly. The constant-multiple case is the same argument with the constant-multiple limit law instead of the sum law.

Because f is decreasing, on each unit interval [k,k+1] the function never exceeds its left value: f(x) \le f(k) = a_k for x in [k,k+1], so \int_k^{k+1} f(x)\,dx \le a_k. Summing from k=N to k=M-1 gives \int_N^{M} f(x)\,dx \le \sum_{n=N}^{M-1} a_n. The same decreasing property also gives f(x) \ge f(k+1) = a_{k+1} on [k,k+1], so a_{k+1} \le \int_k^{k+1} f(x)\,dx, and summing that from k=N to M-1 gives \sum_{n=N+1}^{M} a_n \le \int_N^{M} f(x)\,dx. Together, the partial sums and the running integral are trapped against each other closely enough that one grows without bound exactly when the other does — since f\ge 0 makes both \int_N^t f(x)\,dx and the partial sums nondecreasing in the first place, "trapped and nondecreasing" forces them to converge or diverge together as M\to\infty.

For p\le 0 the terms 1/n^p don't even shrink to zero (they're constant or growing), so the series diverges by the divergence test alone. For p>0, f(x)=1/x^p is continuous, positive, and decreasing on [1,\infty), so the Integral Test applies: \int_1^{\infty} x^{-p}\,dx converges exactly when p>1 (a direct antiderivative computation — for p=1 the antiderivative is \ln x, unbounded as x\to\infty; for p\ne 1 it's \frac{x^{1-p}}{1-p}, which has a finite limit at infinity only when the exponent 1-p is negative, i.e. p>1). The series and the integral share that same threshold.

R_n = \sum_{k=n+1}^{\infty} a_k is itself a tail series with the same integral-test bounding argument applied starting at N=n+1: the same inequalities used to prove the Integral Test give \int_{n+1}^{\infty} f(x)\,dx \le \sum_{k=n+1}^{\infty} a_k (the lower integral bound on each unit strip) and \sum_{k=n+1}^{\infty} a_k \le \int_{n}^{\infty} f(x)\,dx (the upper bound, re-indexed one step earlier since a_{n+1}\le\int_n^{n+1}f). Both tail integrals converge whenever the original one does, since a convergent improper integral's tail beyond any finite point is itself a convergent improper integral.

Since every term is nonnegative, both series' partial sums are nondecreasing sequences, so each either converges or grows without bound — there's no third option. If \sum b_n converges to L, every partial sum of \sum a_n satisfies \sum_{n=1}^{N} a_n \le \sum_{n=1}^{N} b_n \le L, so \sum a_n's partial sums are nondecreasing and bounded above, forcing convergence. The second bullet is the contrapositive of the first.

Since the ratio a_n/b_n\to L>0, eventually \frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2}, which rearranges to \frac{L}{2}\,b_n < a_n < \frac{3L}{2}\,b_n for all large enough n. If \sum b_n converges, the right inequality and the Direct Comparison Test give \sum a_n converges too (comparing against \frac{3L}{2}b_n, itself a convergent series since it's a constant multiple of one). If \sum b_n diverges, the left inequality and Direct Comparison give \sum a_n diverges too (comparing \frac{L}{2}b_n \le a_n, and a constant multiple of a divergent nonnegative series still diverges). Swapping the roles of a_n and b_n (using 1/L as the limiting ratio the other way) gives the same conclusion starting from either series, so the two series stand or fall together.

Consider the even partial sums S_{2n} of \sum(-1)^{n+1}b_n. Grouping consecutive pairs, S_{2n} = (b_1-b_2)+(b_3-b_4)+\cdots+(b_{2n-1}-b_{2n}) — every grouped pair is nonnegative since b_n is decreasing, so S_{2n} is nondecreasing in n. Regrouping the same sum as S_{2n}=b_1-(b_2-b_3)-(b_4-b_5)-\cdots-b_{2n} shows S_{2n}\le b_1 for every n, since each parenthesized pair is again nonnegative and is being subtracted. A nondecreasing sequence bounded above converges, so S_{2n}\to S for some S. The odd partial sums satisfy S_{2n+1}=S_{2n}+b_{2n+1}, and b_{2n+1}\to 0, so S_{2n+1}\to S as well — both subsequences converge to the same limit, which forces the whole sequence of partial sums to converge to S. The other sign pattern is the same series multiplied by -1, which converges whenever the original one does.

From the proof of the Alternating Series Test, S lies between any two consecutive partial sums: S_n and S_{n+1}=S_n+(-1)^{n+2}b_{n+1} straddle S, since the even partial sums increase to S from below while the odd ones decrease to S from above (or vice versa, depending on parity). Being trapped between S_n and S_{n+1} means |S-S_n| is at most the gap between those two consecutive partial sums, which is exactly |S_{n+1}-S_n| = b_{n+1}.

For every n, 0 \le a_n + |a_n| \le 2|a_n| (this holds whether a_n is positive, negative, or zero: if a_n\ge 0 the middle expression is 2a_n; if a_n<0 it's 0). Since \sum|a_n| converges, so does \sum 2|a_n|, and the Direct Comparison Test then gives that \sum (a_n+|a_n|) converges too. Writing a_n = (a_n+|a_n|) - |a_n| expresses the original series as a difference of two convergent series, and a difference of convergent series converges (term-by-term subtraction of two convergent partial sums converges to the difference of their limits).

Suppose L<1. Pick r with L<r<1. Since the ratio's limit is L, eventually (say for n\ge N) |a_{n+1}/a_n| < r, so |a_{n+1}| < r|a_n|. Applying this repeatedly from n=N gives |a_{N+k}| < r^k|a_N| for every k\ge 0 — the tail is dominated term by term by a convergent geometric series (ratio r<1) scaled by |a_N|. The Direct Comparison Test then gives \sum|a_n| converges, i.e. absolute convergence. If L>1, the same argument with r between 1 and L shows |a_{n+1}|>|a_n| eventually, so the terms are eventually increasing in magnitude and cannot shrink to 0 — the Divergence Test then rules out convergence. L=1 is inconclusive because both a convergent p-series (p=2) and the divergent harmonic series (p=1) produce a ratio limit of exactly 1, so the test alone cannot distinguish them.

Suppose L<1. Pick r with L<r<1. Eventually \sqrt[n]{|a_n|}<r, so |a_n|<r^n for all large n — the tail is dominated term by term by a convergent geometric series with ratio r<1. The Direct Comparison Test then gives \sum|a_n| converges. If L>1, eventually \sqrt[n]{|a_n|}>1, so |a_n|>1 infinitely often (in fact eventually always), meaning the terms cannot shrink to 0 — the Divergence Test then rules out convergence. L=1 is inconclusive for the same reason as the Ratio Test: both a convergent p-series and the divergent harmonic series give a root limit of exactly 1.

Split the terms into their positive part p_n (the nonnegative terms, in order) and negative part q_n (the negative terms, in order, as positive magnitudes). Since \sum a_n converges but \sum|a_n| diverges, both \sum p_n and \sum q_n must diverge to +\infty — if either were finite, the other would have to be finite too (their difference converges and their sum is \sum|a_n|), contradicting \sum|a_n|=\infty. With both piles individually unbounded, a rearrangement can always take just enough positive terms to push the running total just past M, then just enough negative terms to pull it back just under M, and repeat forever. Because a_n\to 0 (convergence of the original series forces this), each overshoot and undershoot shrinks toward 0, so the running total is forced to converge to M exactly. The same construction with "past M" replaced by "past n" for every n produces a rearrangement diverging to +\infty instead.

Substitute x = i\theta into the Maclaurin series for e^x and use i^2=-1, i^3=-i, i^4=1 (the pattern repeats every four powers): e^{i\theta} = \sum_{n=0}^{\infty}\frac{(i\theta)^n}{n!} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + i\frac{\theta^5}{5!} - \cdots Collecting the terms with no factor of i and the terms with one factor of i into two separate sums gives e^{i\theta} = \underbrace{\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right)}_{\cos\theta} + i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)}_{\sin\theta}. The first parenthesis is exactly the Maclaurin series for \cos\theta and the second is exactly the Maclaurin series for \sin\theta, so e^{i\theta} = \cos\theta + i\sin\theta.