Formula Sheet Some Trig Identities

cos2(x) + sin2(x) = 1 , 1 + tan2(x) = sec2(x) , 1 cos2(x) = (1 + cos(2x)) , 2 1 sin2(x) = (1 − cos(2x)) , 2 The next 3 formulae are not required for the test 1 sin(A) cos(B) = (sin(A − B) + sin(A + B)) 2 1 sin(A) sin(B) = (cos(A − B) − cos(A + B)) 2 1 cos(A) cos(B) = (cos(A − B) + cos(A + B)) . 2 Arc- and Volumes of Revolution y = f(x), a ≤ x ≤ b x = g(y), c ≤ y ≤ d R b p 02 R d p 02 Arc Length L = a 1 + f(x) dx L = c 1 + g(y) dy Axis of Rotation x-axis y-axis x-axis y-axis R b 2 R b R d R d 2 Volume of Revolution V = a πf(x) dx V = a 2πxf(x) dx V = c 2πyg(y) dy c πg(y) dy

R b a f(x) dx

Method Formula Error Bound1 K(b−a)3 Midpoint Mn = h(f(x ¯1) + f(x ¯2) + ··· + f(x ¯n)) |EM | ≤ 24n2 h K(b−a)3 Trapezoidal Tn = 2 (f(x0) + 2f(x1) + 2f(x2) + ··· + 2f(xn−1) + f(xn)) |ET | ≤ 12n2 5 2 h K1(b−a) Simpsons Sn = 3 (f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + ··· + 2f(xn−2) + 4f(xn−1) + f(xn)) |ET | ≤ 180n4 b−a xi+xi−1 00 (4) Here, h = n , xi = a + ih, x¯i = 2 ,K = maxa≤x≤b |f (x)| and K1 = maxa≤x≤b |f (x)|. Parametric (x, y) = (x(t), y(t)), a ≤ t ≤ b

dy y0(t) • of curve dx = x0(t)

2 d dy d y dt ( dx ) • dx2 = x0(t)

R b p 0 2 0 2 • Arc-length L = a (x (t)) + (y (t)) dt Polar Curves r = f(θ), a ≤ θ ≤ b

dy f 0(θ) cos(θ)+f(θ) sin(θ) • Slope of dx = f 0(θ) cos(θ)−f(θ) sin(θ) .

R b 1 2 • Area between θ = a and θ = b, A = a 2 [f(θ)] dθ.

3 R b p 2 0 2 • Arclength L = a [f(θ)] + [f (θ)] dθ.

1You don’t have to memorize these, but you must know how to use them. 2n must be even 3Not required for the test.

1 Important we know so far

R n 1 n+1 R 1 x dx = n+1 x + c, n 6= 1 x dx = ln |x| + c

R x x R x 1 x e dx = e + c a dx = ln a a + c

R sin x dx = − cos x + c R cos x dx = sin x + c

R sec2 x dx = tan x + c R csc2 x dx = − cot x + c

R sec x tan x dx = sec x + c R csc x cot x dx = − csc x + c

R 1 −1 R 1+x2 dx = tan x + c tan x dx = ln | sec x| + c

R cot x dx = ln | sin x| + c R sec x dx = ln | sec x + tan x| + c

R R 2 1 1 csc x dx = ln | csc x − cot x| + c sin x dx = 2 x − 4 sin 2x + c

R 2 1 1 R cos x dx = 2 x + 4 sin 2x + c ln x dx = x ln x − x + c

R √ 1 dx = sin−1 x + c R u dv = uv − R v du 1−x2 Special Series Converges Diverges P∞ n−1 Geometric n=1 ar |r| < 1 |r| ≥ 1 P∞ 1 p-series n=1 np p > 1 p ≤ 1 P∞ Theorem: If n=1 an converges then limn→∞ an = 0. Be careful: This theorem gives a requirement for convergence, but not a guarantee of convergence. P∞ Test: If limn→∞ an 6= 0 then n=1 an will diverge.

2 List of Tests Notes Series with positive terms P P Comparison Test Suppose that we have two series an and bn with If the series is similar to a p-series, ge- an, bn ≥ 0 and an ≤ bn for all n. Then, ometric series of any series of know be- P P haviour. 1. If bn is convergent then so is an. P P 2. If an is divergent then so is bn.

P P Comparison Test Suppose that we have two series an and bn with If the series is a rational expression in- an an, bn ≥ 0 for all n. Let c = limn→∞ . If c is volving only polynomials or polynomi- bn finite and positive then either both series converge als under radicals. or diverge. Test Suppose that f(x) is a positive decreasing function If an = f(n) for some positive decreas- R ∞ on the interval [k, ∞) and that an = f(n), then, ing function and a f(x) dx is easy to R ∞ P∞ evaluate then the Integral Test may 1. If k f(x) dx is convergent then so is n=k an. work. R ∞ P∞ 2. If k f(x) dx is divergent then so is n=k an. P∞ Suppose we have the series an. Let L = If the series contains a of con- n=1 an+1 stants raised to powers involving n. limn→∞ . Then an 1. If L < 1 the series is absolutely convergent (and hence convergent). 2. If L > 1 the series is divergent. 3. If L = 1 the test in inconclusive.

Alternating Series P Test Suppose that we have a series an and either an = We note that for these series the magni- n n+1 (−1) bn or an = (−1) bn where bn ≥ 0 for all n. tude of the error due to truncating after Then if, the nth term is smaller then the magni- tude of the (n + 1)st term. 1. limn→∞ bn = 0 and

2. {bn} is an eventually decreasing sequence P∞ the series n=1 an converges. If f(x) has a expansion at the a, then it is given by,

∞ X f(a)(n) f(x) = (x − a)n . n! n=0

If we truncate the series after N terms, then the N th Taylor polynomial of f(x) is,

N X f(a)(n) T (x) = (x − a)n . N n! n=0

We can bound the magnitude of the error, RN (x) caused by this truncation by, M |R (x)| ≤ |x − a|N+1 , for |x − a| ≤ r , N (N + 1)! where, M = max |f (n+1)(x) . |x−a|≤r

3 Separable Differential Equations

A separable differential equation is of the form, dy = f(y)g(x) , y(a) = y . dx 0 The solution is given implicitly by, Z dy Z = g(x) dx . f(y)

The is used to satisfy the condition y(a) = y0.

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