MATH 141 Fundamental Theorem of Calculus (FTC) Let f : [a, b] → R
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MATH 141 Fundamental Theorem of Calculus (FTC) Let f : [a, b] → R be a continuous function. Let F : [a, b] → R be an antiderivative of f (recall, this means F 0 (x) = f (x)). Let x ∈ (a, b). Then : Z Z b a f (x) dx ≡ and a ·Z Dx b F 0 (x) dx = F (b) − F (a) ¸ x f (t)dt = f (x) . a Basic Differentiation Rules If the functions y = f (x) and y = g(x) are differentiable at x and a and b are constants, then: 1. Dx [af (x) + bg(x)] = af 0 (x) + bg 0 (x) 2. Dx [f (x) · g(x)] = f 0 (x) · g(x) + f (x) · g 0 (x) h 3. Dx f (x) g(x) i 4. Dx [f (x)] = r f 0 (x) · g(x) − f (x) · g 0 (x) [g(x)]2 r−1 = r [f (x)] provided g(x) 6= 0 f 0 (x) If f is differentiable at x and g is differentiable at f (x), then: 5. Dx [g(f (x))] = g 0 (f (x)) f 0 (x) Exp and Log derivatives Dx ax = ax ln a 1 Dx loga |x| = x ln a (loga b)(logb c) = loga c FTC =⇒ integrals Z Z ax dx = ax + C ln a 1 dx = ln |x| + C x ln x loga x = ln a Trig tan x = sin x cos x cot x = cos x sin x sec x = 1 − cos(2x) 2 1 + cos(2x) cos2 x = 2 sin2 x = 1 cos x csc x = 1 sin x sin(2x) = 2 sin x cos x cos(2x) = cos2 x − sin2 x FTC =⇒ derivatives Z du dx du Dx tan u = sec2 u dx integrals cos u du = sin u + C Dx sin u = cos u Z sec2 u du = tan u + C Z du Dx sec u = sec u tan u dx du Dx cos u = − sin u dx du Dx cot u = − csc2 u dx du Dx csc u = − csc u cot u dx sec u tan u du = sec u + C Z Z Z sin u du = − cos u + C csc2 u du = − cot u + C csc u cot u du = − csc u + C more integrals Z tan u du = − ln |cos u| + C = ln |sec u| + C cot u du = ln |sin u| + C = − ln |csc u| + C sec u du = ln |sec u + tan u| + C = e − ln |sec u − tan u| + C csc u du = − ln |csc u + cot u| + C = e ln |csc u − cot u| + C Z Z Z 2 inverse trig functions y = sin θ ⇔ θ = sin−1 y where −1≤y ≤1 and y = cos θ ⇔ θ = cos−1 y where −1≤y ≤1 and y = tan θ ⇔ θ = tan−1 y where y∈R and y = cot θ ⇔ θ = cot−1 y where y∈R and y = sec θ ⇔ θ = sec−1 y where |y| ≥ 1 and y = csc θ ⇔ θ = csc−1 y where |y| ≥ 1 and FTC =⇒ derivatives 1 du 1 − u2 dx du 1 Dx tan−1 u = 2 1 + u dx 1 du √ Dx sec−1 u = 2 |u| u − 1 dx −1 du Dx cos−1 u = √ 1 − u2 dx −1 du Dx cot−1 u = 1 + u2 dx −1 du √ Dx csc−1 u = |u| u2 − 1 dx Dx sin−1 u = √ |u| ≤ 1 |u| ≥ 1 |u| ≤ 1 |u| ≥ 1 3 π −π ≤θ≤ 2 2 0≤θ≤π π −π <θ< 2 2 0<θ<π π 0 ≤ θ ≤ π , θ 6= 2 π −π ≤ θ ≤ , θ 6= 0 2 2 integrals 1 √ du = sin−1 u + C 1 − u2 Z 1 du = tan−1 u + C 1 + u2 Z 1 √ du = sec−1 |u| + C u u2 − 1 Z
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