Calculus I Integration: A Very Short Summary Definition
Transcription
Calculus I Integration: A Very Short Summary Definition: Antiderivative The function F (x) is an antiderivative of the function f (x) on an interval I if F 0 (x) = f (x) for all x in I. Notice, a function may have infinitely many antiderivatives. For example, the function f (x) = 2x has antiderivatives such as x2 , x2 + 3, x2 − π, and x2 + .002, just to name a few. Definition: General Antiderivative The function F (x) + C is the General Antiderivative of the function f (x) on an interval I if F 0 (x) = f (x) for all x in I and C is an arbitrary constant. The function x2 + C where C is an arbitrary constant, is the General Antiderivative of 2x. This is actually a family of functions, each with its own value of C. Definition: Indefinite Integral The Indefinite Integral of f (x) is the General Antiderivative of f (x). Z Z f (x) dx = F (x) + C 2x dx = x2 + C Definition: Riemann Sum The Riemann Sum is a sum of the areas of n rectangles formed over n subintervals in [a, b] . Here the subintervals are of equal length, but they need not be. The height of the ith rectangle, is the value of f (x) at a chosen sample point in the ith subinterval. The width of each rectangle is ∆x = (b−a) and the height of the rectangle in the ith subinterval is given by f (x∗i ) where x∗i is a n sample point in the ith subinterval. The Riemann Sum used to approximate Z b n X f (x∗i )∆x = f (x∗1 )∆x + f (x∗2 )∆x + · · · + f (x∗n )∆x f (x) dx is given by a i=1 Here xi ∗ is the sample point in the ith subinterval. If the sample points are the midpoints of the subintervals, we call the Riemann Sum the Midpoint Rule. Definition: Definite Integral The Definite integral of f from a to b, written Z b f (x) dx, is defined to be the limit of a a Riemann sum as n → ∞ , if the limit exists (for all choices of sample points x∗1 , x∗2 , . . . x∗n in the n subintervals). Z b n X f (x)dx = lim f (x∗i )∆x = lim f (x∗1 )∆x + f (x∗2 )∆x + · · · + f (x∗n )∆x Thus, n→∞ a n→∞ i=1 The First Fundamental Theorem of Calculus: Let f be continuous on the closed interval Z b [a, b], then f (x)dx = F (b) − F (a) where F is any antiderivative of f on [a, b]. a Z 3 2 2x dx = x 1 3 1 = 32 − 12 = 8 The Second Fundamental Theorem of Calculus: Let f be continuous " Z on the#closed interval Z x x d [a, b], and define G(x) = f (t)dt where a ≤ x ≤ b. Then G0 (x) = f (t)dt = f (x). dx a a Z x Z x3 2 2 0 G(x) = sin (t) dt G (x) = sin (x) H(x) = sin2 (t) dt H 0 (x) = 3x2 sin2 (x3 ) 0 0 1 Integration by Substitution: Let uZ= g(x) and F (x) be the antiderivative of f (x). Then Z du = g 0 (x)dx and f g(x) g 0 (x) dx = f (u) du = F (u) + C Z b Z g(b) 0 Also, f g(x) g (x)dx = f (u) du = F g(b) − F g(a) a g(a) 2 Ex: u = g(x) = x , 0 du = g (x)dx = 2x dx Z 5 x2 2xe dx = 2 Integration Rules Z k du = ku + C Z ur+1 ur du = + C for r 6= −1 r+1 Z Z du = u−1 du = ln |u| + C for r = −1 u Z au au du = +C ln a Z eu du = eu + C Z cos u du = sin u + C Z sin u du = − cos u + C Z sec2 u du = tan u + C Z sec u tan u du = sec u + C Z csc2 u du = − cot u + C Z csc u cot u du = − csc u + C Z Z kf (u) du = k f (u) du Z Z Z [f (u) ± g(u)] du = f (u) du ± g(u) du a b f (x) dx = − a Z a b f (x) dx = Z Z a a f (x) dx (exchange integration limits) b c f (x) dx + Z 52 =25 u u e du = e 22 =4 25 4 = e25 − e4 Examples Z Z 3 du = 3u + C π dt = πt + C Z u6 u5 du = + C for 5 6= −1 6 Z 2x + 3 dx = ln |x2 + 3x| + C x2 + 3x Z 4t 4t dt = +C ln 4 Z esin x cos x dx = esin x + C Z 3x2 cos x3 dx = sin x3 + C Z 7t6 sin t7 dt = − cos t7 + C Z 20x3 sec2 5x4 dx = tan 5x4 + C Z 4x3 sec x4 tan x4 dx = sec x4 + C Z 8 csc2 8x dx = − cot 8x + C Z 5x4 csc x5 cot x5 dx = − csc x5 + C Z Z 4 cos x dx = 4 cos x dx = 4 sin x + C Z [4x3 ± sec2 x] dx = x4 ± tan x + C Properties of the Definite Integral Z a f (x) dx = 0 (same integration limits) Z Z b f (x) dx where a < c < b. c 2 Notice, rules 4 through 6 below are simply negatives of rules 1 through 3. Inverse T rig Integration Rules Examples 1. Z du √ = sin−1 u + C 2 1−u Z 3x2 dx √ = sin−1 x3 + C 6 1−x 2. Z du = tan−1 u + C 1 + u2 Z 4x3 dx = tan−1 x4 + C 1 + x8 3. Z du √ = sec−1 u + C |u| u2 − 1 Z dx p = sec−1 (ln x) + C 2 x| ln x| (ln x) − 1 4. Z −du √ = cos−1 u + C = − sin−1 u + C 2 1−u Z −3x2 dx √ = cos−1 x3 + C = − sin−1 x3 + C 6 1−x 5. Z −du = cot−1 u + C = − tan−1 u + C 1 + u2 Z −4x3 dx = cot−1 x4 + C = − tan−1 x4 + C 1 + x8 6. Z −du √ = csc−1 u + C = − sec−1 u + C |u| u2 − 1 Z −dx p = csc−1 (ln x) + C x| ln x| (ln x)2 − 1 = − sec−1 (ln x) + C 3
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