Section5_4_review

Section5_4_review - Section 5.4: Indefinite Integrals ′ )...

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Unformatted text preview: Section 5.4: Indefinite Integrals ′ ) f xs xd = ( , x ( The indefinite integral is ∫f()x F )where Fx= () o F(x) is the most general antiderivative of f (x). An indefinite integral represents a family of functions. An indefinite integral is assumed to be valid on an interval so discontinuities associated with the integrand f (x) are not an issue. Table of Indefinite Integrals(where a, c, and k are constants) f x =∫ x d d xd d x [ x+( ] = x + g d ∫a()x a f()x ∫ f() g)x ∫f()x ∫ ()x xx k c ∫kd = x+ i xx − x c o ∫snd = cs + e d= x c a ∫scxx t n + sc t nd= x c e ∫ exaxx sc + 2 n1 x+ n xd = x + , n≠ 1 c − ∫ n1 + oxx s + n ∫cs d =i x c 2 c d= ct + c ∫ s xx −ox c c ct d=c x c c s ∫ sxoxx − c + Example. Evaluate Solution. φ ∫ (3 cos φ + φ )dφ = 3sin φ + ∫ (3 cos φ + φ )dφ . 2 Example. Evaluate ∫ sec θ tan θdθ . Solution. 4 2 +c ∫ sec θ tan θdθ = secθ + c x Example. Evaluate ∫x1 2 ) . (+ x d x Solution. ∫x1 2 ) =∫(+ x) = + x c c (+ x d x 2 d x 2x + =x +x + 2 6 23 4 2 6 2 5 5 Example. Evaluate Solution. Since s2 2 xs i x s cx n = no, i sn c s 2i x ox sin2x d 2 o d 2i x c ∫ sinx dx ∫ sn x= ∫csxx= sn + ix 3 ∫ sinx dx. sin2x Example. Evaluate ∫y y +) y (2 1 d . Solution. Although there is any easier way to evaluate this integral, we do not know the necessary technique yet. As a result, we need to expand the integrand before evaluating. 3 y 1 y 34 32 1 (2+)= 6+y +y+ 6 5 y 2 1=( +y+y+ =7 3 +y+ y) ( + yy 34 32 1 y+y 33 y ) 3 2 3 7 5 3 8 6 Therefore, y ∫y y +)d = ( +y+y+) =y +3y +3y +y +c ( 1 y ∫ y 3 3 yd 8 6 4 2 4 2 8 6 2 y y 34 y y =++ ++ c 8242 ...
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This note was uploaded on 01/12/2010 for the course MATH 44245 taught by Professor Famiglietti during the Fall '07 term at UC Irvine.

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