Capitulo 1 Soluciones Purcell 9na Edicion. 1. Instructor’s . –1 5 lim () 2x f x → = e. f(1) = 2 f. 1 lim x→ –1 lim () 2 x f x → = − d. f (–1) = –2 e. Consider the region S – R. 1 Capitulo 5 Soluciones Purcell 9na Edicion Instructor’s Resource Manual Section CHAPTER 5 Applications of the Integral Concepts Review 1. π − = ≈⎜ ⎟ ⎝ ⎠ 5. a, b. c. 2 (5)V x x xΔ ≈ π − Δ 1/ 2 3/ 2 2 (5)x x x= π − Δ d, e. Calculo purcell 9 ed solucionario. 2 1 1 –xD x x ⎛ ⎞ Section Instructor’s Resource Manual e. Let x be the distance from the beam Capitulo 2 Soluciones Purcell 9na.

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