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Calculus II: Week One notes

by: Alison Holden

Calculus II: Week One notes M 145

Marketplace > University of Hartford > M 145 > Calculus II Week One notes
Alison Holden
University of Hartford
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About this Document

This includes u-substitution, particle motion, difference between distance, and displacement, and integral rules. Will be on exam 1.
Calculus II
Dr. Hadad
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This 4 page Class Notes was uploaded by Alison Holden on Tuesday September 13, 2016. The Class Notes belongs to M 145 at University of Hartford taught by Dr. Hadad in Fall 2016. Since its upload, it has received 28 views.


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Date Created: 09/13/16
Calculus II Week One Notes Calculus I Review Function: relationship between an independent and dependent variables. Every independent has 1 dependent Derivative: rate of change (slope) at any point of a line Y= f(x) Derivative can be represented by: dy/dx, y’, or f’(x) Constant of Integration 2 2 y = 6x + 3x + 1 vs y = 6x + 3x +2 The original functions are different due to the constant, however when taking the derivative, the answer is the same for both: y = 12x + 3. This creates a dilemma when going the opposite direction. When integrating a function, you must always account for the constant that may or may not be in the function by adding a + C. ∫ 12x+3 )ⅆx Ex: 6x +3x+C = ∫ f x dx Symbol for Integration: Properties of Integrals  If there is a constant multiplied by the function, you can move it outside the integral and multiply after o ∫Cf (x)ⅆx=¿ C ∫ f(x)ⅆx 2 2 o ∫6x ⅆx=6∫ x ⅆx  If there is more than one function inside an integral separated by addition or subtraction, you can separate them o ∫ f(x)+g x ⅆx=∫ f (x)ⅆx+∫g (x)ⅆx o ∫sin +cosx ⅆx = ∫sin x ⅆx+∫cos (x)ⅆx  The power rule of integration: xn+1 o ∫ x ⅆx= +C n+1 7+1 8 7 x x o ∫ x ⅆx= 7+1 +C= 8 +C Integration by substitution: ∫ x cos(x +2 )ⅆx Ex: x +2 Let u = Take the derivative: du = 4 x ⅆx 1 Rewrite integral in terms of u: 4 ∫cos u)ⅆu The u term covers the part inside the trigonometry and du covers both x and dx. However, you must add ¼ in front of integral to compensate for the 4 inside of du. 1 1 Integrate: ∫cosuⅆu= sinu+C 4 4 1 sin(x +2)+C Replace u: 4 Ex 2: ∫ √x+1ⅆx = ∫ (2x + 1)1/dx u= 2x + 1 du=2du 1 1 ∫( )ⅆx 2 3 1 2 2 1 3 ∕ 2 = 2 3 u = 3(2x+1 ) +c Definite Integrals b ∫ f (x)d(x=F b ⋅F (a) a Integral: Area under the curve to the x axis 3 ∫ (x −6x )ⅆx Ex: 0 4 x 2 Integrate: 4 −3x 34 2 0 4 2 Put in bounds: −3 3 )− [ ) −3 (0)] 4 4 81 Simplify: −27 = -6.75 4 Particle Motion S (t) t= time S= displacement s't) V (t) = V = instantaneous velocity Instantaneous velocity is the velocity a one single point on a graph. Average velocity is velocity from t = a to t = b A = dv/dt = s’’(t) A = Acceleration Displacement vs Distance Distance from origin. Total distance travel in time = t Ex: Find the displacement of the particle when 1≤t≤4 V(t=t −t−6 You must integrate in order to find the s(t) function:t −t−6 )ⅆx 3 2 t − −6x = 3 2 from 1 to 4 1 1 43 42 = 3− 26 1 ( ) – (3 − 2 −6 ( )−4.5 meters


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