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## Calculus II Week 1 Notes

by: Blanche Langenbach

62

7

10

# Calculus II Week 1 Notes MAC 2312

Marketplace > University of South Florida > Mathematics > MAC 2312 > Calculus II Week 1 Notes
Blanche Langenbach
USF
GPA 4.0

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Covers 5.4, 5.5, 6.1 material: Fundamental Theorem of Calculus, Substitution Rule, and Integration by Parts 08/22/16-08/26/16
COURSE
Calculus II
PROF.
David Lee Milligan
TYPE
Class Notes
PAGES
10
WORDS
CONCEPTS
Math, Calculus, calculusii, Substitution, Integration, FTC
KARMA
Free

## Popular in Mathematics

This 10 page Class Notes was uploaded by Blanche Langenbach on Friday August 26, 2016. The Class Notes belongs to MAC 2312 at University of South Florida taught by David Lee Milligan in Fall 2016. Since its upload, it has received 62 views. For similar materials see Calculus II in Mathematics at University of South Florida.

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Date Created: 08/26/16
Week 1 Day 1 Notes August 22, 2016 5. 4 The Fundamental Theorem of Calculus  Calculus II The FTC says: x 1. g(x)=  ∫ f t)dt a f is a continuous function on [a,b] and x varies between a and b g depends only on x If x is a fixed number, the integral is definite g(x) is the area under graph of f from a to x.  x ∫ f t)dt Ex: 1      If g(x)=  a  , find values of g(0), g(1), g(2), g(3), g(4), g(5) 0 g(0)= ∫ f (t)dt=0 *use graph of f (when given the function) and find  0 the  f(t)dt=¿ g(1)= 1    area under the curve from 0  to each point* ∫ ¿ 0 1 2 ∫ f (t)dt+∫ f (tdt g(2)= 0 1    Example Graph of f(t): Finding g(2) by Area Under the Curve If g is defined as an integral of f, then g is the antiderivative.  g’(x)=f(x) g(x+h)−g(x) f(x)= h Leibniz Notation for Derivatives of Integrals Basically, the equation says if we first integrate the function, then differentiate, we should get  back to the original function.  d x ∫ f(t)dt= f (x) dx a x4 d Ex: 2   dx ∫ sec(t)dt 1 Chain Rule w/ FTC 4  u=x (substitution) u d dx ∫ sec(t)dt 1 4 3 ¿sec (x )(4 x ) Average Value of a Function y +y +y +…y 1 2 3 n yave n   b 1 ∫ f (x)dx b−a a Week 1 Day 2 Notes August 24, 2016 5.5 The Substitution Rule Calculus II The Substitution Rule for Indefinite Integrals ∫ 2 x√1+x dx Ex: 1  u=1+x 2 du=2xdx ∫ √1+x 2xdx ∫ √udu 2 22 (1+x ) +C *Substitute what u equal for u* 3 2 x+1dx Ex: 2    √ u=2x+1 du=2dx  dx=(1/2)du 1 ∫ √u du 2 1 1 ∫ u du 2 3 2u 2 3 1 +C] ¿ 2 3 1u +C 3 *Substitute what u equals for u* 3 1(2 x+1) +C 3 Ex: 3   ∫ tan(x)dx sinx ∫ dx cosx u= cosx du= ­sinx dx 1du 1 du=ln ∨u∨¿ ­ ∫ u Remember that  ∫ u ­1[ ln|u+C −1 1 ­ ln∨cos?(x)∨+C=ln c|sx | )+C=ln +C=ln |ecx |C (|cos| ^Could be left as this or this^ *Differentiate to check solutions* The Substitution Rule for Definite Integrals Two methods: 1. Use Evaluation Theorem 2. Change limits of integration when variable is change Ex: 1 Using Method 2 2 ∫ dx 1 (3−5x) 2 u=3­5x du=­5dx  dx= (­1/5)du New Limits:  x=1, u=3­5(1)=­2 x=2, u=3­5(2)=­7 −7 −1 ∫ du 5 −2 u2 *Because of new limits, we don’t have to go back 1 u −¿ ¿ and substitute x for u* −1 ¿ 5 −7 1 5u −2 1 −1 −1 1 5 7 − ( 2 ) 14 Symmetry      Suppose f is continuous [­a, a] a a a) If f is even [f(­x)=f(x)], then )dx=2 ∫ f x)dx −a 0 a ∫ f(x)dx=0 b) If f is odd [f(­x)=­f(x)],−ahen  6 ta2x Ex: 1   f(x)=x +1 f(­x)=f(x) even Ex:2 f(x)= 1+x +4 6 (x +1)dx=¿ 2 f(­x)=­f(x) odd ∫ ¿ −2 2 6 1 tanx 2∫ (x +1)dx ∫ 2 dx=0 0 −1 1+x +4 +x 7 2[ x ¿¿2 7 0 ⌈ 128+2 −0⌉ 2 ( 7 ) 284 7 Week 1 Day 3 Notes August 26, 2016 6.1 Integration By Parts Calculus II Product Rule says:  d ' dx [f(x)g(x)]= f(x)g (x+g (x) f '(x) Formula for Integration by Parts ' ' ∫ f( ) ( )dx= f x( ) ( ) ∫ g ( ) ( )x Alternate Notation Let u=f(x) v=g(x) f(x)dx dv=g 'x dx du= ∫ udv=uv− v∫u Ex: 1   ∫ xsinxdx u=x dv=sinx dx du=dx v=­cosx u dv=uv− v du 1. ∫ ∫ 2. ∫ xsinxdx=x −(osx − ) ∫ (cosx )x 3. −xcosx+ ∫ cosxdx 4. −xcosx+sinx+C Ex: 2   ∫ e sinx dx x u= e dv=sinx x du= e dxv=−cosx *It gets a little complicated  because you have to integrate by parts multiple times to get  the solution. * For definite integrals: Use same concept of integration by parts, but use the evaluation theorem or change the limits,  similar to definite integration with substitution.  b b f (x)g'(x)dx= f (x)g(x) ] g x )f'(x)dx a a a Reduction Formula: n 1 n−1 n−1 n−2 ∫ s¿ x dx= cnsxs¿ x+ n ∫ sin xdx Integer n ≥2

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