Lecture2-slides-substitution - Lecture 2 Warm-Up Problems 20 Z Suppose that f(x = Find f 0(x p 1 \u2212 5u 3 du x Find the following indefinite

# Lecture2-slides-substitution - Lecture 2 Warm-Up...

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Lecture 2 Warm-Up Problems Suppose thatf(x) =1-5u3du.Findf0(x).Find the following indefinite integrals(Hint: simplify using the distributive property)1.Zsec2(x)(1 + 2 cos2(x))dx2.Z(x2+ 3)22x dx3.Z(x2+ 3)1002x dx Z 20 x p Announcements: I Quiz 1 on Tuesday, Sept 11: FTC and Substitution I Homework 0 (optional) and Homework 1 are posted. HW0 is due on Sept 12, HW1 is due on Sept 13, both at 8:45am.
Recall from last class Theorem (The Fundamental Theorem of Calculus) Suppose that f ( x ) is continuous on [ a , b ] . Then, 1. if g ( x ) = R x a f ( t ) dt for all x [ a , b ] , then g 0 ( x ) = f ( x ) . In other words, d dx Z x a f ( t ) dt = f ( x ) . 2. Suppose F ( x ) is an antiderivative of f ( x ) (i.e. F 0 ( x ) = f ( x ) ), then Z b a f ( x ) dx = F ( b ) - F ( a ) .
Recall from last class Theorem (The Fundamental Theorem of Calculus) Suppose that f ( x ) is continuous on [ a , b ] . Then, 1. if g ( x ) = R x a f ( t ) dt for all x [ a , b ] , then g 0 ( x ) = f ( x ) . In other words, d dx Z x a f ( t ) dt = f ( x ) . 2. Suppose F ( x ) is an antiderivative of f ( x ) (i.e. F 0 ( x ) = f ( x ) ), then Z b a f ( x ) dx = F ( b ) - F ( a ) . “Integrals undo derivatives”
Recall from last class Theorem (The Fundamental Theorem of Calculus) Suppose that f ( x ) is continuous on [ a , b ] . Then, 1. if g ( x ) = R x a f ( t ) dt for all x [ a , b ] , then g 0 ( x ) = f ( x ) . In other words, d dx Z x a f ( t ) dt = f ( x ) . 2. Suppose F ( x ) is an antiderivative of f ( x ) (i.e. F 0 ( x ) = f ( x ) ), then Z b a f ( x ) dx = F ( b ) - F ( a ) . “Integrals undo derivatives” Properties of Definite Integrals 1. Z b a f ( x ) dx = - Z a b f ( x ) dx
Recall from last class Theorem (The Fundamental Theorem of Calculus) Suppose that f ( x ) is continuous on [ a , b ] . Then, 1. if g ( x ) = R x a f ( t ) dt for all x [ a , b ] , then g 0 ( x ) = f ( x ) . In other words, d dx Z x a f ( t ) dt = f ( x ) . 2. Suppose F ( x ) is an antiderivative of f ( x ) (i.e. F 0 ( x ) = f ( x ) ), then Z b a f ( x ) dx = F ( b ) - F ( a ) . “Integrals undo derivatives” Properties of Definite Integrals 1. Z b a f ( x ) dx = - Z a b f ( x ) dx 2. Z b a f ( x ) dx = Z c a f ( x ) dx + Z b c f ( x ) dx
Example Worksheet 1, Problem 3b: Find the derivative of h ( z ) = Z 2 z tan 1 u 2 + 1 du
Example Worksheet 1, Problem 3b: Find the derivative of h ( z ) = Z 2 z tan 1 u 2 + 1 du h 0 ( z ) = d dz Z 2 z tan 1 u 2 + 1 du = d dz - Z z 2 tan 1 u 2 + 1 du = - tan 1 z 2 + 1
FTC (part 1) + Chain Rule Worksheet 1, Problem 3c: Find the derivative of k ( x ) = Z x - 2 x sec 2 ( s ) ds
FTC (part 1) + Chain Rule Worksheet 1, Problem 3c: Find the derivative of k ( x ) = Z x - 2 x sec 2 ( s ) ds k ( x ) = Z 0 - 2 x sec 2 ( s ) ds + Z x 0 sec 2 ( s ) ds
FTC (part 1) + Chain Rule Worksheet 1, Problem 3c: Find the derivative of k ( x ) = Z x - 2 x sec 2 ( s ) ds k ( x ) = Z 0 - 2 x sec 2 ( s ) ds + Z x 0 sec 2 ( s ) ds = - Z - 2 x 0 sec 2 ( s ) ds + Z x 0 sec 2 ( s ) ds
FTC (part 1) + Chain Rule Worksheet 1, Problem 3c: Find the derivative of k ( x ) = Z x - 2 x sec 2 ( s ) ds k ( x ) = Z 0 - 2 x sec 2 ( s ) ds + Z x 0 sec 2 ( s ) ds = - Z - 2 x 0 sec 2 ( s ) ds + Z x 0 sec 2 ( s ) ds k 0 ( x ) = - sec 2 ( - 2 x ) d dx [ - 2 x ] + sec 2 ( x )
FTC (part 1) + Chain Rule Worksheet 1, Problem 3c: Find the derivative of k ( x ) = Z x - 2 x sec 2 ( s ) ds k ( x ) = Z 0 - 2 x sec 2 ( s ) ds + Z x 0 sec 2 ( s ) ds = - Z - 2 x 0 sec 2 ( s ) ds + Z x 0 sec 2 ( s ) ds k 0 ( x ) = - sec 2 ( - 2 x ) d dx [ - 2 x ] + sec 2 ( x ) = 2 sec 2 ( - 2 x ) + sec 2 ( x )
FTC (part 1) + Chain Rule d dx " Z h ( x ) a f ( t ) dt # = f ( h ( x ) ) h 0 ( x )
FTC (part 1) + Chain Rule Example: Find the derivative of ( θ ) = Z sin( θ ) 2 1 t 3 - t dt .