SlideShare a Scribd company logo
11.1 Antiderivatives and
     Indefinite Integrals
 A. Let’s figure out the formula.
B. Those with rational or negative
              exponents
  C. Those that need expanding
  D. Those that need simplifying
      E. Word Problems
A. Let’s figure out the formula.
Recall the power rule for derivatives went like this :
x n → nx n −1
Consider these basic ones :
x 4 + 13 → 4 x 3       x 4 + 7 → 4 x3      x 4 + 122 → 4 x 3
Since it doesn' t matter what the constant term is, I could say :
x 4 + C → 4 x 3 , where C is any constant.


Our next task will be trying to take the 4 x 3 and get the
" indefinite integral" of it x 4 + C.
How can we get from 4 x to x + C ?
                             3      4


 First add one to its exponent.
 Then, divide the term by its new exponent.
 Last, add the + C.
          3+1
        4x
 4x →
    3
              +C
         3 +1
      4
   4x
 =      +C
    4
 = x +C
    4
This is sort of undoing differentiation!

• It’s like we are answering the riddle: “What is the
  function whose derivative is this?”

∫ 4 x dx = " indefinite integral of 4 x         = x +C
        3                                   3     4




              n +1
         x
∫ x dx = n + 1 + C
    n
∫ 6 x dx   You try : ∫ 12 x dx
    2                     3
Term by term :

∫(x                         )
          + 4 x 3 − 3 x 2 + 7 dx
      5



is the same as ∫ x 5 dx + ∫ 4 x 3dx − ∫ 3 x 2 dx + ∫ 7 dx
i.e., take the integral of each term, like 4 smaller tasks.
Notice the last term is ∫ 7 dx. What function would have
a derivative of just plain 7? Answer 7x.
x 6 4 x 4 3x 3
    +       −     + 7x + C
 6     4       3
   x6
=     + x 4 − x3 + 7 x + C
   6
 8               5
                             2x      
You try : ∫  3x − 2 x + x −
            
                      7   6
                                + 14 dx
                                     
                             3      
B. Those with rational or negative
                exponents
• As you have seen before, sometimes it is
  necessary to REWRITE the problem in
  exponential form without negative exponents
  BEFORE using the rules.
• Examples of rewriting:
       3           4 3       6          dx
∫  4 x − x 2 dx
             
                    ∫  3 ⋅ x + 4 x 3 dx
                      
                      
                                      
                                      
                                            ∫ x3
You try :
        7        3 2     2           du
∫
 
   6 t − 2 dt
        t       ∫
                  4⋅ x +
                  
                          3
                            x2
                                 dx
                                 
                                 
                                       ∫ u4
C. Those that need expanding
• You know that you can integrate term-by-
  term. Sometimes you have to EXPAND things
  (F.O.I.L., distribute, etc.) before you can do
  the term-by-term thing.
• “Terms” are only connected to each other by
  plus signs. Nothing else.
∫ ( x + 2) 3 dx   You try : ∫ ( x − 1) dx
                                     3
∫ ( x + 2) 2   x dx   You try : ∫ ( x − 1)
                                             23
                                                  x dx
4 x 4 − x 3 + 6 x 2 + 14 x                  5x5 + x 4 − 4 x 2 + 6 x
∫              x
                             dx   You try : ∫
                                                        x
                                                                      dx
D. Those that need simplifying

  ( x + 3)( x + 5) dx = ( x + 3) dx = x   2

∫     x+5              ∫              2
                                              + 3x + C


•FACTOR TOP AND BOTTOM
•CANCEL VERTICALLY IF POSSIBLE
•INTEGRATE TERM BY TERM
3x − x − 2
              2
You try : ∫            dx
               x −1
E. Word Problems
Differentiation does this :
profit/revenue/cost → marginal profit/revenue/cost
curve → slope of the tangent line
distance → velocity
velocity → acceleration
any quantity → rate of change
So Integration does the opposite :
marginals → profit/revenue/cost functions
slope of the tangent line → the equation of the curve
velocity → distance
acceleration → velocity
rate of change → the function for the quantity
A company's marginal cost function is MC ( x ) = 6 x and the
fixed cost is $1000. Find the cost function.
A company's marginal revenue function is MR = 12 x + 83 x ,
where x is the number of units. Find the revenue function.
[Hint for finding C : How much revenue is brought in for zero
production? zero.]
3
A person can memorize words at the rate of      words per minute.
                                              t
Find a formula for the total number of words that can be memorized
in t minutes. [Hint for finding C : How many words can be memorized
in zero minutes? zero.]
dC
The marginal cost for producing x units of a product is modeled by       = 32 − 0.04 x.
                                                                      dx
It costs $50 to produce one unit. Find the total cost of producing 200 units.

More Related Content

What's hot

January 23
January 23January 23
January 23
khyps13
 
Adding and subtracting polynomials
Adding and subtracting polynomialsAdding and subtracting polynomials
Adding and subtracting polynomials
holmsted
 
1201 ch 12 day 1
1201 ch 12 day 11201 ch 12 day 1
1201 ch 12 day 1
festivalelmo
 
1202 ch 12 day 2
1202 ch 12 day 21202 ch 12 day 2
1202 ch 12 day 2
festivalelmo
 
Quadraticequation
QuadraticequationQuadraticequation
Quadraticequation
Allanna Unias
 
Integral calculus
Integral calculusIntegral calculus
Integral calculus
IndiraDevi24
 
March 6, 2014
March 6, 2014March 6, 2014
March 6, 2014
khyps13
 
Lesson 27: Integration by Substitution (worksheet solutions)
Lesson 27: Integration by Substitution (worksheet solutions)Lesson 27: Integration by Substitution (worksheet solutions)
Lesson 27: Integration by Substitution (worksheet solutions)
Matthew Leingang
 
Lesson 27: Integration by Substitution (worksheet)
Lesson 27: Integration by Substitution (worksheet)Lesson 27: Integration by Substitution (worksheet)
Lesson 27: Integration by Substitution (worksheet)
Matthew Leingang
 
Polynomials and factoring
Polynomials and factoringPolynomials and factoring
Polynomials and factoring
Shilpi Singh
 
Multiplying polynomials
Multiplying polynomialsMultiplying polynomials
Multiplying polynomials
chrystal_brinson
 
Lesson 28: Integration by Substitution (worksheet solutions)
Lesson 28: Integration by Substitution (worksheet solutions)Lesson 28: Integration by Substitution (worksheet solutions)
Lesson 28: Integration by Substitution (worksheet solutions)
Matthew Leingang
 
P6 factoring
P6 factoringP6 factoring
P6 factoring
salamhello
 
Calculus 08 techniques_of_integration
Calculus 08 techniques_of_integrationCalculus 08 techniques_of_integration
Calculus 08 techniques_of_integration
tutulk
 
Feb6
Feb6Feb6
Feb6
khyps13
 
Polynomials
PolynomialsPolynomials
Polynomials
Ver Louie Gautani
 
3.2 factoring polynomials
3.2   factoring polynomials3.2   factoring polynomials
3.2 factoring polynomials
Nuch Pawida
 
Engr 213 midterm 1b sol 2010
Engr 213 midterm 1b sol 2010Engr 213 midterm 1b sol 2010
Engr 213 midterm 1b sol 2010
akabaka12
 
Multiplying Polynomials I
Multiplying Polynomials IMultiplying Polynomials I
Multiplying Polynomials I
Iris
 
1203 ch 12 day 3
1203 ch 12 day 31203 ch 12 day 3
1203 ch 12 day 3
festivalelmo
 

What's hot (20)

January 23
January 23January 23
January 23
 
Adding and subtracting polynomials
Adding and subtracting polynomialsAdding and subtracting polynomials
Adding and subtracting polynomials
 
1201 ch 12 day 1
1201 ch 12 day 11201 ch 12 day 1
1201 ch 12 day 1
 
1202 ch 12 day 2
1202 ch 12 day 21202 ch 12 day 2
1202 ch 12 day 2
 
Quadraticequation
QuadraticequationQuadraticequation
Quadraticequation
 
Integral calculus
Integral calculusIntegral calculus
Integral calculus
 
March 6, 2014
March 6, 2014March 6, 2014
March 6, 2014
 
Lesson 27: Integration by Substitution (worksheet solutions)
Lesson 27: Integration by Substitution (worksheet solutions)Lesson 27: Integration by Substitution (worksheet solutions)
Lesson 27: Integration by Substitution (worksheet solutions)
 
Lesson 27: Integration by Substitution (worksheet)
Lesson 27: Integration by Substitution (worksheet)Lesson 27: Integration by Substitution (worksheet)
Lesson 27: Integration by Substitution (worksheet)
 
Polynomials and factoring
Polynomials and factoringPolynomials and factoring
Polynomials and factoring
 
Multiplying polynomials
Multiplying polynomialsMultiplying polynomials
Multiplying polynomials
 
Lesson 28: Integration by Substitution (worksheet solutions)
Lesson 28: Integration by Substitution (worksheet solutions)Lesson 28: Integration by Substitution (worksheet solutions)
Lesson 28: Integration by Substitution (worksheet solutions)
 
P6 factoring
P6 factoringP6 factoring
P6 factoring
 
Calculus 08 techniques_of_integration
Calculus 08 techniques_of_integrationCalculus 08 techniques_of_integration
Calculus 08 techniques_of_integration
 
Feb6
Feb6Feb6
Feb6
 
Polynomials
PolynomialsPolynomials
Polynomials
 
3.2 factoring polynomials
3.2   factoring polynomials3.2   factoring polynomials
3.2 factoring polynomials
 
Engr 213 midterm 1b sol 2010
Engr 213 midterm 1b sol 2010Engr 213 midterm 1b sol 2010
Engr 213 midterm 1b sol 2010
 
Multiplying Polynomials I
Multiplying Polynomials IMultiplying Polynomials I
Multiplying Polynomials I
 
1203 ch 12 day 3
1203 ch 12 day 31203 ch 12 day 3
1203 ch 12 day 3
 

Viewers also liked

crossing a moat
crossing a moatcrossing a moat
crossing a moat
Jeneva Clark
 
119 powerpoint 1.3
119 powerpoint 1.3119 powerpoint 1.3
119 powerpoint 1.3
Jeneva Clark
 
euler characteristic theorem
euler characteristic theoremeuler characteristic theorem
euler characteristic theorem
Jeneva Clark
 
Moseley Math125 Syllabus Spring 2013
Moseley Math125 Syllabus Spring 2013Moseley Math125 Syllabus Spring 2013
Moseley Math125 Syllabus Spring 2013
Jeneva Clark
 
125 2.5
125 2.5125 2.5
125 2.5
Jeneva Clark
 
美崎栄一郎さんをソーシャルメディア人脈育成術で新潟へ呼ぼう
美崎栄一郎さんをソーシャルメディア人脈育成術で新潟へ呼ぼう美崎栄一郎さんをソーシャルメディア人脈育成術で新潟へ呼ぼう
美崎栄一郎さんをソーシャルメディア人脈育成術で新潟へ呼ぼう
新潟コンサルタント横田秀珠
 
We three sisters
We three sistersWe three sisters
We three sisters
Jeneva Clark
 
The Efficient Waterline
The Efficient WaterlineThe Efficient Waterline
The Efficient Waterline
Jeneva Clark
 
modeling division
modeling divisionmodeling division
modeling division
Jeneva Clark
 
College algebra p3
College algebra p3College algebra p3
College algebra p3
Jeneva Clark
 

Viewers also liked (11)

crossing a moat
crossing a moatcrossing a moat
crossing a moat
 
119 powerpoint 1.3
119 powerpoint 1.3119 powerpoint 1.3
119 powerpoint 1.3
 
euler characteristic theorem
euler characteristic theoremeuler characteristic theorem
euler characteristic theorem
 
Moseley Math125 Syllabus Spring 2013
Moseley Math125 Syllabus Spring 2013Moseley Math125 Syllabus Spring 2013
Moseley Math125 Syllabus Spring 2013
 
125 2.5
125 2.5125 2.5
125 2.5
 
美崎栄一郎さんをソーシャルメディア人脈育成術で新潟へ呼ぼう
美崎栄一郎さんをソーシャルメディア人脈育成術で新潟へ呼ぼう美崎栄一郎さんをソーシャルメディア人脈育成術で新潟へ呼ぼう
美崎栄一郎さんをソーシャルメディア人脈育成術で新潟へ呼ぼう
 
125 3.1
125 3.1125 3.1
125 3.1
 
We three sisters
We three sistersWe three sisters
We three sisters
 
The Efficient Waterline
The Efficient WaterlineThe Efficient Waterline
The Efficient Waterline
 
modeling division
modeling divisionmodeling division
modeling division
 
College algebra p3
College algebra p3College algebra p3
College algebra p3
 

Similar to 125 11.1

125 5.3
125 5.3125 5.3
125 5.3
Jeneva Clark
 
125 4.1 through 4.5
125 4.1 through 4.5125 4.1 through 4.5
125 4.1 through 4.5
Jeneva Clark
 
AP Calculus Project
AP Calculus ProjectAP Calculus Project
AP Calculus Project
kaitlinbianchi
 
Lesson 29: Integration by Substition
Lesson 29: Integration by SubstitionLesson 29: Integration by Substition
Lesson 29: Integration by Substition
Matthew Leingang
 
Lesson 29: Integration by Substition (worksheet solutions)
Lesson 29: Integration by Substition (worksheet solutions)Lesson 29: Integration by Substition (worksheet solutions)
Lesson 29: Integration by Substition (worksheet solutions)
Matthew Leingang
 
Lesson 29: Integration by Substition (worksheet)
Lesson 29: Integration by Substition (worksheet)Lesson 29: Integration by Substition (worksheet)
Lesson 29: Integration by Substition (worksheet)
Matthew Leingang
 
Lesson 29: Integration by Substition (worksheet solutions)
Lesson 29: Integration by Substition (worksheet solutions)Lesson 29: Integration by Substition (worksheet solutions)
Lesson 29: Integration by Substition (worksheet solutions)
Matthew Leingang
 
Integration intro
Integration introIntegration intro
Integration intro
Shaun Wilson
 
125 7.7
125 7.7125 7.7
125 7.7
Jeneva Clark
 
Paige's DEV
Paige's DEVPaige's DEV
Paige's DEV
ps114
 
U1 04 factorizacion
U1   04 factorizacionU1   04 factorizacion
U1 04 factorizacion
UNEFA Zulia
 
Math project
Math projectMath project
Math project
phyatt216
 
Lesson 7: Limits at Infinity
Lesson 7: Limits at InfinityLesson 7: Limits at Infinity
Lesson 7: Limits at Infinity
Matthew Leingang
 
Notes solving polynomial equations
Notes   solving polynomial equationsNotes   solving polynomial equations
Notes solving polynomial equations
Lori Rapp
 
Anti derivatives
Anti derivativesAnti derivatives
Anti derivatives
canalculus
 
Math refresher
Math refresherMath refresher
Math refresher
delilahnan
 
Integral Calculus Anti Derivatives reviewer
Integral Calculus Anti Derivatives reviewerIntegral Calculus Anti Derivatives reviewer
Integral Calculus Anti Derivatives reviewer
JoshuaAgcopra
 
Jackson d.e.v.
Jackson d.e.v.Jackson d.e.v.
Jackson d.e.v.
Dougfield32
 
Sample2
Sample2Sample2
Sample2
Nima Rasekh
 
Lesson 26: The Fundamental Theorem of Calculus (slides)
Lesson 26: The Fundamental Theorem of Calculus (slides)Lesson 26: The Fundamental Theorem of Calculus (slides)
Lesson 26: The Fundamental Theorem of Calculus (slides)
Matthew Leingang
 

Similar to 125 11.1 (20)

125 5.3
125 5.3125 5.3
125 5.3
 
125 4.1 through 4.5
125 4.1 through 4.5125 4.1 through 4.5
125 4.1 through 4.5
 
AP Calculus Project
AP Calculus ProjectAP Calculus Project
AP Calculus Project
 
Lesson 29: Integration by Substition
Lesson 29: Integration by SubstitionLesson 29: Integration by Substition
Lesson 29: Integration by Substition
 
Lesson 29: Integration by Substition (worksheet solutions)
Lesson 29: Integration by Substition (worksheet solutions)Lesson 29: Integration by Substition (worksheet solutions)
Lesson 29: Integration by Substition (worksheet solutions)
 
Lesson 29: Integration by Substition (worksheet)
Lesson 29: Integration by Substition (worksheet)Lesson 29: Integration by Substition (worksheet)
Lesson 29: Integration by Substition (worksheet)
 
Lesson 29: Integration by Substition (worksheet solutions)
Lesson 29: Integration by Substition (worksheet solutions)Lesson 29: Integration by Substition (worksheet solutions)
Lesson 29: Integration by Substition (worksheet solutions)
 
Integration intro
Integration introIntegration intro
Integration intro
 
125 7.7
125 7.7125 7.7
125 7.7
 
Paige's DEV
Paige's DEVPaige's DEV
Paige's DEV
 
U1 04 factorizacion
U1   04 factorizacionU1   04 factorizacion
U1 04 factorizacion
 
Math project
Math projectMath project
Math project
 
Lesson 7: Limits at Infinity
Lesson 7: Limits at InfinityLesson 7: Limits at Infinity
Lesson 7: Limits at Infinity
 
Notes solving polynomial equations
Notes   solving polynomial equationsNotes   solving polynomial equations
Notes solving polynomial equations
 
Anti derivatives
Anti derivativesAnti derivatives
Anti derivatives
 
Math refresher
Math refresherMath refresher
Math refresher
 
Integral Calculus Anti Derivatives reviewer
Integral Calculus Anti Derivatives reviewerIntegral Calculus Anti Derivatives reviewer
Integral Calculus Anti Derivatives reviewer
 
Jackson d.e.v.
Jackson d.e.v.Jackson d.e.v.
Jackson d.e.v.
 
Sample2
Sample2Sample2
Sample2
 
Lesson 26: The Fundamental Theorem of Calculus (slides)
Lesson 26: The Fundamental Theorem of Calculus (slides)Lesson 26: The Fundamental Theorem of Calculus (slides)
Lesson 26: The Fundamental Theorem of Calculus (slides)
 

More from Jeneva Clark

161 course information spring 2016
161 course information spring 2016161 course information spring 2016
161 course information spring 2016
Jeneva Clark
 
Summer 2014 moseley 098 syllabus addendum
Summer 2014 moseley 098 syllabus addendumSummer 2014 moseley 098 syllabus addendum
Summer 2014 moseley 098 syllabus addendum
Jeneva Clark
 
2.1 homework
2.1 homework2.1 homework
2.1 homework
Jeneva Clark
 
2.6 homework
2.6 homework2.6 homework
2.6 homework
Jeneva Clark
 
2.2 homework b
2.2 homework b2.2 homework b
2.2 homework b
Jeneva Clark
 
2.2 homework a
2.2 homework a2.2 homework a
2.2 homework a
Jeneva Clark
 
College algebra hw 1.4
College algebra hw 1.4College algebra hw 1.4
College algebra hw 1.4
Jeneva Clark
 
1.3 homework
1.3 homework1.3 homework
1.3 homework
Jeneva Clark
 
Math 111 College Algebra 1.2
Math 111 College Algebra 1.2Math 111 College Algebra 1.2
Math 111 College Algebra 1.2
Jeneva Clark
 
P.5 homework a
P.5 homework aP.5 homework a
P.5 homework a
Jeneva Clark
 
P.3 homework
P.3 homeworkP.3 homework
P.3 homework
Jeneva Clark
 
P 2 homework b
P 2 homework bP 2 homework b
P 2 homework b
Jeneva Clark
 
P.2 homework A
P.2 homework AP.2 homework A
P.2 homework A
Jeneva Clark
 
P.1 hw
P.1 hwP.1 hw
P.1 hw
Jeneva Clark
 
Moseley schedule
Moseley scheduleMoseley schedule
Moseley schedule
Jeneva Clark
 
Moseley m113 syllabus_summer2013
Moseley m113 syllabus_summer2013Moseley m113 syllabus_summer2013
Moseley m113 syllabus_summer2013
Jeneva Clark
 
Magic Money
Magic MoneyMagic Money
Magic Money
Jeneva Clark
 
Which Road to Travel
Which Road to TravelWhich Road to Travel
Which Road to Travel
Jeneva Clark
 
Triple beer rings
Triple beer ringsTriple beer rings
Triple beer rings
Jeneva Clark
 

More from Jeneva Clark (20)

161 course information spring 2016
161 course information spring 2016161 course information spring 2016
161 course information spring 2016
 
Summer 2014 moseley 098 syllabus addendum
Summer 2014 moseley 098 syllabus addendumSummer 2014 moseley 098 syllabus addendum
Summer 2014 moseley 098 syllabus addendum
 
2.1 homework
2.1 homework2.1 homework
2.1 homework
 
2.6 homework
2.6 homework2.6 homework
2.6 homework
 
2.2 homework b
2.2 homework b2.2 homework b
2.2 homework b
 
2.2 homework a
2.2 homework a2.2 homework a
2.2 homework a
 
College algebra hw 1.4
College algebra hw 1.4College algebra hw 1.4
College algebra hw 1.4
 
1.3 homework
1.3 homework1.3 homework
1.3 homework
 
Math 111 College Algebra 1.2
Math 111 College Algebra 1.2Math 111 College Algebra 1.2
Math 111 College Algebra 1.2
 
P.5 homework a
P.5 homework aP.5 homework a
P.5 homework a
 
P.4 homework
P.4 homeworkP.4 homework
P.4 homework
 
P.3 homework
P.3 homeworkP.3 homework
P.3 homework
 
P 2 homework b
P 2 homework bP 2 homework b
P 2 homework b
 
P.2 homework A
P.2 homework AP.2 homework A
P.2 homework A
 
P.1 hw
P.1 hwP.1 hw
P.1 hw
 
Moseley schedule
Moseley scheduleMoseley schedule
Moseley schedule
 
Moseley m113 syllabus_summer2013
Moseley m113 syllabus_summer2013Moseley m113 syllabus_summer2013
Moseley m113 syllabus_summer2013
 
Magic Money
Magic MoneyMagic Money
Magic Money
 
Which Road to Travel
Which Road to TravelWhich Road to Travel
Which Road to Travel
 
Triple beer rings
Triple beer ringsTriple beer rings
Triple beer rings
 

125 11.1

  • 1. 11.1 Antiderivatives and Indefinite Integrals A. Let’s figure out the formula. B. Those with rational or negative exponents C. Those that need expanding D. Those that need simplifying E. Word Problems
  • 2. A. Let’s figure out the formula. Recall the power rule for derivatives went like this : x n → nx n −1 Consider these basic ones : x 4 + 13 → 4 x 3 x 4 + 7 → 4 x3 x 4 + 122 → 4 x 3 Since it doesn' t matter what the constant term is, I could say : x 4 + C → 4 x 3 , where C is any constant. Our next task will be trying to take the 4 x 3 and get the " indefinite integral" of it x 4 + C.
  • 3. How can we get from 4 x to x + C ? 3 4  First add one to its exponent.  Then, divide the term by its new exponent.  Last, add the + C. 3+1 4x 4x → 3 +C 3 +1 4 4x = +C 4 = x +C 4
  • 4. This is sort of undoing differentiation! • It’s like we are answering the riddle: “What is the function whose derivative is this?” ∫ 4 x dx = " indefinite integral of 4 x = x +C 3 3 4 n +1 x ∫ x dx = n + 1 + C n
  • 5. ∫ 6 x dx You try : ∫ 12 x dx 2 3
  • 6. Term by term : ∫(x ) + 4 x 3 − 3 x 2 + 7 dx 5 is the same as ∫ x 5 dx + ∫ 4 x 3dx − ∫ 3 x 2 dx + ∫ 7 dx i.e., take the integral of each term, like 4 smaller tasks. Notice the last term is ∫ 7 dx. What function would have a derivative of just plain 7? Answer 7x. x 6 4 x 4 3x 3 + − + 7x + C 6 4 3 x6 = + x 4 − x3 + 7 x + C 6
  • 7.  8 5 2x  You try : ∫  3x − 2 x + x −  7 6 + 14 dx   3 
  • 8. B. Those with rational or negative exponents • As you have seen before, sometimes it is necessary to REWRITE the problem in exponential form without negative exponents BEFORE using the rules. • Examples of rewriting:
  • 9. 3  4 3 6  dx ∫  4 x − x 2 dx   ∫  3 ⋅ x + 4 x 3 dx     ∫ x3
  • 10. You try :  7  3 2 2  du ∫  6 t − 2 dt t  ∫ 4⋅ x +  3 x2 dx   ∫ u4
  • 11. C. Those that need expanding • You know that you can integrate term-by- term. Sometimes you have to EXPAND things (F.O.I.L., distribute, etc.) before you can do the term-by-term thing. • “Terms” are only connected to each other by plus signs. Nothing else.
  • 12. ∫ ( x + 2) 3 dx You try : ∫ ( x − 1) dx 3
  • 13. ∫ ( x + 2) 2 x dx You try : ∫ ( x − 1) 23 x dx
  • 14. 4 x 4 − x 3 + 6 x 2 + 14 x 5x5 + x 4 − 4 x 2 + 6 x ∫ x dx You try : ∫ x dx
  • 15. D. Those that need simplifying ( x + 3)( x + 5) dx = ( x + 3) dx = x 2 ∫ x+5 ∫ 2 + 3x + C •FACTOR TOP AND BOTTOM •CANCEL VERTICALLY IF POSSIBLE •INTEGRATE TERM BY TERM
  • 16. 3x − x − 2 2 You try : ∫ dx x −1
  • 17. E. Word Problems Differentiation does this : profit/revenue/cost → marginal profit/revenue/cost curve → slope of the tangent line distance → velocity velocity → acceleration any quantity → rate of change So Integration does the opposite : marginals → profit/revenue/cost functions slope of the tangent line → the equation of the curve velocity → distance acceleration → velocity rate of change → the function for the quantity
  • 18. A company's marginal cost function is MC ( x ) = 6 x and the fixed cost is $1000. Find the cost function.
  • 19. A company's marginal revenue function is MR = 12 x + 83 x , where x is the number of units. Find the revenue function. [Hint for finding C : How much revenue is brought in for zero production? zero.]
  • 20. 3 A person can memorize words at the rate of words per minute. t Find a formula for the total number of words that can be memorized in t minutes. [Hint for finding C : How many words can be memorized in zero minutes? zero.]
  • 21. dC The marginal cost for producing x units of a product is modeled by = 32 − 0.04 x. dx It costs $50 to produce one unit. Find the total cost of producing 200 units.