Lesson 23: Antiderivatives (slides)

Sec on 4.7An deriva ves

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

April 19, 2011

Announcements

I Quiz 5 on Sec ons4.1–4.4 April 28/29

I Final Exam Thursday May12, 2:00–3:50pm

I I am teaching Calc II MW2:00pm and Calc III TR2:00pm both Fall ’11 andSpring ’12

ObjectivesI Given a ”simple“ elementaryfunc on, find a func on whosederiva ve is that func on.

I Remember that a func onwhose deriva ve is zero alongan intervalmust be zero alongthat interval.

I Solve problems involvingrec linear mo on.

OutlineWhat is an an deriva ve?

Tabula ng An deriva vesPower func onsCombina onsExponen al func onsTrigonometric func onsAn deriva ves of piecewise func ons

Finding An deriva ves Graphically

Rec linear mo on

What is an antiderivative?

Defini onLet f be a func on. An an deriva ve for f is a func on F such thatF′ = f.

Who cares?Ques on

Why would we want the an deriva ve of a func on?

Answers

I For the challenge of itI For applica ons when the deriva ve of a func on is known butthe original func on is not

I Biggest applica on will be a er the Fundamental Theorem ofCalculus (Chapter 5)

Hard problem, easy check

Example

Find an an deriva ve for f(x) = ln x.

Solu on???

Example

Find an an deriva ve for f(x) = ln x.

Solu on???

Example

is F(x) = x ln x− x an an deriva ve for f(x) = ln x?

Solu on

(x ln x− x) = 1 · ln x+ x · 1x− 1 = ln x

Example

is F(x) = x ln x− x an an deriva ve for f(x) = ln x?

Solu on

(x ln x− x) = 1 · ln x+ x · 1x− 1 = ln x"

Why the MVT is the MITCMost Important Theorem In Calculus!

TheoremLet f′ = 0 on an interval (a, b). Then f is constant on (a, b).

Proof.Pick any points x and y in (a, b) with x < y. By MVT there exists apoint z in (x, y) such that

f(y) = f(x) + f′(z)(y− x)

But f′(z) = 0, so f(y) = f(x). Since this is true for all x and y in (a, b),then f is constant.

Functions with the same derivativeTheoremSuppose f and g are two differen able func ons on (a, b) withf′ = g′. Then f and g differ by a constant. That is, there exists aconstant C such that f(x) = g(x) + C.

Proof.

I Let h(x) = f(x)− g(x)I Then h′(x) = f′(x)− g′(x) = 0 on (a, b)I So h(x) = C, a constantI This means f(x)− g(x) = C on (a, b)

Rec linear mo on

Antiderivatives of power functionsRecall that the deriva ve of apower func on is a powerfunc on.Fact (The Power Rule)

If f(x) = xr, then f′(x) = rxr−1.

So in looking foran deriva ves of powerfunc ons, try powerfunc ons!

f(x) = x2

f′(x) = 2x

F(x) = ?

f(x) = x2

f′(x) = 2x

F(x) = ?

f(x) = x2

f′(x) = 2x

F(x) = ?

f(x) = x2

f′(x) = 2x

F(x) = ?

Antiderivatives of power functionsExample

Find an an deriva ve for the func on f(x) = x3.

Solu on

I Try a power func on F(x) = axr

I Then F′(x) = arxr−1, so we want arxr−1 = x3.

I r− 1 = 3 =⇒ r = 4

, and ar = 1 =⇒ a =14.

I So F(x) =14x4 is an an deriva ve.

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3"

I Any others?

Yes, F(x) =14x4 + C is the most general form.

Solu on

I r− 1 = 3 =⇒ r = 4

, and ar = 1 =⇒ a =14.

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3"

I Any others?

Solu on

I r− 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =14.

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3"

I Any others?

Solu on

I r− 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =14.

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3"

I Any others?

Solu on

I r− 1 = 3 =⇒ r = 4, and ar = 1 =⇒ a =14.

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3"

I Any others?

Solu on

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3"

I Any others?

Solu on

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3"

I Any others?

Solu on

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3"

I Any others?

Solu on

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3"

I Any others? Yes, F(x) =14x4 + C is the most general form.

General power functionsFact (The Power Rule for an deriva ves)

If f(x) = xr, then

F(x) =1

r+ 1xr+1

is an an deriva ve for f…

as long as r ̸= −1.

If f(x) = x−1 =1x, then F(x) = ln |x|+ C is an an deriva ve for f.

If f(x) = xr, then

F(x) =1

r+ 1xr+1

is an an deriva ve for f as long as r ̸= −1.

If f(x) = xr, then

F(x) =1

r+ 1xr+1

is an an deriva ve for f as long as r ̸= −1.

What’s with the absolute value?

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I The domain of F is all nonzero numbers, while ln x is onlydefined on posi ve numbers.

I If x > 0,ddx

ln |x| = ddx

ln(x) =1x"

I If x < 0,ddx

ln |x| = ddx

ln(−x) =1−x

· (−1) =1x"

I We prefer the an deriva ve with the larger domain.

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x|

ln(x) =1x"

I If x < 0,ddx

ln |x| = ddx

ln(−x) =1−x

· (−1) =1x"

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x| = ddx

I If x < 0,ddx

ln |x| = ddx

ln(−x) =1−x

· (−1) =1x"

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x| = ddx

ln(x) =1x

I If x < 0,ddx

ln |x| = ddx

ln(−x) =1−x

· (−1) =1x"

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x| = ddx

ln(x) =1x"

I If x < 0,ddx

ln |x| = ddx

ln(−x) =1−x

· (−1) =1x"

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x| = ddx

ln(x) =1x"

I If x < 0,ddx

ln |x|

ln(−x) =1−x

· (−1) =1x"

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x| = ddx

ln(x) =1x"

I If x < 0,ddx

ln |x| = ddx

ln(−x)

=1−x

· (−1) =1x"

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x| = ddx

ln(x) =1x"

I If x < 0,ddx

ln |x| = ddx

ln(−x) =1−x

· (−1)

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x| = ddx

ln(x) =1x"

I If x < 0,ddx

ln |x| = ddx

ln(−x) =1−x

· (−1) =1x

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x| = ddx

ln(x) =1x"

I If x < 0,ddx

ln |x| = ddx

ln(−x) =1−x

· (−1) =1x"

F(x) = ln |x| =

{ln(x) if x > 0;ln(−x) if x < 0.

I If x > 0,ddx

ln |x| = ddx

ln(x) =1x"

I If x < 0,ddx

ln |x| = ddx

ln(−x) =1−x

· (−1) =1x"

Graph of ln |x|

. f(x) = 1/x

Graph of ln |x|

. f(x) = 1/x.

F(x) = ln(x)

Graph of ln |x|

. f(x) = 1/x.

F(x) = ln |x|

Combinations of antiderivatives

Fact (Sum and Constant Mul ple Rule for An deriva ves)

I If F is an an deriva ve of f and G is an an deriva ve of g, thenF+ G is an an deriva ve of f+ g.

I If F is an an deriva ve of f and c is a constant, then cF is anan deriva ve of cf.

Combinations of antiderivativesProof.These follow from the sum and constant mul ple rule forderiva ves:

I If F′ = f and G′ = g, then

(F+ G)′ = F′ + G′ = f+ g

I Or, if F′ = f,(cF)′ = cF′ = cf

Antiderivatives of PolynomialsExample

Find an an deriva ve for f(x) = 16x+ 5.

Solu on

The expression12x2 is an an deriva ve for x, and x is an

an deriva ve for 1. So

F(x) = 16 ·(12x2)+ 5 · x+ C = 8x2 + 5x+ C

is the an deriva ve of f.

Antiderivatives of PolynomialsExample

Find an an deriva ve for f(x) = 16x+ 5.

Solu on

The expression12x2 is an an deriva ve for x, and x is an

an deriva ve for 1. So

F(x) = 16 ·(12x2)+ 5 · x+ C = 8x2 + 5x+ C

is the an deriva ve of f.

Antiderivatives of Polynomials

Ques on

Do we need two C’s or just one?

AnswerJust one. A combina on of two arbitrary constants is s ll anarbitrary constant.

Antiderivatives of Polynomials

Ques on

Do we need two C’s or just one?

AnswerJust one. A combina on of two arbitrary constants is s ll anarbitrary constant.

Exponential FunctionsFactIf f(x) = ax, f′(x) = (ln a)ax.

Accordingly,Fact

If f(x) = ax, then F(x) =1ln a

ax + C is the an deriva ve of f.

Proof.Check it yourself.

In par cular,FactIf f(x) = ex, then F(x) = ex + C is the an deriva ve of f.

Accordingly,Fact

Exponential Functions

In par cular,

FactIf f(x) = ex, then F(x) = ex + C is the an deriva ve of f.

Exponential Functions

In par cular,

FactIf f(x) = ex, then F(x) = ex + C is the an deriva ve of f.

Logarithmic functions?I Remember we found F(x) = x ln x− x is an an deriva ve off(x) = ln x.

I This is not obvious. See Calc II for the full story.I However, using the fact that loga x =

ln xln a

, we get:

FactIf f(x) = loga(x)

F(x) =1ln a

(x ln x− x) + C = x loga x−1ln a

is the an deriva ve of f(x).

I This is not obvious. See Calc II for the full story.

I However, using the fact that loga x =ln xln a

, we get:

F(x) =1ln a

I This is not obvious. See Calc II for the full story.I However, using the fact that loga x =

ln xln a

, we get:

F(x) =1ln a

Trigonometric functionsFact

sin x = cos xddx

cos x = − sin x

So to turn these around,Fact

I The func on F(x) = − cos x+ C is the an deriva ve off(x) = sin x.

I The func on F(x) = sin x+ C is the an deriva ve off(x) = cos x.

sin x = cos xddx

cos x = − sin x

sin x = cos xddx

cos x = − sin x

More TrigExample

Find an an deriva ve of f(x) = tan x.

AnswerF(x) = ln | sec x|.

sec x· ddx

sec x· sec x tan x = tan x"

More about this later.

More TrigExample

Solu on???

sec x· ddx

More TrigExample

sec x· ddx

More TrigExample

sec x· ddx

More TrigExample

sec x· ddx

More TrigExample

sec x· ddx

sec x =1

sec x· sec x tan x

= tan x"

More TrigExample

sec x· ddx

sec x =1

sec x· sec x tan x = tan x

More TrigExample

sec x· ddx

sec x =1

More TrigExample

sec x· ddx

sec x =1

Antiderivatives of piecewise functions

Example

f(x) =

{x if 0 ≤ x ≤ 1;1− x2 if 1 < x.

Find the an deriva ve of f with F(0) = 1.

Antiderivatives of piecewise functions

Solu onWe can an differen ate each piece:

F(x) =

12x2 + C1 if 0 ≤ x ≤ 1;

x− 13x3 + C2 if 1 < x.

The constants need to be chosen so that F(0) = 1 and F iscon nuous (at 1).

F(x) =

12x2 + C1 if 0 ≤ x ≤ 1;

x− 13x3 + C2 if 1 < x.

I Note F(0) =1202 + C1 = C1, so if F(0) is to be 1, C1 = 1.

I This means limx→1−

F(x) =1212 + 1 =

I On the other hand,

limx→1+

F(x) = 1− 13+ C2 =

23+ C2

So for F to be con nuous we need32=

23+ C2. Solving, C2 =

F(x) =

12x2 + C1 if 0 ≤ x ≤ 1;

x− 13x3 + C2 if 1 < x.

F(x) =1212 + 1 =

limx→1+

F(x) = 1− 13+ C2 =

23+ C2

F(x) =

12x2 + C1 if 0 ≤ x ≤ 1;

x− 13x3 + C2 if 1 < x.

F(x) =1212 + 1 =

limx→1+

F(x) = 1− 13+ C2 =

23+ C2

Rec linear mo on

Finding Antiderivatives Graphically

ProblemPictured is the graph of afunc on f. Draw the graph ofan an deriva ve for f.

.......y = f(x)

Using f to make a sign chart for FAssuming F′ = f, we can make a sign chart for f and f′ to find theintervals of monotonicity and concavity for F:

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

The only ques on le is: What are the func on values?

......

.. f = F′.F

. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +

. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −

. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −

. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

......

.. f = F′.F

. +. +. −. −. +.↗

f′ = F′′

−−

Could you repeat the question?ProblemBelow is the graph of a func on f. Draw the graph of thean deriva ve for f with F(1) = 0.

Solu on

I We start with F(1) = 0.I Using the sign chart, we draw arcswith the specified monotonicity andconcavity

I It’s harder to tell if/when F crossesthe axis; more about that later.

.......f

......

Solu on

I We start with F(1) = 0.

I Using the sign chart, we draw arcswith the specified monotonicity andconcavity

.......f

Solu on

.......f

Solu on

.......f

Solu on

.......f

Solu on

.......f

Solu on

.......f

Solu on

.......f

Solu on

.......f

Solu on

.......f

Solu on

.......f

......

Solu on

.......f

......

Rec linear mo on

Say what?

I “Rec linear mo on” just means mo on along a line.I O en we are given informa on about the velocity oraccelera on of a moving par cle and we want to know theequa ons of mo on.

Application: Dead Reckoning

ProblemSuppose a par cle of mass m is acted upon by a constant force F.Find the posi on func on s(t), the velocity func on v(t), and theaccelera on func on a(t).

Solu on

I By Newton’s Second Law (F = ma) a constant force induces aconstant accelera on. So a(t) = a =

I Since v′(t) = a(t), v(t)must be an an deriva ve of theconstant func on a. So

v(t) = at+ C = at+ v0

where v0 is the ini al velocity.I Since s′(t) = v(t), s(t)must be an an deriva ve of v(t),meaning

s(t) =12at2 + v0t+ C =

12at2 + v0t+ s0

Solu on

v(t) = at+ C = at+ v0

s(t) =12at2 + v0t+ C =

12at2 + v0t+ s0

Solu on

v(t) = at+ C = at+ v0

where v0 is the ini al velocity.

I Since s′(t) = v(t), s(t)must be an an deriva ve of v(t),meaning

s(t) =12at2 + v0t+ C =

12at2 + v0t+ s0

Solu on

v(t) = at+ C = at+ v0

s(t) =12at2 + v0t+ C =

12at2 + v0t+ s0

An earlier HatsumonExample

Drop a ball off the roof of the Silver Center. What is its velocity whenit hits the ground?

Solu onAssume s0 = 100m, and v0 = 0. Approximate a = g ≈ −10. Then

s(t) = 100− 5t2

So s(t) = 0 when t =√20 = 2

√5. Then

v(t) = −10t,

so the velocity at impact is v(2√5) = −20

√5m/s.

An earlier HatsumonExample

Drop a ball off the roof of the Silver Center. What is its velocity whenit hits the ground?

Solu onAssume s0 = 100m, and v0 = 0. Approximate a = g ≈ −10. Then

s(t) = 100− 5t2

So s(t) = 0 when t =√20 = 2

√5. Then

v(t) = −10t,

so the velocity at impact is v(2√5) = −20

√5m/s.

Finding initial velocity fromstopping distance

Example

The skid marks made by an automobile indicate that its brakes werefully applied for a distance of 160 before it came to a stop.Suppose that the car in ques on has a constant decelera on of20 ft/s2 under the condi ons of the skid. How fast was the cartraveling when its brakes were first applied?

Solu on (Setup)

I While braking, the car has accelera on a(t) = −20I Measure me 0 and posi on 0 when the car starts braking. Sos(0) = 0.

I The car stops at me some t1, when v(t1) = 0.I We know that when s(t1) = 160.I We want to know v(0), or v0.

Example

Solu on (Setup)

I While braking, the car has accelera on a(t) = −20

I Measure me 0 and posi on 0 when the car starts braking. Sos(0) = 0.

Example

Solu on (Setup)

I The car stops at me some t1, when v(t1) = 0.

I We know that when s(t1) = 160.I We want to know v(0), or v0.

Example

Solu on (Setup)

Implementing the SolutionIn general,

s(t) = s0 + v0t+12at2

Since s0 = 0 and a = −20, we have

s(t) = v0t− 10t2

v(t) = v0 − 20t

for all t.

Plugging in t = t1,

160 = v0t1 − 10t210 = v0 − 20t1

We need to solve these two equa ons.

Implementing the SolutionIn general,

s(t) = s0 + v0t+12at2

Since s0 = 0 and a = −20, we have

s(t) = v0t− 10t2

v(t) = v0 − 20t

for all t. Plugging in t = t1,

160 = v0t1 − 10t210 = v0 − 20t1

We need to solve these two equa ons.

SolvingWe have

v0t1 − 10t21 = 160 v0 − 20t1 = 0

I The second gives t1 = v0/20, so subs tute into the first:

v0 ·v020

− 10( v020

)2= 160

I Solve:

− 10v20400

2v20 − v20 = 160 · 40 = 6400

So v0 = 80 ft/s ≈ 55mi/hr

SolvingWe have

v0t1 − 10t21 = 160 v0 − 20t1 = 0

v0 ·v020

− 10( v020

)2= 160

I Solve:

− 10v20400

2v20 − v20 = 160 · 40 = 6400

SolvingWe have

v0t1 − 10t21 = 160 v0 − 20t1 = 0

v0 ·v020

− 10( v020

)2= 160

I Solve:

− 10v20400

2v20 − v20 = 160 · 40 = 6400

Summary of Antiderivatives so farf(x) F(x)

xr, r ̸= 11

r+ 1xr+1 + C

1x= x−1 ln |x|+ Cex ex + C

ax1ln a

ax + Cln x x ln x− x+ C

loga xx ln x− x

ln a+ C

sin x − cos x+ Ccos x sin x+ Ctan x ln | tan x|+ C

Final Thoughts

I An deriva ves are auseful concept, especiallyin mo on

I We can graph anan deriva ve from thegraph of a func on

I We can computean deriva ves, but notalways

....... f.......F

f(x) = e−x2

f′(x) = ???

Lesson 23: Antiderivatives (slides)

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