Lesson 21: Antiderivatives (slides)

Section 4.7Antiderivatives

V63.0121.006/016, Calculus I

New York University

April 8, 2010

Announcements

I Quiz April 16 on §§4.1–4.4I Final Exam: Monday, May 10, 10:00am

..Image credit: Ian Hampton . . . . . .

http://www.flickr.com/photos/ianhampton/65178598/

. . . . . .

Announcements

I Quiz April 16 on §§4.1–4.4I Final Exam: Monday, May 10, 10:00am

V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 2 / 32

. . . . . .

Outline

What is an antiderivative?

Tabulating AntiderivativesPower functionsCombinationsExponential functionsTrigonometric functions

Finding Antiderivatives Graphically

Rectilinear motion


. . . . . .

Objectives

I Given an expression forfunction f, find adifferentiable function Fsuch that F′ = f (F is calledan antiderivative for f).

I Given the graph of afunction f, find adifferentiable function Fsuch that F′ = f

I Use antiderivatives tosolve problems inrectilinear motion


. . . . . .

Hard problem, easy check

Example

Find an antiderivative for f(x) = ln x.

Solution???

Example

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

Solution

ddx

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


. . . . . .


Example


Solution???

Example


Solution

ddx

(x ln x− x)

= 1 · ln x+ x · 1x− 1 = ln x"


. . . . . .


Example


Solution???

Example


Solution

ddx

(x ln x− x) = 1 · ln x+ x · 1x− 1

= ln x"


. . . . . .


Example


Solution???

Example


Solution

ddx

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

"


. . . . . .


Example


Solution???

Example


Solution

ddx

(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. Then f is continuous on[x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y)such that

f(y)− f(x)y− x

= f′(z) =⇒ 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.


. . . . . .

When two functions have the same derivative

TheoremSuppose f and g are two differentiable functions on (a,b) with f′ = g′.Then f and g differ by a constant. That is, there exists a constant Csuch 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)


. . . . . .

Outline




Rectilinear motion


. . . . . .

Antiderivatives of power functions

Recall that the derivative of apower function is a powerfunction.

Fact (The Power Rule)

If f(x) = xr, then f′(x) = rxr−1.

So in looking for antiderivativesof power functions, try powerfunctions!

..x

.y.f(x) = x2

.f′(x) = 2x

.F(x) = ?


. . . . . .

Example

Find an antiderivative for the function f(x) = x3.

Solution

I Try a power function 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 antiderivative.

I Check:ddx

(14x4)

= 4 · 14x4−1 = x3 "

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


. . . . . .

Example


Solution



I r− 1 = 3 =⇒ r = 4

, and ar = 1 =⇒ a =14.


I Check:ddx

(14x4)

= 4 · 14x4−1 = x3 "



. . . . . .

Example


Solution



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


I Check:ddx

(14x4)

= 4 · 14x4−1 = x3 "



. . . . . .

Example


Solution



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


I Check:ddx

(14x4)

= 4 · 14x4−1 = x3

"



. . . . . .

Example


Solution



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


I Check:ddx

(14x4)

= 4 · 14x4−1 = x3 "



. . . . . .

Example


Solution



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


I Check:ddx

(14x4)

= 4 · 14x4−1 = x3 "

I Any others?

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


. . . . . .

Example


Solution



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


I Check:ddx

(14x4)

= 4 · 14x4−1 = x3 "



. . . . . .

Fact (The Power Rule for antiderivatives)

If f(x) = xr, then

F(x) =1

r+ 1xr+1

is an antiderivative for f…

as long as r ̸= −1.

Fact

If f(x) = x−1 =1x, then

F(x) = ln |x|+ C

is an antiderivative for f.


. . . . . .

Fact (The Power Rule for antiderivatives)

If f(x) = xr, then

F(x) =1

r+ 1xr+1

is an antiderivative for f as long as r ̸= −1.

Fact

If f(x) = x−1 =1x, then

F(x) = ln |x|+ C

is an antiderivative for f.


. . . . . .

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 only definedon positive 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 antiderivative 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|

=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"



. . . . . .


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"



. . . . . .


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"



. . . . . .


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"

I We prefer the antiderivative with the larger domain.V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32

. . . . . .

Graph of ln |x|

. .x

.y

.f(x) = 1/x

.F(x) =


. . . . . .

Graph of ln |x|

. .x

.y

.f(x) = 1/x

.F(x) = ln(x)


. . . . . .

Graph of ln |x|

. .x

.y

.f(x) = 1/x

.F(x) = ln |x|


. . . . . .

Combinations of antiderivatives

Fact (Sum and Constant Multiple Rule for Antiderivatives)

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

I If F is an antiderivative of f and c is a constant, then cF is anantiderivative of cf.

Proof.These follow from the sum and constant multiple rule for derivatives:

I If F′ = f and G′ = g, then

(F+G)′ = F′ +G′ = f+ g

I Or, if F′ = f,(cF)′ = cF′ = cf


. . . . . .

Antiderivatives of Polynomials

Example

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

Solution

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

QuestionWhy do we not need two C’s?

AnswerA combination of two arbitrary constants is still an arbitrary constant.


. . . . . .


Example


Solution

The expression12x2 is an antiderivative for x, and x is an antiderivative

for 1. So

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

is the antiderivative of f.




. . . . . .


Example


Solution

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




. . . . . .

Exponential Functions

FactIf f(x) = ax, f′(x) = (ln a)ax.

Accordingly,

Fact

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

ax + C is the antiderivative of f.

Proof.Check it yourself.

In particular,

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


. . . . . .

Logarithmic functions?

I Remember we found

F(x) = x ln x− x

is an antiderivative of f(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

x+ C

is the antiderivative of f(x).


. . . . . .

Logarithmic functions?

I Remember we found

F(x) = x ln x− x

is an antiderivative of f(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

x+ C

is the antiderivative of f(x).


. . . . . .

Trigonometric functions

Fact

ddx

sin x = cos xddx

cos x = − sin x

So to turn these around,

Fact

I The function F(x) = − cos x+C is the antiderivative of f(x) = sin x.I The function F(x) = sin x+ C is the antiderivative of f(x) = cos x.


. . . . . .


Fact

ddx

sin x = cos xddx

cos x = − sin x


Fact

I The function F(x) = − cos x+C is the antiderivative of f(x) = sin x.

I The function F(x) = sin x+ C is the antiderivative of f(x) = cos x.


. . . . . .


Fact

ddx

sin x = cos xddx

cos x = − sin x


Fact

I The function F(x) = − cos x+C is the antiderivative of f(x) = sin x.I The function F(x) = sin x+ C is the antiderivative of f(x) = cos x.


. . . . . .

More Trig

Example

Find an antiderivative of f(x) = tan x.

Solution???

AnswerF(x) = ln(sec x).

Check

ddx

=1

sec x· ddx

sec x =1

sec x· sec x tan x = tan x"

More about this later.


. . . . . .

More Trig

Example


Solution???


Check

ddx

=1

sec x· ddx

sec x

=1




. . . . . .

More Trig

Example


Solution???


Check

ddx

=1

sec x· ddx

sec x =1

sec x· sec x tan x

= tan x"



. . . . . .

More Trig

Example


Solution???


Check

ddx

=1

sec x· ddx

sec x =1

sec x· sec x tan x = tan x

"



. . . . . .

More Trig

Example


Solution???


Check

ddx

=1

sec x· ddx

sec x =1




. . . . . .

More Trig

Example


Solution???


Check

ddx

=1

sec x· ddx

sec x =1


More about this later.V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32

. . . . . .

Outline




Rectilinear motion


. . . . . .

ProblemBelow is the graph of a function f. Draw the graph of an antiderivativefor F.

..x

.y

..1

..2

..3

..4

..5

..6

.

.

.

. .

. .y = f(x)


. . . . . .

Using f to make a sign chart for F

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

..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .− .+.↗ .↗ .↘ .↘ .↗. max .

min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?

The only question left is: What are the function values?


. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+

.+ .− .− .+.↗ .↗ .↘ .↘ .↗. max .

min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+

.− .− .+.↗ .↗ .↘ .↘ .↗. max .

min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .−

.− .+.↗ .↗ .↘ .↘ .↗. max .

min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .−

.+.↗ .↗ .↘ .↘ .↗. max .

min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .− .+

.↗ .↗ .↘ .↘ .↗. max .min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .− .+.↗

.↗ .↘ .↘ .↗. max .min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .− .+.↗ .↗

.↘ .↘ .↗. max .min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .− .+.↗ .↗ .↘

.↘ .↗. max .min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .− .+.↗ .↗ .↘ .↘

.↗. max .min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .− .+.↗ .↗ .↘ .↘ .↗

. max .min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .− .+.↗ .↗ .↘ .↘ .↗. max

.min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

.++ .−− .−− .++ .++.⌣ .⌢ .⌢ .⌣ .⌣

.IP

.IP

.F

.shape..1

..2

..3

..4

..5

..6. " ." . . . ".? .? .? .? .? .?



. . . . . .



..x

.y

..1

..2

..3

..4

..5

..6

.

.

.. .

.

. .f = F′

.F..1

..2

..3

..4

..5

..6

.+ .+ .− .− .+.↗ .↗ .↘ .↘ .↗. max .

min

.f′ = F′′

.F..1

..2

..3

..4

..5

..6

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. . . . . .

Could you repeat the question?

ProblemBelow is the graph of a function f. Draw the graph of the antiderivativefor F with F(1) = 0.

Solution

I We start with F(1) = 0.I Using the sign chart, we

draw arcs with thespecified monotonicity andconcavity

I It’s harder to tell if/when Fcrosses the axis; moreabout that later.

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. . . . . .



Solution

I We start with F(1) = 0.

I Using the sign chart, wedraw arcs with thespecified monotonicity andconcavity


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. . . . . .



Solution




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Solution




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Solution




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Solution




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Solution




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. . . . . .

Outline




Rectilinear motion


. . . . . .

Say what?

I “Rectilinear motion” just means motion along a line.I Often we are given information about the velocity or acceleration

of a moving particle and we want to know the equations of motion.


. . . . . .

Application: Dead Reckoning


. . . . . .

ProblemSuppose a particle of mass m is acted upon by a constant force F.Find the position function s(t), the velocity function v(t), and theacceleration function a(t).

Solution

I By Newton’s Second Law (F = ma) a constant force induces a

constant acceleration. So a(t) = a =Fm.

I Since v′(t) = a(t), v(t) must be an antiderivative of the constantfunction a. So

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

where v0 is the initial velocity.I Since s′(t) = v(t), s(t) must be an antiderivative of v(t), meaning

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

12at2 + v0t+ s0


. . . . . .


Solution




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

where v0 is the initial velocity.

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

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

12at2 + v0t+ s0


. . . . . .


Solution




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

where v0 is the initial velocity.I Since s′(t) = v(t), s(t) must be an antiderivative of v(t), meaning

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

12at2 + v0t+ s0


. . . . . .

An earlier Hatsumon

Example

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

SolutionAssume 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.


. . . . . .

Example

The skid marks made by an automobile indicate that its brakes werefully applied for a distance of 160 ft before it came to a stop. Supposethat the car in question has a constant deceleration of 20 ft/s2 under theconditions of the skid. How fast was the car traveling when its brakeswere first applied?

Solution (Setup)

I While breaking, the car has acceleration a(t) = −20I Measure time 0 and position 0 when the car starts braking. So

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


. . . . . .

Example


Solution (Setup)

I While breaking, the car has acceleration a(t) = −20

I Measure time 0 and position 0 when the car starts braking. Sos(0) = 0.

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


. . . . . .

Example


Solution (Setup)


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

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


. . . . . .

Example


Solution (Setup)


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


. . . . . .

Implementing the Solution

In 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 equations.


. . . . . .

Implementing the Solution

In 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 equations.


. . . . . .

Solving

We havev0t1 − 10t21 = 160 v0 − 20t1 = 0

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

v0 ·v020

− 10( v020

)2= 160

or

v2020

−10v20400

= 160

2v20 − v20 = 160 · 40 = 6400

I So v0 = 80 ft/s ≈ 55mi/hr


. . . . . .

What have we learned today?

I Antiderivatives are a usefulconcept, especially inmotion

I We can graph anantiderivative from thegraph of a function

I We can computeantiderivatives, but notalways

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.

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f(x) = e−x2

f′(x) = ???


Lesson 21: Antiderivatives (slides)

Education

Transcript of Lesson 21: Antiderivatives (slides)