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An antiderivative of a function is a function whose derivative is the given function. The problem of antidifferentiation is interesting, complicated, and useful, especially when discussing motion.This is the slideshow version from class.
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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 . . . . . .
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 3 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 4 / 32
. . . . . .
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"
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
. . . . . .
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"
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
. . . . . .
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"
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
. . . . . .
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"
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
. . . . . .
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"
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
. . . . . .
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
"
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
. . . . . .
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"
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 5 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 6 / 32
. . . . . .
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)
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 7 / 32
. . . . . .
Outline
What is an antiderivative?
Tabulating AntiderivativesPower functionsCombinationsExponential functionsTrigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 8 / 32
. . . . . .
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) = ?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
. . . . . .
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) = ?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
. . . . . .
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) = ?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
. . . . . .
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) = ?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 9 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 10 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 11 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 11 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 11 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 12 / 32
. . . . . .
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.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) =
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32
. . . . . .
Graph of ln |x|
. .x
.y
.f(x) = 1/x
.F(x) = ln(x)
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32
. . . . . .
Graph of ln |x|
. .x
.y
.f(x) = 1/x
.F(x) = ln |x|
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 13 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 14 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 14 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
. . . . . .
Antiderivatives of Polynomials
Example
Find an antiderivative for f(x) = 16x+ 5.
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.
QuestionWhy do we not need two C’s?
AnswerA combination of two arbitrary constants is still an arbitrary constant.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
. . . . . .
Antiderivatives of Polynomials
Example
Find an antiderivative for f(x) = 16x+ 5.
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.
QuestionWhy do we not need two C’s?
AnswerA combination of two arbitrary constants is still an arbitrary constant.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 15 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 16 / 32
. . . . . .
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).
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 17 / 32
. . . . . .
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).
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 17 / 32
. . . . . .
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).
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 17 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 18 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 18 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 18 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
. . . . . .
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.V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 19 / 32
. . . . . .
Outline
What is an antiderivative?
Tabulating AntiderivativesPower functionsCombinationsExponential functionsTrigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 20 / 32
. . . . . .
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)
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 21 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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?
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 22 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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, wedraw arcs with thespecified monotonicity andconcavity
I It’s harder to tell if/when Fcrosses the axis; moreabout that later.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
.
.
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
.
.
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
.
.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
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.
..x
.y
..1
..2
..3
..4
..5
..6
.
.
.. .
. .f
.F
.shape..1
..2
..3
..4
..5
..6. " ." . . . "
.IP .max
.IP .min
.
..
.
..
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 23 / 32
. . . . . .
Outline
What is an antiderivative?
Tabulating AntiderivativesPower functionsCombinationsExponential functionsTrigonometric functions
Finding Antiderivatives Graphically
Rectilinear motion
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 24 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 25 / 32
. . . . . .
Application: Dead Reckoning
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 26 / 32
. . . . . .
Application: Dead Reckoning
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 26 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 27 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 28 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 28 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
. . . . . .
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) = −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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 29 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 30 / 32
. . . . . .
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.
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 30 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 31 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 31 / 32
. . . . . .
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
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 31 / 32
. . . . . .
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
..x
.y
..1
..2
..3
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.
.
.. .
. .f.
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.
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f(x) = e−x2
f′(x) = ???
V63.0121, Calculus I (NYU) Section 4.7 Antiderivatives April 8, 2010 32 / 32