antiderivatives and initial value problems - dartmouth … · antiderivatives and initial value...
TRANSCRIPT
Antiderivatives and Initial Value Problems
Warm up
If
d
dx
f (x) = 2x , what is f (x)?
f (x) = x
2
Can you think of another function that f (x) could be?
Some other candidates:
f (x) = x
2+ 1, x
2 � 2, x
2+ 13⇡, x
2 � 143.7
If
d
dx
f (x) = 3x
2+ 1, what is f (x)?
f (x) = x
3+ x
Can you think of another function that f (x) could be?
Some other candidates:
f (x) = x
3+ x + 1, x
3+ x � 2, x
3+ x + 13⇡, x
3+ x � 143.7
DefinitionAn antiderivative of a function f on an interval I is another
function F such that F
0(x) = f (x) for all x 2 I .
Examples:
1. An antiderivative of f (x) = 2x is F (x) = x
2.
2. Another antiderivative of f (x) = 2x is F (x) = x
2+ 1.
3. There are lots of antiderivatives of f (x) = 2x which look like
F (x) = x
2+ C .
Suppose that h is di↵erentiable in an interval I ,
and h
0(x) = 0 for all x in I .
Then h is a constant function!
i.e. h(x) = C for all x 2 I , where C is a constant.
So, if F (x) is one antiderivative of f (x), then any other
antiderivative must be of the form F (x) + C .
Example: All of the antiderivatives of f (x) = 2x look like
F (x) = x
2+ C
for some constant C .
Every function f that has at least one antiderivative F has
infinitely many antiderivatives
F (x) + C .
We refer to F (x) + C as the
general antiderivative or the indefinite integral
and denote it by
F (x) + C =
Zf (x)dx .
Example: Z2x dx = x
2+ C .
Examples
Zx
2dx =
13x
3+ C , because
d
dx
(
13x
3+ C ) =
13 ⇤ 3x2 = x
2
Zx
3dx =
14x
4+ C , because
d
dx
(
14x
4+ C ) =
14 ⇤ 4x3 = x
3
Zx
5dx =
16x
6+ C , because
d
dx
(
16x
6+ C ) =
16 ⇤ 6x5 = x
5
Zx
�3dx =
1�2x
�2+ C , because
d
dx
(
1�2x
�2+ C ) = x
�3
Zx
k
dx =
1k+1x
k+1+ C , because
d
dx
(
1k+1x
k+1+ C ) = x
k
Except!! What if k = �1?
Some important basic integrals
Zx
k
dx =
1
k + 1
x
k+1+ C if k 6= �1
Zx
�1dx =
ln(x) + C b/c
d
dx
ln(x) = x
�1
Zsin(x) dx =
� cos(x) + C
Zcos(x) dx =
sin(x) + C
Ze
x
dx =
e
x
+ C
Zsec
2(x) dx =
tan(x) + C
Z1p
1� x
2dx =
arcsin(x) + C
Theorem (Opposite of sum and constant rules)
Suppose the functions f and g both have antiderivatives on the
interval I . Then for any constants a and b, the function af + bg
has an antiderivative on I and
Z �a ⇤ f (x) + b ⇤ g(x)
�dx = a
Zf (x)dx + b
Zg(x)dx
Di↵erential equations
A di↵erential equation is an equation involving derivatives.
The goal is usually to solve for y .
Just like you could use algebra to solve
y
2+ x
2= 1
for y , you can use calculus (and algebra) to solve things like
dy
dx
� 5y = 0 for y .
A solution to a di↵erential equation is a function you can plug in
that satisfies the equation.
For example, y = e
5xis a solution to the di↵erential equation
above since
d
dx
e
5x= 5e
5x ,
so
dy
dx
� 5y = (5e
5x)� 5(e
5x) = 0 X
.
Simplest di↵erential equations: antiderivatives
Finding an antiderivative can also be thought of as solving a
di↵erential equation:
“Solve the di↵erential equation
d
dx
y = x
2.”
Answer: y =
Zx
2dx =
1
3
x
3+ C .
Check:
d
dx
1
3
x
3+ C . =
1
3
⇤ 3 ⇤ x2 + 0 = x
2 X
Examples
(1) Solve the di↵erential equation y
0= 2x + sin(x).
y = x
2 � cos(x) + C
(2) Check that cos(x) + sin(x) is a solution to
d
2y
dx
2+ y = 0.
d
dx
y =
d
dx
(cos(x) + sin(x)) = � sin(x) + cos(x), so
d
2
dx
2y = � cos(x)� sin(x) = �(cos(x) + sin(x)).
Therefore,
d
2y
dx
2+y = �(cos(x)+sin(x))+(cos(x)+sin(x)) = 0 X
Initial value problems
Find a solution to the di↵erential equation
d
dx
y = x
2+ 1 which
also satisfies y(2) = 8/3.
general solution: y =
13x
3+ x + C
particular solution: y =
13x
3+ x � 2
Each color corresponds to a choice of C .
Red cuve is the particular solution.
Initial value problems
Find a solution to the di↵erential equation
d
dx
y = x
2+ 1 which
also satisfies y(2) = 8/3.
general solution: y =
13x
3+ x + C
particular solution: y =
13x
3+ x � 2
Each color corresponds to a choice of C .
Red cuve is the particular solution.
Definition
An initial-value problem is a di↵erential equation together with
enough additional conditions to specify the constants of
integration that appear in the general solution.
The particular solution of the problem is then a function that
satisfies both the di↵erential equation and also the additional
conditions.
Solve the initial value problem
dy
dx
= 2x + sin(x)
subject to y(0) = 0.
general solution: y = x
2 � cos(x) + C
Algebraically: get a particular solution by solving
0 = y(0) = (0)
2 � cos(0) + C = �1 + C (for C )
C = 1, so y = x
2 � cos(x) + 1.
Solve the initial-value problem y
00= cos x , y
0(
⇡2 ) = 2, y(
⇡2 ) = 3⇡.
Step 1: Calculate the antiderivative of cos(x) to find the general
solution for y
0.
Ans: y
0= sin(x) + C
Step 2: Plug in the values y
0(
⇡2 ) = 2 to calculate C .
Ans: 2 = sin(⇡/2) + C = 1 + C , so C = 1
Step 3: Write down the particular solution for y
0.
Ans: y
0= sin(x) + 1
Step 4: Calculate the antiderivative of your particular solution in Step
3 to find the general solution for y .
Ans: y = � cos(x) + x + D
Step 5: Plug in the values y(
⇡2 ) = 3⇡ to solve for the new constant.
Ans: 3⇡ = � cos(⇡/2) + ⇡/2 + D = ⇡/2 + D so D = 5⇡/2
Step 6: Write down the particular solution for y .
Ans: y
0= � cos(x) + x + 5⇡/2
Word problem:An object dropped from a cli↵ has acceleration a = �9.8 m/sec2
under the influence of gravity. What is the function s(t) that
models its height at time t?
Initial value problem:Solve
d
2s
dt
2= �9.8, s(0) = s0, s
0(0) = 0.
Word problem:Suppose that a baseball is thrown upward from the roof of a 100
meter high building. It hits the street below eight seconds later.
What was the initial velocity of the baseball, and how high did it
rise above the street before beginning its descent?
Initial value problem:Solve
d
2s
dt
2= �9.8, s(0) = 100, s(8) = 0.
Use your solution to
(1) calculate s
0(0), and
(2) solve s
0(t1) = 0 for t1 and calculate s(t1).
Extra practice: Basic antiderivatives
Problem A. Indefinite integrals.
1.
Zx
7dx
2.
Zx
�7dx
3.
Zx
�1dx
4.
Zx
5/3dx
5.
Zx
�5/4dx
6.
Z3px
2dx
7.
Z1
4px
3dx
8.
Z2
x
2dx
9.
Z(8� x+ 2x
3 � 6/x
3+ 2x
�5+ 5x
�1) dx
10.
Z(2� 5x)(3 + 2x)(1� x) dx
11.
Z px(ax
2+ bx+ c) dx
12.
Z(x
2 � 1/x
2)
3dx
13.
Z(
px� 1/
px) dx
14.
Z ✓px+
1px
◆2
dx
15.
Z(1 + 2x)
3
x
4dx
16.
Z(1 + x)
3
px
dx
17.
Z2x
2+ x� 2
x� 2
dx
18. If
df
dx
= x� 1/x
2and f(1) = 1/2 find f(x).
Problem B. Indefinite integrals with trigonometric functions.
1.
Z ✓9 sinx� 7 cosx� 6
cos
2x
+
2
sin
2x
+ cot
2x
◆dx
2.
Z ✓cotx
sinx
� tan
2x� tanx
cosx
+
2
cos
2x
◆dx
3.
Zsecx(secx+ tanx) dx
4.
Zcscx(cscx� cotx) dx
5.
Z(tanx+ cotx)
2dx
6.
Z1 + 2 sinx
cos
2x
dx
7.
Z3 cosx+ 4
sin
2x
dx
8.
Z1
1� cosx
dx
[hint: mult and divide by (1 + cos(x)) and
rewrite using csc(x), etc.]
9.
Z1
1 + cosx
dx
10.
Ztanx
secx+ tanx
dx
11.
Zcscx
cscx� cotx
dx
12.
Zcosx
1 + cosx
dx
13.
Zsinx
1� sinx
dx
14.
Z p1 + cos 2x dx
15.
Z p1� cos 2x dx
16.
Z1
1 + cos 2x
dx
17.
Z1
1� cos 2x
dx
18.
Z p1 + sin 2x dx
19.
Zsin
3x+ cos
3x
sin
2x cos
2x
dx
Problem C. Integrals with exponential functions and inverse functions.
1.
Z2
x
dx
2.
Z(6x
5 � 2x
�4 � 7x+ 3/x� 5 + 4e
x
+ 7
x
) dx
3.
Z(x/a+ a/x+ x
a
+ a
x
+ ax) dx
4.
Z ✓px� 3
px
4+
7
3px
2� 6e
x
+ 1
◆dx
5.
Z ✓1 +
1
1 + x
2� 2p
1� x
2+
5
x
px
2 � 1
+ a
x
◆dx
6.
Zx
2
1 + x
2dx
[hint: add and subtract 1 in the numerator]
7.
Zx
2 � 1
x
2+ 1
dx
8.
Zx
6 � 1
x
2+ 1
dx
9.
Zx
4
1 + x
2dx
10.
Zcos
�1(sinx) dx