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