An antiderivative of the function f is a function F so that

The Integral
• Indefinite Integral………………...………….…2
• Integration by Substitution………………..…4
• Definite Integral……………………………......7
• The Fundamental Theorem of Calculus….13

Indefinite Integral
An antiderivative of the function f is a function F so that
F ′( x ) = f ( x ), ∀ x ∈ I
1 3
Example: F ( x) = x + C is antiderivative of f ( x) = x 2
3
because of F ′( x) = f ( x) .

The antiderivative of the function f is not unique, but
they are different by constant.
The collection of antiderivatives of function f is called the
indefinite integral of f with respect to x and denote by
f ( x) dx = F ( x) + C
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2

Some of Integral
Formulas

x r +1
1. x dx =
+ C , r ≠ −1
r +1
r

2. sin x dx = − cos x + C
3. cos x dx = sin x + C
4. sec 2 x dx = tan x + C
5. csc 2 x dx = − cot x + C

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3

Integration
by Substitution
Let u = g ( x), du = g ′( x)dx , and F antiderivative of f ,
then

f ( g ( x)) g ′( x) dx =

f (u ) du = F (u ) + C = F ( g ( x)) + C

Example:
sin ( 2 x + 1) dx
1. Find
1
Answer: Let u = 2 x + 1, du = 2dx → dx = du
2
1
sin ( 2 x + 1) dx =
sin u du
2
1

1
= − cos u + C = − cos ( 2 x + 1) + C
2
2
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4

The Integral
2. Find

chapter

6

( x 3 + 1)10 x 5 dx

1
Answer: Let u = x + 1, du = 3x dx → dx = 2 du
3x

1 10 3
3
10 5
10 5 du
u x du
=
( x + 1) x dx = u x
then
2
3x
3
3

2

3
3
x
= u − 1 , thus
u

=
x
+
1
because
then

(x3 +1)10 x5dx = 13 u10 (u −1)du = 13 u11 −u10du
= 361 u12 − 331 u11 + C
= 361 (x3 +1)12 − 331 (x3 +1)11 + C
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

5

Sigma Notation
n

n

k = k + k + ... + k = nk

ai = a1 + a2 + ... + an and

i =1

i =1

n term

Algebra rules for finite sums:
n

n

( k ai + lbi ) = k

1.
i =1

n

ai + l
i =1

bi
i =1

n

n(n + 1)
2
i =1
n
n(n + 1)(2n + 1)
3.
i2 =
6
i =1
i=

2.

n

i3 =

4.
i =1

n(n + 1)
2

2

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6

Definite Integral
Let a function f defined on [ a,b ]. We will construct

definite integral as limit of Riemann sum.
Steps :
1. Divided [a,b] into n subinterval
with a = x0 < x1 < ... < xn = b
ck
point of cut off are
P = { a = x0 , x1, x2 ,...,b = xn}
xk −1 xk b
a x1
is called partition of [a,b].
∆xk
2. Length of partition P is

|| P ||= max | ∆xk |, ∆xk = xk − xk −1
1≤ k ≤ n

3. Choose ck ∈ [ xk −1 , xk ], k = 1, 2, , n.
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7

The Integral

chapter

6

4. Defined Riemann sum
f (ck )

ck

a x2 xk −1 xk

b

n

f (ck ) ∆xk
k =1

∆xk

If || P || → 0 , then we get limit of Riemann sum
n

lim

|| P||→0

k =1

f (ck ) ∆xk

If this limit exists, then f is integrable on [a,b], denoted
by
b
n
n
f (c )∆x
f ( x) dx = lim
f (c ) ∆x = lim
k
k
n→∞ k =1 k k
|P||→0 k =1
a
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8

Example
2

Use Riemann sum to evaluate

( x − 2) dx
0

Answer:

(i) Divide interval [0,2] into n subinterval all
of the same length, that is
∆x = n2
∆x ∆x

0 x1 x2

∆x

∆x

xi−1

xi

xn−1 2

x0 = 0
x1 = 0 + ∆x = n2
x2 = 0 + 2∆x = 2n.2

xi = 0 + i∆x = 2ni

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9

The Integral

chapter

6

(ii) Choose ci = xi
(iii) Defined Riemann sum
n

n

f ( ci ) ∆xi =
i =1

i =1

( 2ni − 2) n2 =

n
i =1

(

4i
n2

− n4

)

4
= 2
n

n

4
i−
n
i =1

n

1
i =1

4 n(n +1) 4
2
= 2
− n = −2 +
n
n
n
2

(iv) If n → ∞ then
2

( x − 2)dx = lim ( −2 + n2 ) = −2
0

n→∞

Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

10

The Integral

chapter

6

Remark: If function y = f(x) positive on interval [a,b]
then definite integral = area of region under the graph
of y = f(x) over the interval [a,b].
Some properties of definite integral
1. Linear
b

b

[ p f ( x) + q g( x)] dx = p

a

b

f ( x) dx + q g( x) dx

a

a

2. If a < b < c, then
c

b

c

f (x) dx = f (x) dx + f (x) dx
a

a

b

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11

The Integral
a

3.

b

f (x) dx = 0

a

6

f ( x) dx = − f (x) dx

and
a

a

chapter

b
a

4. If f(x) is an odd function, then

f ( x) dx = 0
−a

a

5. If f(x) is an even function, then

a

f ( x) dx = 2 f ( x) dx
−a

0

3

x x 4 + x 2 + 7 dx

Example: Evaluate
Answer:
Because of

−3

f (− x) = − x (− x)4 + (− x)2 + 7 = − x x4 + x2 + 7 = − f ( x)
3

x x4 + x2 + 7 dx = 0

then, f(x) is a odd function, therefore
−3

Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

12

.

The Fundamental
Theorem of Calculus Part 1
The Fundamental Theorem of Calculus Part 1
x

If f is continuous on [a,b] then F ( x) = f (t ) dt is
a

continuous on [a,b] and differentiable and its
derivative is f(x);
x

d
′
F ( x) =
f (t ) dt = f ( x)
dx a

Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

13

The Integral

chapter

6

From the fundamental theorem of calculus part 1,
we can derive:
d
dx

u( x)

d
dx

v( x)

f (t )dt = f (u( x))u′( x)
a

f (t )dt = f (v( x))v′( x) − f (u( x))u′( x)
u ( x)

Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

14

The Integral

chapter

6

Example: Evaluate G (x) of
x2

x

a. G ( x) =

1 + t 3 dt

b. G ( x) =
4

1
.

1 + t 3 dt

Answer:
3
a. f (t ) = 1 + t

G′( x) = 1 + x3

3
f
(
t
)
=
1
+
t
b.

u( x) = x

2

G′( x) = 1 + ( x2 )3

d 2
(x )
dx

= 2 x 1 + x6

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15

The Fundamental
Theorem of Calculus Part 2
The Fundamental Theorem of Calculus Part 2
If f is continuous at every point of [a,b] and F is
any antiderivative of f on [a,b] then
b

f ( x) dx = F (b) − F (a)
a

Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

16

Example
π

sin ( 2x) dx

1. Evaluate

π

2

Answer: Let u = 2x

du = 2 dx

then,
1
sin 2 x dx = − cos 2 x
2

therefore,
π

π

1
−1
= (cos2π − cosπ ) = −1
sin(2x)dx = − cos2x
2
2
π
π /2
2

Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

17

The Integral

chapter

6

5

2. Evaluate

| x − 2 | dx
1

Answer:

f ( x ) =| x − 2 |=

5

2

x − 2, x ≥ 2
− ( x − 2 ),x < 2
5

| x − 2 | dx = − ( x − 2) dx +
1

1

=

−1
2

( x − 2) dx
2

2

2

1
2

2

x + 2x + x − 2x
1

5
2

= ((−2 + 4) − (− 1 2 + 2)) + ((25 2 −10) − (2 − 4))
1 9
= + =5
2 2
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

18

Problem Set 1
For problems 1-5, find the antiderivative F(x) + C of f(x).
1.

f ( x) = 3x2 + 10x + 5

2.

f ( x) = x2 (20x7 − 7 x5 + 6)

3.
4.
5.

1

6

f (x ) = 3 + 7
x
x
f (x ) =

2x3 − 3x2 + 1

f ( x) = x

x2
− 34

Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

19

Problem Set 2
For problems 1-5, evaluate each integral given.
3

6.

(x − 4) 2x dx

7.

(x

8.

3x 3x2 + 7 dx

9.

2

(

2

)

− 3x + 2 ( 2x − 3) dx

)

2

3

5x + 1 5x + 3x − 2 dx
3y

10.
11.

2

dy

2

2y + 5

(

4

)

cos 2x ( − 2 sin 2x) dx
Calculus I (MA 1114) - Faculty of Science Telkom Institute of Technology

20

Problem Set 3
5

For problem 12 - 15, evaluate

f (x) dx
0

.

x+2 ,0≤ x < 2
12. f (x) = 6 − x , 2 ≤ x ≤ 5
x

13.

, 0≤ x