Benginning Calculus Lecture notes 12 - anti derivatives indefinite and definite integrals

Beginning Calculus
- Antiderivatives and The De…nite Integrals -
Shahrizal Shamsuddin Norashiqin Mohd Idrus
Department of Mathematics,
FSMT - UPSI
(LECTURE SLIDES SERIES)
VillaRINO DoMath, FSMT-UPSI
(I1) Antiderivatives and The De…nite Integrals 1 / 37

Anti Derivatives Area Under a Curve Riemann Sums Net Area Properties of the De…nite Integral
Learning Outcomes
Use substitution and advanced guessing methods to evaluate anti
derivatives.
Compute Riemann Sums.
Compute areas under the curve and net areas.
State and apply properties of the de…nite integrals.

Anti Derivatives
G (x) =
Z
g (x) dx
G (x) is called the anti derivative of g, or the inde…nite integral
of g.
G0 (x) = g (x)

Example
R
sin xdx = cos x + C

Example
R
xadx =
xa+1
a + 1
+ C, for a 6= 1

Example
R dx
x
= ln jxj + C

More Examples
R
sec2 xdx = tan x + C
R dx
p
1 x2
= sin 1 x + C
R dx
1 + x2
= tan 1 x + C
R
ex dx = ex + C

Uniqueness of anti derivatives up to a constant
Theorem 1
If F0 = G0, then F (x) = G (x) + C.
Proof.
Suppose F0 = G0. Then,
(F G)0
= F0
G0
= 0
F (x) G (x) = C
) F (x) = G (x) + C

Method of Substitution - For Di¤erential Notation
R
x3 x4 + 2
5
dx

Example
R xdx
p
1 + x2

Example
R
e6x dx

Example - Advanced Guessing
R
xe x2
dx

Example
R dx
x ln x

Area Under a Curve
b
( )dxxf
b
a∫
a
( )xfy =
Area under a curve =
R b
a f (x) dx

Area Under a Curve
To compute the area under a curve:
b
L
a
1 Divide into n rectangles
2 Add up the areas
3 Take the limit as n ! ∞ (the rectangles get thinner and thinner).

Example
f (x) = x2; a = 0, b = arbitrary
a = 0 nb/n
f(x) = x2
b/n 2b/n
f(x)L
L3b/n
divide into n rectangles
each rectangle has equal base-length =
b
n
.

Example - continue
Base x
b
n
2b
n
3b
n
b =
nb
n
Height f (x)
b
n
2
2b
n
2
3b
n
2
b2
The sum of the areas of the rectangles
b
n
b
n
2
+
b
n
2b
n
2
+
b
n
3b
n
2
+ +
b
n
nb
n
2
=
b
n
3
12
+ 22
+ 32
+ + (n 1)2
+ n2
=
b
n
3 n
∑
i 1
i2

Example - continue
n
∑
i=1
i2 =
n (n + 1) (2n + 1)
6
.
b3
n3
n (n + 1) (2n + 1)
6
=
b3 2n3 + 3n2 + n
6n3
=
2b3 +
3
n
+
1
n2
6
Take the limit as n ! ∞.
lim
n!∞
2b3 +
3
n
+
1
n2
6
=
b3
3
So the sum of the areas of the rectangles:
Z b
0
x2
dx =
b3
3

Examples
f (x) = x. The area under the curve:
Z b
0
xdx =
b2
2
f (x) = 1. The area under the curve:
Z b
0
1dx =
b1
1
= b
In general, f (x) = xn. The area under the curve:
Z b
0
xn
dx =
bn+1
n + 1

General Procedures for De…nite Integrals
Divide the base into n intervals with equal length ∆x.
x∆
ix
a b
( )xfy =
( )ixf
∆x =
b a
n
; xi = a + i∆x
The Riemann sum:
n
∑
i=1
f (xi ) ∆x :
Z b
a
f (x) dx = lim
n!∞
n
∑
i=1
f (xi ) 4x

General Procedures for De…nite Integrals - continue
Note that: Z b
a
f (x) dx = lim
n!∞
n
∑
i=1
f (xi ) 4x
can also be written as
Z b
a
f (x) dx = lim
n!∞
b a
n
n
∑
i=1
f a +
i (b a)
n

Example - continue
To evaluate
Z 1
0
x2dx : 4x =
1 0
n
=
1
n
; xi = 0 + i4x =
i
n
.
So, the de…nite integral is
Z 1
0
x2
dx = lim
n!∞
n
∑
i=1
f
i
n
1
n
= lim
n!∞
1
n
n
∑
i=1
f
i
n
= lim
n!∞
1
n
n
∑
i=1
i2
n2
= lim
n!∞
1
n3
n
∑
i=1
i2

Example - continue
= lim
n!∞
1
n3
n
∑
i=1
i2
= lim
n!∞
1
n3
n (n + 1) (2n + 1)
6
= lim
n!∞
2n2 + 3n + 1
6n2
= lim
n!∞
2 +
3
n
+
1
n2
6
=
2
6
=
1
3

Example
f (x) = x3 6x is a bounded function on [0, 3] . To evaluate the
Riemann sum with n = 6,
4x =
3 0
6
= 0.5
x1 = 0 + 0.5 = 0.5, x2 = 1.0, x3 = 1.5, x4 = 2.0, x5 = 2.5, x6 = 3.0.
So, the Riemann sum is
n
∑
i=1
f (xi ) 4x
=
1
2
[f (0.5) + f (1.0) + f (1.5) + f (2.0) + f (2.5) + f (3.0)]
=
1
2
( 2.875 5 5.625 4 + 0.625 + 9)
= 3.9375

Example - continue
To evaluate the de…nite integral
Z 3
0
x3 6x dx : 4x =
3 0
n
=
3
n
; xi = 0 + i4x =
3i
n
.
So, the de…nite integral is
Z 3
0
x3
6x dx = lim
n!∞
n
∑
i=1
f (xi ) 4x
= lim
n!∞
n
∑
i=1
f
3i
n
3
n
= lim
n!∞
3
n
n
∑
i=1
"
3i
n
3
6
3i
n
#

Example - continue
= lim
n!∞
3
n
n
∑
i=1
27
n3
i3 18
n
i
= lim
n!∞
3
n
"
27
n3
n
∑
i=1
i3 18
n
n
∑
i=1
i
#
= lim
n!∞
"
81
n4
n
∑
i=1
i3 54
n2
n
∑
i=1
i
#
= lim
n!∞
"
81
n4
n (n + 1)
2
2
54
n2
n (n + 1)
2
#

Example - continue
= lim
n!∞
81
4
n4 + 2n3 + n2
n4
54
2
n2 + n
n2
= lim
n!∞
81
4
1 +
2
n
+
1
n2
27 1 +
1
n
= lim
n!∞
"
81
4
1 +
1
n
2
27 1 +
1
n
#
=
81
4
27 =
27
4

Net Area
Geometrically the value of the de…nite integral represents the area
bounded by y = f (x) , the x axis and the ordinates at x = a and
x = b only if f (x) 0.
If f (x) is sometimes positive and sometimes negatives, the de…nite
integral represents the algebraic sum of the area above and below
the x axis (the net area).

Area Under The Curve and Net Area
b
x∆
kxa
( )xfy =
y
x
If f (x) 0, the Riemann
sum
n
∑
k=1
f (xk ) 4x is the
sum of the areas of rectangles.
ba
y
x
( )xfy =
If f (x) 0, the Integral
Z b
a
f (x) dx is the area under
the curve from a to b.

Area Under The Curve and Net Area
x
y
)(xfy =
+ +
-
ba
Z b
a
f (x) dx is the net area

Extend Integration to the Case f < 0 - Example
Z 2π
0
sin xdx
x
y
Z 2π
0
sin xdx = ( cos x)j2π
0
= ( cos 2π) ( cos 0) = 1 + 1 = 0

Total Distance and Net Distance
Total distance travelled:
Z b
a
jv (t)j dt
Net distance travelled: Z b
a
v (t) dt

Monotonicity, Continuity and Integral
Theorem 2
Every monotonic function f on [a, b] is integrable.
Theorem 3
Every continuous function f on [a, b] is integrable.

Properties of the De…nite Integral
Let f and g be integrable functions on [a, b], and c is a constant. Then,
1.
Z b
a
cdx = c (b a)
2.
Z a
a
f (x) dx = 0
3.
Z b
a
f (x) dx =
Z a
b
f (x) dx
4. cf is integrable and
Z b
a
cf (x) dx = c
Z b
a
f (x) dx.
5. f g is integrable and
Z b
a
(f g) (x) dx =
Z b
a
f (x) dx
Z b
a
g (x) dx.

Properties of the De…nite Integral - continue
6.
Z b
a
f (x) dx =
Z c
a
f (x) dx +
Z b
c
f (x) dx provided that f is integral
on [a, c] and [c, b] . (works without ordering a, b, c )
7. (Estimation) If f (x) g (x) for x 2 [a, b] , then
Z b
a
f (x) dx
Z b
a
g (x) dx. (a < b )
8. jf j is integrable and
Z b
a
f (x) dx
Z b
a
jf (x)j dx.

Example - Illustration of Property (6).
ex 1, x 0
Z b
0
ex dx
Z b
0
1dx
Z b
0
ex
dx = (ex
)jb
0 = eb
1
Z b
0
1dx = b
eb
1 + b, b 0

Example - continue
Repeat:
ex 1 + x, x 0
Z b
0
ex dx
Z b
0
(1 + x) dx
Z b
0
ex
dx = (ex
)jb
0 = eb
1
Z b
0
(1 + x) dx = x +
x2
2
b
0
= b +
b2
2
eb
1 + b +
b2
2
, b 0
Repeat: Gives a good approximation of ex .

Benginning Calculus Lecture notes 12 - anti derivatives indefinite and definite integrals

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