Download Calculus is divided into two main branches: • Differential Calculus

Calculus is divided into two main branches:
• Differential Calculus
the procedures and applications of the derivative
• Integral Calculus
the inverse of differentiation
A derivative originates from a function.
How easy is it to work backwards from the derivative?
How would you use your knowledge of differentiation
to help you match a derivative and its function?
1
Antiderivatives
The original function from which the
derivative was obtained is termed the
antiderivative, integral, or primitive.
Antidifferentiation is the inverse operation
of differentiation.
2
The question in this lesson is:
How do we determine the original function
if we are given the derivative?
If
dy
 3x 2
dx
what is y?
Here, the function we are seeking is called an
antiderivative or an integral
3
Find the derivatives of the following functions.
f(x) = x2
f ′ (x) = 2x
f (– 1) = (– 1)2 = 1
f ′ (– 1) = – 2
g (x) = x2 – 3
g′ (x) = 2x
g (– 1) = – 2
g′ (– 1) = – 2
h(x) = x2 + 1
h′ (x) = 2x
h (– 1) = 2
h′ (– 1) = – 2
k(x) = x2 + 5
k′ (x) = 2x
x = -1
k′ (– 1) = – 2
The slopes of the tangent lines at x = -1 for ALL functions is 2(-1) = -2.
k (– 1) = 6
The tangent lines at x = -1 for all the functions are parallel.
4
Find the derivatives of the following functions.
f(x) = – 2x2 + 4x
f ′ (x) = – 4x + 4
g(x) = – 2x2 + 4x – 3
g ′ (x) = – 4x + 4
h(x) = – 2x2 + 4x + 1
h ′ (x) = – 4x + 4
k(x) = – 2x2 + 4x + 5
k′ (x) = – 4x + 4
The slopes of the tangent lines at
x = 1 for all functions is – 4(1) + 4 = 0
x=1
The tangent lines at x = 1 for all the functions are parallel.
5
Derivative
General antiderivative
y ′ = 2x
y = x2 + C
The antidrivative, or integral, is commonly represented
by
𝟐𝒙 𝒅𝒙
This is read as the integral of 2x with respect to x.
The symbol
 .........dx
is the inverse of the symbol
d  ____ 
dx
6
n 1
x
 x dx  n  1  C
n
This is correct since
n n1111
n

1
x
n

1
x
dd xx     
nn
xx


dxnn11
dx
nn11
n n11
This rule will work for all real values of n except -1.
7
EXAMPLE 1
𝟓𝒙𝟑 + 𝟑𝒙𝟐 − 𝟒𝒙 + 𝟐 𝒅𝒙
𝟓𝒙𝟑+𝟏 𝟑𝒙𝟐+𝟏 𝟒𝒙𝟏+𝟏 𝟐𝒙𝟎+𝟏
𝒚=
+
−
+
+𝑪
𝟑+𝟏 𝟐+𝟏 𝟏+𝟏 𝟎+𝟏
𝟓𝒙𝟒 𝟑𝒙𝟑 𝟒𝒙𝟐 𝟐𝒙
𝒚=
+
−
+
+𝑪
𝟒
𝟑
𝟐
𝟏
𝟓 𝟒
𝒚 = 𝒙 + 𝒙𝟑 − 𝟐𝒙𝟐 + 𝟐𝒙 + 𝑪
𝟒
8
2
1
+ 2 + 3 𝑑𝑥
3
𝑥
𝑥
EXAMPLE 2
𝟐𝒙−𝟑 + 𝒙−𝟐 + 𝟑𝒙𝟎 𝒅𝒙
𝟐𝒙−𝟑+𝟏
𝒙−𝟐+𝟏
𝟑𝒙𝟎+𝟏
𝒚=
+
+
+𝑪
−𝟑 + 𝟏 −𝟐 + 𝟏 𝟎 + 𝟏
𝟐𝒙−𝟐 𝒙−𝟏 𝟑𝒙
𝒚=
+
+
+𝑪
−𝟐
−𝟏
𝟏
𝒚 = −𝒙
−𝟐
−𝟏
−𝒙
𝟏 𝟏
+ 𝟑𝒙 + 𝑪 = − 𝟐 − + 𝟑𝒙 + 𝑪
𝒙
𝒙
9
4 𝑥 + 2 𝑑𝑥
EXAMPLE 3
𝟏
𝟒𝒙𝟐
+ 𝟐𝒙𝟎 𝒅𝒙
𝟏
𝟏+𝟐
𝟐𝒙𝟎+𝟏
+
+𝑪
𝟏 𝟎+𝟏
𝟏+
𝟐
𝟒𝒙
𝟑
𝟒𝒙𝟐
𝟐𝒙
+
+𝑪
𝟑
𝟏
𝟐
𝟖 𝒙𝟑
+ 𝟐𝒙 + 𝑪
𝟑
10
𝑓 𝑥 = 𝑥−2
3
𝑓 𝑥 = 5𝑥 − 2
𝒇′ 𝒙 = 𝟑 𝒙 − 𝟐 𝟐 (𝟏)
𝒇′ 𝒙 = 𝟑 𝒙 − 𝟐
𝟐
𝑓 𝑥 =4 𝑥−2
3
𝒇′ 𝒙 = 𝟒(𝟑) 𝒙 − 𝟐 𝟐 (𝟏)
𝒇′ 𝒙 = 𝟏𝟐 𝒙 − 𝟐
𝟐
3
𝒇′ 𝒙 = 𝟑 𝟓𝒙 − 𝟐 𝟐 (𝟓)
𝒇′ 𝒙 = 𝟏𝟓 𝟓𝒙 − 𝟐
𝑓 𝑥 = 6 5𝑥 − 2
𝟐
3
𝒇′ 𝒙 = 𝟔(𝟑) 𝟓𝒙 − 𝟐 𝟐 (𝟓)
𝒇′ 𝒙 = 𝟗𝟎 𝟓𝒙 − 𝟐
𝟐
11
(𝒙 + 𝟒)𝟒 𝒅𝒙 =
EXAMPLE 4
 x  4
41
4 1
EXAMPLE 5
x  4


5
5
(𝟐𝒙
xx44
4411
4411
xx44



55
55
11
5
 xx44
55
1
5
  x  4  C
5
+ 𝟑)𝟒 𝒅𝒙
𝟒+𝟏
(𝟐𝒙
+
𝟑)
𝟏
=
× +𝑪
𝟒+𝟏
𝟐
𝟏
(𝟐𝒙 + 𝟑)𝟓
× +𝑪
=
𝟐
𝟓
𝟏
=
(𝟐𝒙 + 𝟑)𝟓 +𝑪
𝟏𝟎
12
EXAMPLE 6
2 5𝑥 − 4
3
𝑑𝑥
𝟐 𝟓𝒙 − 𝟒 𝟑+𝟏 𝟏
× +𝑪
𝟑+𝟏
𝟓
𝟐 𝟓𝒙 − 𝟒
𝟒
𝟒
𝟓𝒙 − 𝟒
𝟏𝟎
𝟒
𝟏
× +𝑪
𝟓
+𝑪
𝟏
𝟓𝒙 − 𝟒
𝟏𝟎
𝟒
+𝑪
13
7 1 − 6𝑥
EXAMPLE 7
𝟕 𝟏 − 𝟔𝒙 𝟓+𝟏 −𝟏
×
+𝑪
𝟓+𝟏
𝟔
𝟕 𝟏 − 𝟔𝒙
𝟔
𝟔
−𝟏
×
+𝑪
𝟔
5
𝑑𝑥
−𝟕 𝟏 − 𝟔𝒙
𝟑𝟔
−𝟕
𝟏 − 𝟔𝒙
𝟑𝟔
𝟔
+𝑪
𝟔
+𝑪
14
EXAMPLE 8
5
𝑥+2
3
𝑑𝑥
5 𝑥+2
−3
𝑑𝑥
𝟓 𝒙 + 𝟐 −𝟑+𝟏
+𝑪
−𝟑 + 𝟏
𝟓 𝒙+𝟐
−𝟐
−𝟐
𝟓
−
𝟐 𝒙+𝟐
+𝑪
𝟐
+𝑪
15
DEFINITION
Let f be any positive, continuous function on the
interval [a, b], and let F be any antiderivative of f.
a
b
𝒃
𝒇 𝒙 𝒅𝒙
is defined as F(b) – F(a)
𝒂
16
EXAMPLE 1: Evaluate the following definite integral

5
1
(6 x  7)dx
Type equation here.
3xx 
7
7 xx ||
3
3x  72 x |
2
2  7(5)]  [3( 1) 2  7( 1)]

[3(5)
[3(5)
[3(5) 2 7(5)]
7(5)][3(
[3(1)
1) 2 7(
7(1)]
1)]
 [40]
[40] 
 [10]
[10] 
 30
30

 [40]  [10]  30
2
2
2
5
5
51
1
1
Calculator (to be used only for checking!)
Y1 = 6x – 7
2nd CALC
Lower limit then Upper limit
7:
 f ( x)dx
NOTE: The calculator
puts the numbers in
backwards from above.
17
EXAMPLE 2: Evaluate the following definite integral

1
1
5
4
𝟓
𝒙𝟒+𝟏
𝟗
𝒙𝟒
x dx   x dx =
=
0
9
𝟗
𝟓
4 41
+𝟏
𝟒
𝟒
x |
0
9
9
9
4 94 1
4
4
4
4
4
x |
 (1)  (0) 
0
9
9
9
9
4 94 4 94 4
Calculator (tobe (1)
used only
(0)forchecking!)
9
9
 59

Y1 = 4 x5  x 4 
4
5
0
2nd CALC
7:
 f ( x)dx
Lower limit then Upper limit
NOTE: The calculator
puts the numbers in
backwards from above.
18