The document discusses the Fundamental Theorem of Calculus (FTC), which has two parts: FTC1 relates the area under a curve to the antiderivative of a function, stating that the antiderivative of a function is the integral of that function. FTC2 relates the integral of a function to the difference of its antiderivative at the bounds. Examples are provided to illustrate FTC1 and FTC2. The document also covers techniques such as integration by parts and u-substitution that use the FTC to evaluate definite integrals.

1. Beginning Calculus
- The Fundamental Theorem of Calculus -
Shahrizal Shamsuddin Norashiqin Mohd Idrus
Department of Mathematics,
FSMT - UPSI
(LECTURE SLIDES SERIES)
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(I2) FTC1 and FTC2 1 / 15

2. FTC1 FTC2
Learning Outcomes
State and apply FTC1 and FTC2
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3. FTC1 FTC2
The First Fundamental Theorem of Calculus (FTC1)
Theorem 1 (FTC1)
Let f be a continuous and integrable function on [a, b] . For x 2 [a, b] ,
de…ne a function
F (x) =
Z x
a
f (t) dt
Then F is continuous on [a, b] and di¤erentiable on (a, b) , and
F0
(x) = f (x)
That is, F is an antiderivative of f .
VillaRINO DoMath, FSMT-UPSI
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4. FTC1 FTC2
The First Fundamental Theorem of Calculus (FTC1)
The function F depends only
on x.
The variable of integration,
t, is called a dummy variable.
Using Leibniz notation for
derivative, we write
F0
(x) =
d
dx
Z x
a
f (t) dt = f (x)
ba
y
t
()tfy=
area = F(x)
x
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5. FTC1 FTC2
Example
Let F (x) =
Z x
0
f (t) dt where
the function f is given on the
right. Then,
F (0) = 0, F (1) =
2, F (2) = 5, F (3) =
7, F (6) = 3.
F is increasing on (0, 3) .
F has a maximum value at
x = 3.
5
1
x
y
f
0
4
2 3 4 6 7
3
2
1
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6. FTC1 FTC2
Example - continue
Rough sketch of the graph of F
5
1
x
y
f
0
4
2 3 4 6 7
3
2
1
F
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7. FTC1 FTC2
The Second Fundamental Theorem of Calculus (FTC2)
Theorem 2 (FTC2)
If F0 (x) = f (x) , then
Z b
a
F0
(x) dx =
Z b
a
f (x) dx
= F (b) F (a) = F (x)jb
a
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8. FTC1 FTC2
Example
If F (x) =
xn+1
n + 1
, then F0 (x) = xn and so
Z b
a
F0
(x) dx =
Z b
a
xn
dx =
bn+1
n + 1
an+1
n + 1
=
bn+1 an+1
n + 1
If n = 2, then
Z b
a
x2
dx =
b3 a3
3
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9. FTC1 FTC2
Example
Area under one hump of sin x.
x
y
Z π
0
sin xdx = ( cos x)jπ
0 = cos π + cos 0 = 2
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10. FTC1 FTC2
Intuitive Interpretation of FTC2
x (t) is the position at time t.
x0 (t) =
dx
dt
= v (t) is the speed.
Z b
a
v (t)
|{z}
speedometer
dt = x (b) x (a)
| {z }
distance travelled
(odometer)
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(I2) FTC1 and FTC2 10 / 15

11. FTC1 FTC2
Change of Variables - Substitution
Theorem 3 (Change of Variables)
Let u = u (x). Then,
du = u0
(x) dx
Z x2
x1
f [u (x)] u0
(x) dx =
Z u(x2)
u(x1)
f (u) du
Only works when u0 (x) does not change sign. (i.e the function
increase or decrease steadily).
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12. FTC1 FTC2
Example
Z 2
1
x3 + 2
5
x2dx
Let u = x3 + 2, then du = 3x2dx.
x1 = 1, u (1) = 3; x2 = 2, u (2) = 10
Z 2
1
x3
+ 2
5
x2
dx =
1
3
Z 10
3
u5
du
=
1
18
u6
10
3
=
1
18
106
36
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(I2) FTC1 and FTC2 12 / 15

13. FTC1 FTC2
Example - WARNING
Z 1
1
x2dx
Let u = x2, then du = 2xdx
x1 = 1, u ( 1) = 1; x2 = 1, u (1) = 1
Z 1
1
x2
dx =
Z 1
1
u
2
p
u
du = 0, NOT TRUE
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14. FTC1 FTC2
Integration by Parts
Theorem 4 (Integration By Parts)
If u and v are continuous functions on [a, b] and di¤erentiable on (a, b) ,
and if u0 and v0 are integrable on [a, b] , then
Z b
a
u (x) v0
(x) dx +
Z b
a
u0
(x) v (x) dx = u (b) v (b) u (a) v (a) (1)
In Liebniz notation, we normally simplify (1) as
Z
udv +
Z
vdu = uv
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15. FTC1 FTC2
Integration by Parts
Proof:
Let F = uv, then F0 = uv0 + u0v. It can be shown that F0 is integrable.
Then by FTC2,
Z b
a
F0
(x) dx = F (b) F (a) = u (b) v (b) u (a) v (a)
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