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1 | 3818-3821 | Fig 7 2
O
Y
X
X'
Y'
Q
P
C
M
D
L
S
A
B
R
a = x0 x1 x2
xr-1 xr
x =b
n
y
f x
= ( )
332
MATHEMATICS
The region PRSQP under consideration is the sum of n subregions, where each
subregion is defined on subintervals [xr – 1, xr], r = 1, 2, 3, …, n From Fig 7 2, we have
area of the rectangle (ABLC) < area of the region (ABDCA) < area of the rectangle
(ABDM) |
1 | 3819-3822 | 2
O
Y
X
X'
Y'
Q
P
C
M
D
L
S
A
B
R
a = x0 x1 x2
xr-1 xr
x =b
n
y
f x
= ( )
332
MATHEMATICS
The region PRSQP under consideration is the sum of n subregions, where each
subregion is defined on subintervals [xr – 1, xr], r = 1, 2, 3, …, n From Fig 7 2, we have
area of the rectangle (ABLC) < area of the region (ABDCA) < area of the rectangle
(ABDM) (1)
Evidently as xr – xr–1 → 0, i |
1 | 3820-3823 | From Fig 7 2, we have
area of the rectangle (ABLC) < area of the region (ABDCA) < area of the rectangle
(ABDM) (1)
Evidently as xr – xr–1 → 0, i e |
1 | 3821-3824 | 2, we have
area of the rectangle (ABLC) < area of the region (ABDCA) < area of the rectangle
(ABDM) (1)
Evidently as xr – xr–1 → 0, i e , h → 0 all the three areas shown in (1) become
nearly equal to each other |
1 | 3822-3825 | (1)
Evidently as xr – xr–1 → 0, i e , h → 0 all the three areas shown in (1) become
nearly equal to each other Now we form the following sums |
1 | 3823-3826 | e , h → 0 all the three areas shown in (1) become
nearly equal to each other Now we form the following sums sn = h [f(x0) + … + f (xn - 1)] =
1
0
(
)
n
r
r
h
f x
−
=∑ |
1 | 3824-3827 | , h → 0 all the three areas shown in (1) become
nearly equal to each other Now we form the following sums sn = h [f(x0) + … + f (xn - 1)] =
1
0
(
)
n
r
r
h
f x
−
=∑ (2)
and
Sn =
1
2
1
[ ( )
(
)
(
)]
(
)
n
n
r
r
h f x
f x
f x
h
f x
=
+
+…+
= ∑ |
1 | 3825-3828 | Now we form the following sums sn = h [f(x0) + … + f (xn - 1)] =
1
0
(
)
n
r
r
h
f x
−
=∑ (2)
and
Sn =
1
2
1
[ ( )
(
)
(
)]
(
)
n
n
r
r
h f x
f x
f x
h
f x
=
+
+…+
= ∑ (3)
Here, sn and Sn denote the sum of areas of all lower rectangles and upper rectangles
raised over subintervals [xr–1, xr] for r = 1, 2, 3, …, n, respectively |
1 | 3826-3829 | sn = h [f(x0) + … + f (xn - 1)] =
1
0
(
)
n
r
r
h
f x
−
=∑ (2)
and
Sn =
1
2
1
[ ( )
(
)
(
)]
(
)
n
n
r
r
h f x
f x
f x
h
f x
=
+
+…+
= ∑ (3)
Here, sn and Sn denote the sum of areas of all lower rectangles and upper rectangles
raised over subintervals [xr–1, xr] for r = 1, 2, 3, …, n, respectively In view of the inequality (1) for an arbitrary subinterval [xr–1, xr], we have
sn < area of the region PRSQP < Sn |
1 | 3827-3830 | (2)
and
Sn =
1
2
1
[ ( )
(
)
(
)]
(
)
n
n
r
r
h f x
f x
f x
h
f x
=
+
+…+
= ∑ (3)
Here, sn and Sn denote the sum of areas of all lower rectangles and upper rectangles
raised over subintervals [xr–1, xr] for r = 1, 2, 3, …, n, respectively In view of the inequality (1) for an arbitrary subinterval [xr–1, xr], we have
sn < area of the region PRSQP < Sn (4)
As n → ∞ strips become narrower and narrower, it is assumed that the limiting
values of (2) and (3) are the same in both cases and the common limiting value is the
required area under the curve |
1 | 3828-3831 | (3)
Here, sn and Sn denote the sum of areas of all lower rectangles and upper rectangles
raised over subintervals [xr–1, xr] for r = 1, 2, 3, …, n, respectively In view of the inequality (1) for an arbitrary subinterval [xr–1, xr], we have
sn < area of the region PRSQP < Sn (4)
As n → ∞ strips become narrower and narrower, it is assumed that the limiting
values of (2) and (3) are the same in both cases and the common limiting value is the
required area under the curve Symbolically, we write
limSn
n→∞
= lim
n
n
s
→∞
= area of the region PRSQP =
( )
b
∫a f x dx |
1 | 3829-3832 | In view of the inequality (1) for an arbitrary subinterval [xr–1, xr], we have
sn < area of the region PRSQP < Sn (4)
As n → ∞ strips become narrower and narrower, it is assumed that the limiting
values of (2) and (3) are the same in both cases and the common limiting value is the
required area under the curve Symbolically, we write
limSn
n→∞
= lim
n
n
s
→∞
= area of the region PRSQP =
( )
b
∫a f x dx (5)
It follows that this area is also the limiting value of any area which is between that
of the rectangles below the curve and that of the rectangles above the curve |
1 | 3830-3833 | (4)
As n → ∞ strips become narrower and narrower, it is assumed that the limiting
values of (2) and (3) are the same in both cases and the common limiting value is the
required area under the curve Symbolically, we write
limSn
n→∞
= lim
n
n
s
→∞
= area of the region PRSQP =
( )
b
∫a f x dx (5)
It follows that this area is also the limiting value of any area which is between that
of the rectangles below the curve and that of the rectangles above the curve For
the sake of convenience, we shall take rectangles with height equal to that of the
curve at the left hand edge of each subinterval |
1 | 3831-3834 | Symbolically, we write
limSn
n→∞
= lim
n
n
s
→∞
= area of the region PRSQP =
( )
b
∫a f x dx (5)
It follows that this area is also the limiting value of any area which is between that
of the rectangles below the curve and that of the rectangles above the curve For
the sake of convenience, we shall take rectangles with height equal to that of the
curve at the left hand edge of each subinterval Thus, we rewrite (5) as
( )
b
∫a f x dx
=
lim0
[ ( )
(
) |
1 | 3832-3835 | (5)
It follows that this area is also the limiting value of any area which is between that
of the rectangles below the curve and that of the rectangles above the curve For
the sake of convenience, we shall take rectangles with height equal to that of the
curve at the left hand edge of each subinterval Thus, we rewrite (5) as
( )
b
∫a f x dx
=
lim0
[ ( )
(
) (
( – 1) ]
h
h f a
f a
h
f a
n
h
→
+
+
+
+
+
or
( )
b
∫a f x dx
=
( – ) lim1
[ ( )
(
) |
1 | 3833-3836 | For
the sake of convenience, we shall take rectangles with height equal to that of the
curve at the left hand edge of each subinterval Thus, we rewrite (5) as
( )
b
∫a f x dx
=
lim0
[ ( )
(
) (
( – 1) ]
h
h f a
f a
h
f a
n
h
→
+
+
+
+
+
or
( )
b
∫a f x dx
=
( – ) lim1
[ ( )
(
) (
( –1) ]
n
b
a
f a
f a
h
f a
n
h
→∞n
+
+
+
+
+ |
1 | 3834-3837 | Thus, we rewrite (5) as
( )
b
∫a f x dx
=
lim0
[ ( )
(
) (
( – 1) ]
h
h f a
f a
h
f a
n
h
→
+
+
+
+
+
or
( )
b
∫a f x dx
=
( – ) lim1
[ ( )
(
) (
( –1) ]
n
b
a
f a
f a
h
f a
n
h
→∞n
+
+
+
+
+ (6)
where
h =
–
0
b
a
as n
n
→
→ ∞
The above expression (6) is known as the definition of definite integral as the limit
of sum |
1 | 3835-3838 | (
( – 1) ]
h
h f a
f a
h
f a
n
h
→
+
+
+
+
+
or
( )
b
∫a f x dx
=
( – ) lim1
[ ( )
(
) (
( –1) ]
n
b
a
f a
f a
h
f a
n
h
→∞n
+
+
+
+
+ (6)
where
h =
–
0
b
a
as n
n
→
→ ∞
The above expression (6) is known as the definition of definite integral as the limit
of sum Remark The value of the definite integral of a function over any particular interval
depends on the function and the interval, but not on the variable of integration that we
INTEGRALS 333
choose to represent the independent variable |
1 | 3836-3839 | (
( –1) ]
n
b
a
f a
f a
h
f a
n
h
→∞n
+
+
+
+
+ (6)
where
h =
–
0
b
a
as n
n
→
→ ∞
The above expression (6) is known as the definition of definite integral as the limit
of sum Remark The value of the definite integral of a function over any particular interval
depends on the function and the interval, but not on the variable of integration that we
INTEGRALS 333
choose to represent the independent variable If the independent variable is denoted by
t or u instead of x, we simply write the integral as
( )
b
∫a f t dt
or
( )
b
∫a f u du
instead of
( )
b
∫a f x dx |
1 | 3837-3840 | (6)
where
h =
–
0
b
a
as n
n
→
→ ∞
The above expression (6) is known as the definition of definite integral as the limit
of sum Remark The value of the definite integral of a function over any particular interval
depends on the function and the interval, but not on the variable of integration that we
INTEGRALS 333
choose to represent the independent variable If the independent variable is denoted by
t or u instead of x, we simply write the integral as
( )
b
∫a f t dt
or
( )
b
∫a f u du
instead of
( )
b
∫a f x dx Hence, the variable of integration is called a dummy variable |
1 | 3838-3841 | Remark The value of the definite integral of a function over any particular interval
depends on the function and the interval, but not on the variable of integration that we
INTEGRALS 333
choose to represent the independent variable If the independent variable is denoted by
t or u instead of x, we simply write the integral as
( )
b
∫a f t dt
or
( )
b
∫a f u du
instead of
( )
b
∫a f x dx Hence, the variable of integration is called a dummy variable Example 25 Find
2
0 (2
1)
x
dx
+
∫
as the limit of a sum |
1 | 3839-3842 | If the independent variable is denoted by
t or u instead of x, we simply write the integral as
( )
b
∫a f t dt
or
( )
b
∫a f u du
instead of
( )
b
∫a f x dx Hence, the variable of integration is called a dummy variable Example 25 Find
2
0 (2
1)
x
dx
+
∫
as the limit of a sum Solution By definition
( )
b
∫a f x dx
=
( – ) lim1
[ ( )
(
) |
1 | 3840-3843 | Hence, the variable of integration is called a dummy variable Example 25 Find
2
0 (2
1)
x
dx
+
∫
as the limit of a sum Solution By definition
( )
b
∫a f x dx
=
( – ) lim1
[ ( )
(
) (
( –1) ],
n
b
a
f a
f a
h
f a
n
h
→∞n
+
+
+
+
+
where,
h =
b–
a
n
In this example, a = 0, b = 2, f (x) = x2 + 1,
2 – 0
2
h
n
n
=
=
Therefore,
2
2
0 (
1)
x
dx
+
∫
=
1
2
4
2 ( –1)
2 lim
[ (0)
( )
( ) |
1 | 3841-3844 | Example 25 Find
2
0 (2
1)
x
dx
+
∫
as the limit of a sum Solution By definition
( )
b
∫a f x dx
=
( – ) lim1
[ ( )
(
) (
( –1) ],
n
b
a
f a
f a
h
f a
n
h
→∞n
+
+
+
+
+
where,
h =
b–
a
n
In this example, a = 0, b = 2, f (x) = x2 + 1,
2 – 0
2
h
n
n
=
=
Therefore,
2
2
0 (
1)
x
dx
+
∫
=
1
2
4
2 ( –1)
2 lim
[ (0)
( )
( ) (
)]
n
n
f
f
f
f
n
n
n
n
→∞
+
+
+
+
=
2
2
2
2
2
2
1
2
4
(2 – 2)
2 lim
[1
(
1)
(
1) |
1 | 3842-3845 | Solution By definition
( )
b
∫a f x dx
=
( – ) lim1
[ ( )
(
) (
( –1) ],
n
b
a
f a
f a
h
f a
n
h
→∞n
+
+
+
+
+
where,
h =
b–
a
n
In this example, a = 0, b = 2, f (x) = x2 + 1,
2 – 0
2
h
n
n
=
=
Therefore,
2
2
0 (
1)
x
dx
+
∫
=
1
2
4
2 ( –1)
2 lim
[ (0)
( )
( ) (
)]
n
n
f
f
f
f
n
n
n
n
→∞
+
+
+
+
=
2
2
2
2
2
2
1
2
4
(2 – 2)
2 lim
[1
(
1)
(
1) 1 ]
n
n
n
n
n
n
→∞
+
+
+
+
+
+
+
=
2
2
2
-
1
1
2 lim
[(1 1 |
1 | 3843-3846 | (
( –1) ],
n
b
a
f a
f a
h
f a
n
h
→∞n
+
+
+
+
+
where,
h =
b–
a
n
In this example, a = 0, b = 2, f (x) = x2 + 1,
2 – 0
2
h
n
n
=
=
Therefore,
2
2
0 (
1)
x
dx
+
∫
=
1
2
4
2 ( –1)
2 lim
[ (0)
( )
( ) (
)]
n
n
f
f
f
f
n
n
n
n
→∞
+
+
+
+
=
2
2
2
2
2
2
1
2
4
(2 – 2)
2 lim
[1
(
1)
(
1) 1 ]
n
n
n
n
n
n
→∞
+
+
+
+
+
+
+
=
2
2
2
-
1
1
2 lim
[(1 1 1)
(2
4 |
1 | 3844-3847 | (
)]
n
n
f
f
f
f
n
n
n
n
→∞
+
+
+
+
=
2
2
2
2
2
2
1
2
4
(2 – 2)
2 lim
[1
(
1)
(
1) 1 ]
n
n
n
n
n
n
→∞
+
+
+
+
+
+
+
=
2
2
2
-
1
1
2 lim
[(1 1 1)
(2
4 (2 – 2) ]
2
→∞
+ +
+
+
+
+
+
144244
3
n
n terms
n
n
n
=
2
2
2
2
1
2
2 lim
[
(1
2 |
1 | 3845-3848 | 1 ]
n
n
n
n
n
n
→∞
+
+
+
+
+
+
+
=
2
2
2
-
1
1
2 lim
[(1 1 1)
(2
4 (2 – 2) ]
2
→∞
+ +
+
+
+
+
+
144244
3
n
n terms
n
n
n
=
2
2
2
2
1
2
2 lim
[
(1
2 ( –1) ]
n
n
n
n
n2
→∞
+
+
+
+
=
1
4 (
1)
(2 –1)
2 lim
[
]
6
n
n
n
n
nn
n2
→∞
−
+
=
1
2 (
1) (2 –1)
2 lim
[
]
3
n
n
n
nn
n
→∞
−
+
=
2
1
1
2 lim [1
(1
) (2 –
)]
3
n
n
n
→∞
+
−
=
4
2 [1
]
3
+
= 14
3
334
MATHEMATICS
Example 26 Evaluate
2
0
xe dx
∫
as the limit of a sum |
1 | 3846-3849 | 1)
(2
4 (2 – 2) ]
2
→∞
+ +
+
+
+
+
+
144244
3
n
n terms
n
n
n
=
2
2
2
2
1
2
2 lim
[
(1
2 ( –1) ]
n
n
n
n
n2
→∞
+
+
+
+
=
1
4 (
1)
(2 –1)
2 lim
[
]
6
n
n
n
n
nn
n2
→∞
−
+
=
1
2 (
1) (2 –1)
2 lim
[
]
3
n
n
n
nn
n
→∞
−
+
=
2
1
1
2 lim [1
(1
) (2 –
)]
3
n
n
n
→∞
+
−
=
4
2 [1
]
3
+
= 14
3
334
MATHEMATICS
Example 26 Evaluate
2
0
xe dx
∫
as the limit of a sum Solution By definition
2
0
∫xe dx
=
2
4
2
– 2
0
(2 – 0) lim1 |
1 | 3847-3850 | (2 – 2) ]
2
→∞
+ +
+
+
+
+
+
144244
3
n
n terms
n
n
n
=
2
2
2
2
1
2
2 lim
[
(1
2 ( –1) ]
n
n
n
n
n2
→∞
+
+
+
+
=
1
4 (
1)
(2 –1)
2 lim
[
]
6
n
n
n
n
nn
n2
→∞
−
+
=
1
2 (
1) (2 –1)
2 lim
[
]
3
n
n
n
nn
n
→∞
−
+
=
2
1
1
2 lim [1
(1
) (2 –
)]
3
n
n
n
→∞
+
−
=
4
2 [1
]
3
+
= 14
3
334
MATHEMATICS
Example 26 Evaluate
2
0
xe dx
∫
as the limit of a sum Solution By definition
2
0
∫xe dx
=
2
4
2
– 2
0
(2 – 0) lim1 n
n
n
n
n
e
e
e
e
n
→∞
+
+
+
+
Using the sum to n terms of a G |
1 | 3848-3851 | ( –1) ]
n
n
n
n
n2
→∞
+
+
+
+
=
1
4 (
1)
(2 –1)
2 lim
[
]
6
n
n
n
n
nn
n2
→∞
−
+
=
1
2 (
1) (2 –1)
2 lim
[
]
3
n
n
n
nn
n
→∞
−
+
=
2
1
1
2 lim [1
(1
) (2 –
)]
3
n
n
n
→∞
+
−
=
4
2 [1
]
3
+
= 14
3
334
MATHEMATICS
Example 26 Evaluate
2
0
xe dx
∫
as the limit of a sum Solution By definition
2
0
∫xe dx
=
2
4
2
– 2
0
(2 – 0) lim1 n
n
n
n
n
e
e
e
e
n
→∞
+
+
+
+
Using the sum to n terms of a G P |
1 | 3849-3852 | Solution By definition
2
0
∫xe dx
=
2
4
2
– 2
0
(2 – 0) lim1 n
n
n
n
n
e
e
e
e
n
→∞
+
+
+
+
Using the sum to n terms of a G P , where a = 1,
n2
r
=e
, we have
2
0
∫xe dx
=
2
2
1
–1
2 lim
[
]
1
n
n
n
n
ne
e
→∞
−
=
2
2
1
–1
2 lim
–1
n
n
n ee
→∞
=
2
2
2 (
–1)
–1
lim
2
2
n
n
e
e
n
→∞
⋅
= e2 – 1
[using
0
(
1)
lim
1
h
h
e
h
→
−
= ]
EXERCISE 7 |
1 | 3850-3853 | n
n
n
n
n
e
e
e
e
n
→∞
+
+
+
+
Using the sum to n terms of a G P , where a = 1,
n2
r
=e
, we have
2
0
∫xe dx
=
2
2
1
–1
2 lim
[
]
1
n
n
n
n
ne
e
→∞
−
=
2
2
1
–1
2 lim
–1
n
n
n ee
→∞
=
2
2
2 (
–1)
–1
lim
2
2
n
n
e
e
n
→∞
⋅
= e2 – 1
[using
0
(
1)
lim
1
h
h
e
h
→
−
= ]
EXERCISE 7 8
Evaluate the following definite integrals as limit of sums |
1 | 3851-3854 | P , where a = 1,
n2
r
=e
, we have
2
0
∫xe dx
=
2
2
1
–1
2 lim
[
]
1
n
n
n
n
ne
e
→∞
−
=
2
2
1
–1
2 lim
–1
n
n
n ee
→∞
=
2
2
2 (
–1)
–1
lim
2
2
n
n
e
e
n
→∞
⋅
= e2 – 1
[using
0
(
1)
lim
1
h
h
e
h
→
−
= ]
EXERCISE 7 8
Evaluate the following definite integrals as limit of sums 1 |
1 | 3852-3855 | , where a = 1,
n2
r
=e
, we have
2
0
∫xe dx
=
2
2
1
–1
2 lim
[
]
1
n
n
n
n
ne
e
→∞
−
=
2
2
1
–1
2 lim
–1
n
n
n ee
→∞
=
2
2
2 (
–1)
–1
lim
2
2
n
n
e
e
n
→∞
⋅
= e2 – 1
[using
0
(
1)
lim
1
h
h
e
h
→
−
= ]
EXERCISE 7 8
Evaluate the following definite integrals as limit of sums 1 b
∫a x dx
2 |
1 | 3853-3856 | 8
Evaluate the following definite integrals as limit of sums 1 b
∫a x dx
2 5
0 (
1)
x
dx
+
∫
3 |
1 | 3854-3857 | 1 b
∫a x dx
2 5
0 (
1)
x
dx
+
∫
3 3
2
2 x dx
∫
4 |
1 | 3855-3858 | b
∫a x dx
2 5
0 (
1)
x
dx
+
∫
3 3
2
2 x dx
∫
4 4
2
1 (
)
x
−x dx
∫
5 |
1 | 3856-3859 | 5
0 (
1)
x
dx
+
∫
3 3
2
2 x dx
∫
4 4
2
1 (
)
x
−x dx
∫
5 1
1
x
e dx
−∫
6 |
1 | 3857-3860 | 3
2
2 x dx
∫
4 4
2
1 (
)
x
−x dx
∫
5 1
1
x
e dx
−∫
6 4
2
0 (
x)
x
e
dx
+
∫
7 |
1 | 3858-3861 | 4
2
1 (
)
x
−x dx
∫
5 1
1
x
e dx
−∫
6 4
2
0 (
x)
x
e
dx
+
∫
7 8 Fundamental Theorem of Calculus
7 |
1 | 3859-3862 | 1
1
x
e dx
−∫
6 4
2
0 (
x)
x
e
dx
+
∫
7 8 Fundamental Theorem of Calculus
7 8 |
1 | 3860-3863 | 4
2
0 (
x)
x
e
dx
+
∫
7 8 Fundamental Theorem of Calculus
7 8 1 Area function
We have defined
( )
b
∫a f x dx
as the area of
the region bounded by the curve y = f(x),
the ordinates x = a and x = b and x-axis |
1 | 3861-3864 | 8 Fundamental Theorem of Calculus
7 8 1 Area function
We have defined
( )
b
∫a f x dx
as the area of
the region bounded by the curve y = f(x),
the ordinates x = a and x = b and x-axis Let x
be a given point in [a, b] |
1 | 3862-3865 | 8 1 Area function
We have defined
( )
b
∫a f x dx
as the area of
the region bounded by the curve y = f(x),
the ordinates x = a and x = b and x-axis Let x
be a given point in [a, b] Then
( )
x
a f x dx
∫
represents the area of the light shaded region
Fig 7 |
1 | 3863-3866 | 1 Area function
We have defined
( )
b
∫a f x dx
as the area of
the region bounded by the curve y = f(x),
the ordinates x = a and x = b and x-axis Let x
be a given point in [a, b] Then
( )
x
a f x dx
∫
represents the area of the light shaded region
Fig 7 3
INTEGRALS 335
in Fig 7 |
1 | 3864-3867 | Let x
be a given point in [a, b] Then
( )
x
a f x dx
∫
represents the area of the light shaded region
Fig 7 3
INTEGRALS 335
in Fig 7 3 [Here it is assumed that f (x) > 0 for x ∈ [a, b], the assertion made below is
equally true for other functions as well] |
1 | 3865-3868 | Then
( )
x
a f x dx
∫
represents the area of the light shaded region
Fig 7 3
INTEGRALS 335
in Fig 7 3 [Here it is assumed that f (x) > 0 for x ∈ [a, b], the assertion made below is
equally true for other functions as well] The area of this shaded region depends upon
the value of x |
1 | 3866-3869 | 3
INTEGRALS 335
in Fig 7 3 [Here it is assumed that f (x) > 0 for x ∈ [a, b], the assertion made below is
equally true for other functions as well] The area of this shaded region depends upon
the value of x In other words, the area of this shaded region is a function of x |
1 | 3867-3870 | 3 [Here it is assumed that f (x) > 0 for x ∈ [a, b], the assertion made below is
equally true for other functions as well] The area of this shaded region depends upon
the value of x In other words, the area of this shaded region is a function of x We denote this
function of x by A(x) |
1 | 3868-3871 | The area of this shaded region depends upon
the value of x In other words, the area of this shaded region is a function of x We denote this
function of x by A(x) We call the function A(x) as Area function and is given by
A (x) = ∫
( )
x
a f x dx |
1 | 3869-3872 | In other words, the area of this shaded region is a function of x We denote this
function of x by A(x) We call the function A(x) as Area function and is given by
A (x) = ∫
( )
x
a f x dx (1)
Based on this definition, the two basic fundamental theorems have been given |
1 | 3870-3873 | We denote this
function of x by A(x) We call the function A(x) as Area function and is given by
A (x) = ∫
( )
x
a f x dx (1)
Based on this definition, the two basic fundamental theorems have been given However, we only state them as their proofs are beyond the scope of this text book |
1 | 3871-3874 | We call the function A(x) as Area function and is given by
A (x) = ∫
( )
x
a f x dx (1)
Based on this definition, the two basic fundamental theorems have been given However, we only state them as their proofs are beyond the scope of this text book 7 |
1 | 3872-3875 | (1)
Based on this definition, the two basic fundamental theorems have been given However, we only state them as their proofs are beyond the scope of this text book 7 8 |
1 | 3873-3876 | However, we only state them as their proofs are beyond the scope of this text book 7 8 2 First fundamental theorem of integral calculus
Theorem 1 Let f be a continuous function on the closed interval [a, b] and let A (x) be
the area function |
1 | 3874-3877 | 7 8 2 First fundamental theorem of integral calculus
Theorem 1 Let f be a continuous function on the closed interval [a, b] and let A (x) be
the area function Then A′′′′′(x) = f (x), for all x ∈∈∈∈∈ [a, b] |
1 | 3875-3878 | 8 2 First fundamental theorem of integral calculus
Theorem 1 Let f be a continuous function on the closed interval [a, b] and let A (x) be
the area function Then A′′′′′(x) = f (x), for all x ∈∈∈∈∈ [a, b] 7 |
1 | 3876-3879 | 2 First fundamental theorem of integral calculus
Theorem 1 Let f be a continuous function on the closed interval [a, b] and let A (x) be
the area function Then A′′′′′(x) = f (x), for all x ∈∈∈∈∈ [a, b] 7 8 |
1 | 3877-3880 | Then A′′′′′(x) = f (x), for all x ∈∈∈∈∈ [a, b] 7 8 3 Second fundamental theorem of integral calculus
We state below an important theorem which enables us to evaluate definite integrals
by making use of anti derivative |
1 | 3878-3881 | 7 8 3 Second fundamental theorem of integral calculus
We state below an important theorem which enables us to evaluate definite integrals
by making use of anti derivative Theorem 2 Let f be continuous function defined on the closed interval [a, b] and F be
an anti derivative of f |
1 | 3879-3882 | 8 3 Second fundamental theorem of integral calculus
We state below an important theorem which enables us to evaluate definite integrals
by making use of anti derivative Theorem 2 Let f be continuous function defined on the closed interval [a, b] and F be
an anti derivative of f Then ∫
( )
b
a f x dx = [F( )] =
ab
x
F (b) – F(a) |
1 | 3880-3883 | 3 Second fundamental theorem of integral calculus
We state below an important theorem which enables us to evaluate definite integrals
by making use of anti derivative Theorem 2 Let f be continuous function defined on the closed interval [a, b] and F be
an anti derivative of f Then ∫
( )
b
a f x dx = [F( )] =
ab
x
F (b) – F(a) Remarks
(i)
In words, the Theorem 2 tells us that
( )
b
∫a f x dx
= (value of the anti derivative F
of f at the upper limit b – value of the same anti derivative at the lower limit a) |
1 | 3881-3884 | Theorem 2 Let f be continuous function defined on the closed interval [a, b] and F be
an anti derivative of f Then ∫
( )
b
a f x dx = [F( )] =
ab
x
F (b) – F(a) Remarks
(i)
In words, the Theorem 2 tells us that
( )
b
∫a f x dx
= (value of the anti derivative F
of f at the upper limit b – value of the same anti derivative at the lower limit a) (ii)
This theorem is very useful, because it gives us a method of calculating the
definite integral more easily, without calculating the limit of a sum |
1 | 3882-3885 | Then ∫
( )
b
a f x dx = [F( )] =
ab
x
F (b) – F(a) Remarks
(i)
In words, the Theorem 2 tells us that
( )
b
∫a f x dx
= (value of the anti derivative F
of f at the upper limit b – value of the same anti derivative at the lower limit a) (ii)
This theorem is very useful, because it gives us a method of calculating the
definite integral more easily, without calculating the limit of a sum (iii)
The crucial operation in evaluating a definite integral is that of finding a function
whose derivative is equal to the integrand |
1 | 3883-3886 | Remarks
(i)
In words, the Theorem 2 tells us that
( )
b
∫a f x dx
= (value of the anti derivative F
of f at the upper limit b – value of the same anti derivative at the lower limit a) (ii)
This theorem is very useful, because it gives us a method of calculating the
definite integral more easily, without calculating the limit of a sum (iii)
The crucial operation in evaluating a definite integral is that of finding a function
whose derivative is equal to the integrand This strengthens the relationship
between differentiation and integration |
1 | 3884-3887 | (ii)
This theorem is very useful, because it gives us a method of calculating the
definite integral more easily, without calculating the limit of a sum (iii)
The crucial operation in evaluating a definite integral is that of finding a function
whose derivative is equal to the integrand This strengthens the relationship
between differentiation and integration (iv)
In
( )
b
∫a f x dx
, the function f needs to be well defined and continuous in [a, b] |
1 | 3885-3888 | (iii)
The crucial operation in evaluating a definite integral is that of finding a function
whose derivative is equal to the integrand This strengthens the relationship
between differentiation and integration (iv)
In
( )
b
∫a f x dx
, the function f needs to be well defined and continuous in [a, b] For instance, the consideration of definite integral
1
3
2
2
2 (
x x–1)
dx
−∫
is erroneous
since the function f expressed by f(x) =
1
2
2
(
x x–1)
is not defined in a portion
– 1 < x < 1 of the closed interval [– 2, 3] |
1 | 3886-3889 | This strengthens the relationship
between differentiation and integration (iv)
In
( )
b
∫a f x dx
, the function f needs to be well defined and continuous in [a, b] For instance, the consideration of definite integral
1
3
2
2
2 (
x x–1)
dx
−∫
is erroneous
since the function f expressed by f(x) =
1
2
2
(
x x–1)
is not defined in a portion
– 1 < x < 1 of the closed interval [– 2, 3] 336
MATHEMATICS
Steps for calculating
( )
b
∫a f x dx |
1 | 3887-3890 | (iv)
In
( )
b
∫a f x dx
, the function f needs to be well defined and continuous in [a, b] For instance, the consideration of definite integral
1
3
2
2
2 (
x x–1)
dx
−∫
is erroneous
since the function f expressed by f(x) =
1
2
2
(
x x–1)
is not defined in a portion
– 1 < x < 1 of the closed interval [– 2, 3] 336
MATHEMATICS
Steps for calculating
( )
b
∫a f x dx (i)
Find the indefinite integral
( )
∫f x dx |
1 | 3888-3891 | For instance, the consideration of definite integral
1
3
2
2
2 (
x x–1)
dx
−∫
is erroneous
since the function f expressed by f(x) =
1
2
2
(
x x–1)
is not defined in a portion
– 1 < x < 1 of the closed interval [– 2, 3] 336
MATHEMATICS
Steps for calculating
( )
b
∫a f x dx (i)
Find the indefinite integral
( )
∫f x dx Let this be F(x) |
1 | 3889-3892 | 336
MATHEMATICS
Steps for calculating
( )
b
∫a f x dx (i)
Find the indefinite integral
( )
∫f x dx Let this be F(x) There is no need to keep
integration constant C because if we consider F(x) + C instead of F(x), we get
( )
[F ( )
C]
[F( )
C]– [F( )
C]
F( ) – F( )
b
ab
a f x dx
x
b
a
b
a
=
+
=
+
+
=
∫ |
1 | 3890-3893 | (i)
Find the indefinite integral
( )
∫f x dx Let this be F(x) There is no need to keep
integration constant C because if we consider F(x) + C instead of F(x), we get
( )
[F ( )
C]
[F( )
C]– [F( )
C]
F( ) – F( )
b
ab
a f x dx
x
b
a
b
a
=
+
=
+
+
=
∫ Thus, the arbitrary constant disappears in evaluating the value of the definite
integral |
1 | 3891-3894 | Let this be F(x) There is no need to keep
integration constant C because if we consider F(x) + C instead of F(x), we get
( )
[F ( )
C]
[F( )
C]– [F( )
C]
F( ) – F( )
b
ab
a f x dx
x
b
a
b
a
=
+
=
+
+
=
∫ Thus, the arbitrary constant disappears in evaluating the value of the definite
integral (ii)
Evaluate F(b) – F(a) = [F ( )]b
xa
, which is the value of
( )
b
∫a f x dx |
1 | 3892-3895 | There is no need to keep
integration constant C because if we consider F(x) + C instead of F(x), we get
( )
[F ( )
C]
[F( )
C]– [F( )
C]
F( ) – F( )
b
ab
a f x dx
x
b
a
b
a
=
+
=
+
+
=
∫ Thus, the arbitrary constant disappears in evaluating the value of the definite
integral (ii)
Evaluate F(b) – F(a) = [F ( )]b
xa
, which is the value of
( )
b
∫a f x dx We now consider some examples
Example 27 Evaluate the following integrals:
(i)
3
2
∫2 x dx
(ii)
9
3
4
22
(30–
)
x
dx
x
∫
(iii)
2
1 (
1) (
2)
x dx
x
x
+
+
∫
(iv)
3
4
0 sin 2 cos 2
t
t dt
π
∫
Solution
(i)
Let
3
2
2
I
= ∫x dx |
1 | 3893-3896 | Thus, the arbitrary constant disappears in evaluating the value of the definite
integral (ii)
Evaluate F(b) – F(a) = [F ( )]b
xa
, which is the value of
( )
b
∫a f x dx We now consider some examples
Example 27 Evaluate the following integrals:
(i)
3
2
∫2 x dx
(ii)
9
3
4
22
(30–
)
x
dx
x
∫
(iii)
2
1 (
1) (
2)
x dx
x
x
+
+
∫
(iv)
3
4
0 sin 2 cos 2
t
t dt
π
∫
Solution
(i)
Let
3
2
2
I
= ∫x dx Since
3
2
F ( )
x3
x dx
x
=
=
∫
,
Therefore, by the second fundamental theorem, we get
I =
27
8
19
F (3) – F (2)
–
3
3
3
=
=
(ii)
Let
9
3
4
2
2
I
(30 –
)
x
dx
x
= ∫ |
1 | 3894-3897 | (ii)
Evaluate F(b) – F(a) = [F ( )]b
xa
, which is the value of
( )
b
∫a f x dx We now consider some examples
Example 27 Evaluate the following integrals:
(i)
3
2
∫2 x dx
(ii)
9
3
4
22
(30–
)
x
dx
x
∫
(iii)
2
1 (
1) (
2)
x dx
x
x
+
+
∫
(iv)
3
4
0 sin 2 cos 2
t
t dt
π
∫
Solution
(i)
Let
3
2
2
I
= ∫x dx Since
3
2
F ( )
x3
x dx
x
=
=
∫
,
Therefore, by the second fundamental theorem, we get
I =
27
8
19
F (3) – F (2)
–
3
3
3
=
=
(ii)
Let
9
3
4
2
2
I
(30 –
)
x
dx
x
= ∫ We first find the anti derivative of the integrand |
1 | 3895-3898 | We now consider some examples
Example 27 Evaluate the following integrals:
(i)
3
2
∫2 x dx
(ii)
9
3
4
22
(30–
)
x
dx
x
∫
(iii)
2
1 (
1) (
2)
x dx
x
x
+
+
∫
(iv)
3
4
0 sin 2 cos 2
t
t dt
π
∫
Solution
(i)
Let
3
2
2
I
= ∫x dx Since
3
2
F ( )
x3
x dx
x
=
=
∫
,
Therefore, by the second fundamental theorem, we get
I =
27
8
19
F (3) – F (2)
–
3
3
3
=
=
(ii)
Let
9
3
4
2
2
I
(30 –
)
x
dx
x
= ∫ We first find the anti derivative of the integrand Put
23
3
30 – |
1 | 3896-3899 | Since
3
2
F ( )
x3
x dx
x
=
=
∫
,
Therefore, by the second fundamental theorem, we get
I =
27
8
19
F (3) – F (2)
–
3
3
3
=
=
(ii)
Let
9
3
4
2
2
I
(30 –
)
x
dx
x
= ∫ We first find the anti derivative of the integrand Put
23
3
30 – Then –
2
x
t
x dx
dt
=
=
or
2
–
3
x dx
dt
=
Thus,
3
2
2
2
–2
3
(30 –
)
x
dt
dx
t
x
=
∫
∫
=
2 1
3 t
=
3
2
2
1
F ( )
3 (30 –
)
x
x
=
INTEGRALS 337
Therefore, by the second fundamental theorem of calculus, we have
I =
9
3
2
4
2
1
F(9) – F(4)
3
(30 –
)
x
=
=
2
1
1
3 (30 – 27)
30 –8
−
=
2 1
1
19
3 3
22
99
−
=
(iii)
Let
2
1
I
(
1) (
2)
x dx
x
x
=
+
+
∫
Using partial fraction, we get
–1
2
(
1) (
2)
1
2
x
x
x
x
x
=
+
+
+
+
+
So
(
1) (
2)
x dx
x
x
+
+
∫
= – log
1
2log
2
F( )
x
x
x
+
+
+
=
Therefore, by the second fundamental theorem of calculus, we have
I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3]
= – 3 log 3 + log 2 + 2 log 4 =
32
log
27
(iv)
Let
3
04
I
sin 2 cos2
t
t dt
= ∫π |
1 | 3897-3900 | We first find the anti derivative of the integrand Put
23
3
30 – Then –
2
x
t
x dx
dt
=
=
or
2
–
3
x dx
dt
=
Thus,
3
2
2
2
–2
3
(30 –
)
x
dt
dx
t
x
=
∫
∫
=
2 1
3 t
=
3
2
2
1
F ( )
3 (30 –
)
x
x
=
INTEGRALS 337
Therefore, by the second fundamental theorem of calculus, we have
I =
9
3
2
4
2
1
F(9) – F(4)
3
(30 –
)
x
=
=
2
1
1
3 (30 – 27)
30 –8
−
=
2 1
1
19
3 3
22
99
−
=
(iii)
Let
2
1
I
(
1) (
2)
x dx
x
x
=
+
+
∫
Using partial fraction, we get
–1
2
(
1) (
2)
1
2
x
x
x
x
x
=
+
+
+
+
+
So
(
1) (
2)
x dx
x
x
+
+
∫
= – log
1
2log
2
F( )
x
x
x
+
+
+
=
Therefore, by the second fundamental theorem of calculus, we have
I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3]
= – 3 log 3 + log 2 + 2 log 4 =
32
log
27
(iv)
Let
3
04
I
sin 2 cos2
t
t dt
= ∫π Consider
3
sin 2 cos2
t
t dt
∫
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = 1
2 du
So
3
sin 2 cos2
t
t dt
∫
=
3
1
2
u du
∫
=
4
4
1
1
[
]
sin 2
F ( )say
8
8
u
t
t
=
=
Therefore, by the second fundamental theorem of integral calculus
I =
4
4
1
1
F ( ) – F (0)
[sin
–sin 0]
4
8
2
8
π
π
=
=
338
MATHEMATICS
EXERCISE 7 |
1 | 3898-3901 | Put
23
3
30 – Then –
2
x
t
x dx
dt
=
=
or
2
–
3
x dx
dt
=
Thus,
3
2
2
2
–2
3
(30 –
)
x
dt
dx
t
x
=
∫
∫
=
2 1
3 t
=
3
2
2
1
F ( )
3 (30 –
)
x
x
=
INTEGRALS 337
Therefore, by the second fundamental theorem of calculus, we have
I =
9
3
2
4
2
1
F(9) – F(4)
3
(30 –
)
x
=
=
2
1
1
3 (30 – 27)
30 –8
−
=
2 1
1
19
3 3
22
99
−
=
(iii)
Let
2
1
I
(
1) (
2)
x dx
x
x
=
+
+
∫
Using partial fraction, we get
–1
2
(
1) (
2)
1
2
x
x
x
x
x
=
+
+
+
+
+
So
(
1) (
2)
x dx
x
x
+
+
∫
= – log
1
2log
2
F( )
x
x
x
+
+
+
=
Therefore, by the second fundamental theorem of calculus, we have
I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3]
= – 3 log 3 + log 2 + 2 log 4 =
32
log
27
(iv)
Let
3
04
I
sin 2 cos2
t
t dt
= ∫π Consider
3
sin 2 cos2
t
t dt
∫
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = 1
2 du
So
3
sin 2 cos2
t
t dt
∫
=
3
1
2
u du
∫
=
4
4
1
1
[
]
sin 2
F ( )say
8
8
u
t
t
=
=
Therefore, by the second fundamental theorem of integral calculus
I =
4
4
1
1
F ( ) – F (0)
[sin
–sin 0]
4
8
2
8
π
π
=
=
338
MATHEMATICS
EXERCISE 7 9
Evaluate the definite integrals in Exercises 1 to 20 |
1 | 3899-3902 | Then –
2
x
t
x dx
dt
=
=
or
2
–
3
x dx
dt
=
Thus,
3
2
2
2
–2
3
(30 –
)
x
dt
dx
t
x
=
∫
∫
=
2 1
3 t
=
3
2
2
1
F ( )
3 (30 –
)
x
x
=
INTEGRALS 337
Therefore, by the second fundamental theorem of calculus, we have
I =
9
3
2
4
2
1
F(9) – F(4)
3
(30 –
)
x
=
=
2
1
1
3 (30 – 27)
30 –8
−
=
2 1
1
19
3 3
22
99
−
=
(iii)
Let
2
1
I
(
1) (
2)
x dx
x
x
=
+
+
∫
Using partial fraction, we get
–1
2
(
1) (
2)
1
2
x
x
x
x
x
=
+
+
+
+
+
So
(
1) (
2)
x dx
x
x
+
+
∫
= – log
1
2log
2
F( )
x
x
x
+
+
+
=
Therefore, by the second fundamental theorem of calculus, we have
I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3]
= – 3 log 3 + log 2 + 2 log 4 =
32
log
27
(iv)
Let
3
04
I
sin 2 cos2
t
t dt
= ∫π Consider
3
sin 2 cos2
t
t dt
∫
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = 1
2 du
So
3
sin 2 cos2
t
t dt
∫
=
3
1
2
u du
∫
=
4
4
1
1
[
]
sin 2
F ( )say
8
8
u
t
t
=
=
Therefore, by the second fundamental theorem of integral calculus
I =
4
4
1
1
F ( ) – F (0)
[sin
–sin 0]
4
8
2
8
π
π
=
=
338
MATHEMATICS
EXERCISE 7 9
Evaluate the definite integrals in Exercises 1 to 20 1 |
1 | 3900-3903 | Consider
3
sin 2 cos2
t
t dt
∫
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = 1
2 du
So
3
sin 2 cos2
t
t dt
∫
=
3
1
2
u du
∫
=
4
4
1
1
[
]
sin 2
F ( )say
8
8
u
t
t
=
=
Therefore, by the second fundamental theorem of integral calculus
I =
4
4
1
1
F ( ) – F (0)
[sin
–sin 0]
4
8
2
8
π
π
=
=
338
MATHEMATICS
EXERCISE 7 9
Evaluate the definite integrals in Exercises 1 to 20 1 1
1(
1)
x
dx
−
+
∫
2 |
1 | 3901-3904 | 9
Evaluate the definite integrals in Exercises 1 to 20 1 1
1(
1)
x
dx
−
+
∫
2 3
2
∫x1 dx
3 |
1 | 3902-3905 | 1 1
1(
1)
x
dx
−
+
∫
2 3
2
∫x1 dx
3 2
3
2
1 (4
– 5
6
9)
x
x
x
dx
+
+
∫
4 |
1 | 3903-3906 | 1
1(
1)
x
dx
−
+
∫
2 3
2
∫x1 dx
3 2
3
2
1 (4
– 5
6
9)
x
x
x
dx
+
+
∫
4 4
0 sin 2x dx
∫π
5 |
1 | 3904-3907 | 3
2
∫x1 dx
3 2
3
2
1 (4
– 5
6
9)
x
x
x
dx
+
+
∫
4 4
0 sin 2x dx
∫π
5 2
0 cos 2x dx
∫π
6 |
1 | 3905-3908 | 2
3
2
1 (4
– 5
6
9)
x
x
x
dx
+
+
∫
4 4
0 sin 2x dx
∫π
5 2
0 cos 2x dx
∫π
6 5
4
x
∫e dx
7 |
1 | 3906-3909 | 4
0 sin 2x dx
∫π
5 2
0 cos 2x dx
∫π
6 5
4
x
∫e dx
7 4
0 tanx dx
π
∫
8 |
1 | 3907-3910 | 2
0 cos 2x dx
∫π
6 5
4
x
∫e dx
7 4
0 tanx dx
π
∫
8 4
6
cosec x dx
π
π∫
9 |
1 | 3908-3911 | 5
4
x
∫e dx
7 4
0 tanx dx
π
∫
8 4
6
cosec x dx
π
π∫
9 1
0
2
1–
dx
x
∫
10 |
1 | 3909-3912 | 4
0 tanx dx
π
∫
8 4
6
cosec x dx
π
π∫
9 1
0
2
1–
dx
x
∫
10 1
2
01
dx
+x
∫
11 |
1 | 3910-3913 | 4
6
cosec x dx
π
π∫
9 1
0
2
1–
dx
x
∫
10 1
2
01
dx
+x
∫
11 3
2
2
1
dx
x −
∫
12 |
1 | 3911-3914 | 1
0
2
1–
dx
x
∫
10 1
2
01
dx
+x
∫
11 3
2
2
1
dx
x −
∫
12 2
2
0 cos x dx
∫π
13 |
1 | 3912-3915 | 1
2
01
dx
+x
∫
11 3
2
2
1
dx
x −
∫
12 2
2
0 cos x dx
∫π
13 3
2
2
1
x dx
∫x +
14 |
1 | 3913-3916 | 3
2
2
1
dx
x −
∫
12 2
2
0 cos x dx
∫π
13 3
2
2
1
x dx
∫x +
14 1
2
0
2
3
5
1
x
dx
x
+
+
∫
15 |
1 | 3914-3917 | 2
2
0 cos x dx
∫π
13 3
2
2
1
x dx
∫x +
14 1
2
0
2
3
5
1
x
dx
x
+
+
∫
15 2
1
0
x
x e dx
∫
16 |
1 | 3915-3918 | 3
2
2
1
x dx
∫x +
14 1
2
0
2
3
5
1
x
dx
x
+
+
∫
15 2
1
0
x
x e dx
∫
16 2
2
2
1
5
4
3
x
x
+x
+
∫
17 |
1 | 3916-3919 | 1
2
0
2
3
5
1
x
dx
x
+
+
∫
15 2
1
0
x
x e dx
∫
16 2
2
2
1
5
4
3
x
x
+x
+
∫
17 2
3
4
0 (2sec
2)
x
x
dx
π
+
+
∫
18 |
1 | 3917-3920 | 2
1
0
x
x e dx
∫
16 2
2
2
1
5
4
3
x
x
+x
+
∫
17 2
3
4
0 (2sec
2)
x
x
dx
π
+
+
∫
18 2
2
0 (sin
–cos
)
2
2
x
x dx
π∫
19 |
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