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MATH1013 Calculus I
Integration I (Chap. 5 (§5.1, 5.2)) 1
Edmund Y. M. Chiang
Department of MathematicsHong Kong University of Science & Technology
November 24, 2014
1Based on Stewart, James, “Single Variable Calculus, Early Transcendentals”, 7th edition, Brooks/Coles, 2012
Briggs, Cochran and Gillett: Calculus for Scientists and Engineers: Early Transcendentals, Pearson 2013
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Definite integrals Riemann Sums Integrable functions
Definite integrals
Riemann Sums
Integrable functions
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Definite integrals Riemann Sums Integrable functions
Area under curve
We shall consider continuous functions defined on a closed intervalonly. The aim is to develop a theory that can be used to find areaof the region under the given function.
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Definite integrals Riemann Sums Integrable functions
Example
We consider the problem of finding the area under the straight linef (x) = x for the interval 0 ≤ x ≤ 1.
We divide the interval [0, 1] into five subintervals of equal width.By a partition of [0, 1] with five points: {x0, x1, x2, x3, x4, x5} of[0, 1]. So we have the subintervals:
[x0, x1] = [0, 1/5]
[x1, x2] = [1/5, 2/5]
[x2, x3] = [2/5, 3/5]
[x3, x4] = [3/5, 4/5]
[x4, x5] = [4/5, 5/5]
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Definite integrals Riemann Sums Integrable functions
Right Riemann sumThe f attains maximum values in each of the above intervals:
f (x1) = f (1/5) = 1/5,
f (x2) = f (2/5) = 2/5,
f (x3) = f (3/5) = 3/5,
f (x4) = f (4/5) = 4/5,
f (x5) = f (5/5) = 5/5 = 1.
Let us sum the areas of the five rectangles with height at theright-end point of [xi−1, xi ] and with base 1/n. Thus we have
S5 = f
(1
5
)1
5+ f
(2
5
)1
5+ f
(3
5
)1
5+ f
(4
5
)1
5+ f
(5
5
)1
5
=1
25
(1 + 2 + 3 + 4 + 5
)=
15
25=
3
25,
is called an right Riemann sum of f over [0, 1] with respect to theabove partition. The approximation 3/25 is larger than the actualarea.
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Definite integrals Riemann Sums Integrable functions
Left Riemann sumThe f attains minimum values in each of the above intervals:
f (x0) = f (0/5) = 0/5,
f (x1) = f (1/5) = 1/5,
f (x2) = f (2/5) = 2/5,
f (x3) = f (3/5) = 3/5,
f (x4) = f (4/5) = 4/5.
Let us sum the areas of the five rectangles height at the left-endpoint in [xi−1, xi ] and with base 1/n. Thus we have
S5 = f
(0
5
)1
5+ f
(1
5
)1
5+ f
(2
5
)1
5+ f
(3
5
)1
5+ f
(4
5
)1
5
=1
25
(0 + 1 + 2 + 3 + 4
)=
10
25=
2
5,
is called an left Riemann sum of f over [0, 1] with respect to theabove partition. The approximation 2/25 is smaller than the actualarea.
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Definite integrals Riemann Sums Integrable functions
Left and Right Riemann sums figuresSuppose the value of the area that we are to find is A. Then weclearly see from the figure below that the following inequalitiesmust hold:
2
5= S5 ≤ A ≤ S5 =
3
5.
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Definite integrals Riemann Sums Integrable functions
Left and Right Riemann sums: general case
It is also clear that the above argument that we divide [0, 1] intofive equal intervals is nothing special. Hence the same argumentapplies for any number of intervals. Hence we must have
Sn ≤ A ≤ Sn, (2)
where n is any positive integer. We partition [0, 1] into n equalsubintervals:
{x0, x1, x2, . . . , xn−1, xn} = {0,1
n,
2
n, . . . ,
n − 1n
,n
n}.
and the width of each of the subintervals of 1/n.
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Definite integrals Riemann Sums Integrable functions
Right Riemann sum: general case
Hence the upper sum is
Sn = f(1
n
) 1n
+ f(2
n
) 1n
+ · · ·+ f(n − 1
n
) 1n
+ f(n
n
)1n
=1
n
1
n+
2
n
1
n+ · · ·+ n − 1
n
1
n+
n
n
1
n
=1
n2(1 + 2 + 3 + · · ·+ (n − 1) + n
)=
1
n2n(n + 1)
2
=1
2
(1 +
1
n
).
Note that we have used the fundamental formula that
1 + 2 + 3 + · · ·+ (n − 1) + n = n(n + 1)2
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Definite integrals Riemann Sums Integrable functions
Left Riemann sum: general caseHence the lower sum is
Sn = f(0
n
) 1n
+ f(1
n
) 1n
+ · · ·+ f(n − 2
n
) 1n
+ f(n − 1
n
) 1n
=1
n
0
n+
1
n
1
n+ · · ·+ n − 2
n
1
n+
n − 1n
1
n
=1
n2(0 + 1 + 2 + · · ·+ (n − 2) + n − 1
)=
1
n2(n − 1)n
2=
1
2
(1− 1
n
).
We deduce that
1
2
(1− 1
n
)≤ A ≤ 1
2
(1 +
1
n
), for all n.
Letting n→ +∞ gives that 12 ≤ A ≤12 . That is A = 1/2 and
limn→+∞
Sn = 1/2 = limn→+∞
Sn.
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Definite integrals Riemann Sums Integrable functions
Riemann sums
We consider a closed interval [a, b] and let
P = {x0, x1, x2, . . . , xn−1, xn}, ∆xi = xi−xi−1 =b − a
n, i = 1, · · · , n
to be any points lying inside [a, b]. Let f (x) be a continuousfunction defined on [a, b]. Then we define a Riemann sum of f (x)over [a, b] with respect to partition P to be the sum
f (x∗1 ) ∆x1 + f (x∗2 ) ∆x2 + · · ·+ f (x∗n ) ∆xn
where x∗i is an arbitrary point lying in [xi−1, xi ], i = 1, 2, · · · , n.Depending on the choices of x∗i , we have
1. Left Riemann sum if x∗i = xi−1;
2. Right Riemann sum if x∗i = xi ;
3. Mid-point Riemann sum if x∗i = (xi−1 + xi )/2.
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Definite integrals Riemann Sums Integrable functions
Riemann sum figure
Figure: (Briggs, et al, Figure 5.8)
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Definite integrals Riemann Sums Integrable functions
Left Riemann sum figure
Figure: (Briggs, et al, Figure 5.9)
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Definite integrals Riemann Sums Integrable functions
Right Riemann sum figure
Figure: (Briggs, et al, Figure 5.10)
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Definite integrals Riemann Sums Integrable functions
Mid-point Riemann sum figure
Figure: (Briggs, et al, Figure 5.11)
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Definite integrals Riemann Sums Integrable functions
Riemann sum of SineWe compute various Riemann sums under the sine curve fromx = 0 to x = π/2 with six intervals. So we have
∆x =b − a
n=π/2− 0
6=
π
12.
• Left Riemann sum gives
f (x∗1 ) ∆x1 + f (x∗2 ) ∆x2 + · · ·+ f (x∗n ) ∆xn ≈ 0.863
• Right Riemann sum gives
f (x∗1 ) ∆x1 + f (x∗2 ) ∆x2 + · · ·+ f (x∗n ) ∆xn ≈ 1.125
• Mid-point Riemann sum gives
f (x∗1 ) ∆x1 + f (x∗2 ) ∆x2 + · · ·+ f (x∗n ) ∆xn ≈ 1.003
So we have0.863 ≤ 1.003 ≤ 1.125.
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Definite integrals Riemann Sums Integrable functions
Partition of Riemann sum of Sine
Figure: (Briggs, et al, Figure 5.2)
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Definite integrals Riemann Sums Integrable functions
Partition of Riemann sum of Sine
Figure: (Briggs, et al, Theorem 5.1)
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Definite integrals Riemann Sums Integrable functions
Partition 50
Figure: (Briggs, et al, Figure 5.15)
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Definite integrals Riemann Sums Integrable functions
Example (Briggs, et al, p. 372)
After choosing a partition that divides [0, 2] into 50 subintervals:
∆xk = xk − xk−1 =2− 0
50=
1
25= 0.04.
The Right Riemann Sum is given by
50∑k=1
f (x∗k ) ∆xk =50∑k=1
f (xk) (xk − xk−1)
=50∑k=1
f (k
25) (0.04) =
50∑k=1
[( k25
)3+ 1]
(0.04)
=[ 1
253(50 · 51
2
)2+ 50
](0.04) = 6.1616.
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Definite integrals Riemann Sums Integrable functions
Example (Briggs, et al, p. 372) II
The Left Riemann Sum is given by
49∑k=0
f (x∗k ) ∆xk+1 =49∑k=0
f (xk) (xk+1 − xk)
=49∑k=0
f (k
25) (0.04) =
49∑k=0
[( k25
)3+ 1]
(0.04)
=[ 1
253(49 · 50
2
)2+ 50
](0.04) = 5.8416.
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Definite integrals Riemann Sums Integrable functions
Example (Briggs, et al, p. 388) III
After choosing a partition that divides [0, 2] into n subintervals:
∆xk = xk − xk−1 =2− 0
n=
2
n.
The Right Riemann Sum is given by
n∑k=1
f (x∗k ) ∆xk =n∑
k=1
f (xk)2
n
=2
n
n∑k=1
[(2kn
)3+ 1]
=2
n
(23n3
n∑k=1
k3 +n∑
k=1
1)
=2
n
(23n3· n
2(n + 1)2
4+ n)
= 2[2(
1 +1
n
)2+ 1]→ 6,
as n→∞.
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Definite integrals Riemann Sums Integrable functions
Example (Briggs, et al, p. 389) III
The Left Riemann Sum is given by
n−1∑k=0
f (x∗k ) ∆xk+1 =n−1∑k=0
f (xk) ·2
n
=2
n
n−1∑k=0
[(2kn
)3+ 1]
=2
n
(23n3
n−1∑k=0
k3 +n−1∑k=0
1)
=2
n
(23n3· n
2(n − 1)2
4+ n)
= 2[2(
1− 1n
)2+ 1]→ 6,
In fact, we have
4(
1− 1n
)2+ 2 ≤ A ≤ 4
(1 +
1
n
)2+ 2
to hold for every integer n. So A = 6.
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Definite integrals Riemann Sums Integrable functions
Example (Stewart, p. 362)
• Let f (x) = x2 and find the area A under f over the interval[0, 1].
• Choose the partition
P = {0, 1n,
2
n, · · · , n − 1
n,
n
n= 1}.
• Show that the following inequalities
1
3
(1− 1
n
)(1− 1
2n
)≤ A ≤ 1
3
(1 +
1
n
)(1 +
1
2n
).
by computing the left Riemann sum and right Riemann sum.
• Then show that the area is A = 13 .• We omit the details.
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Definite integrals Riemann Sums Integrable functions
Some left/right Riemann sums
Figure: (Stewart, Figures 5.1.8-9)
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Definite integrals Riemann Sums Integrable functions
Exercises
• (Briggs, et al: p. 376) Write the following sum in summationnotation:
4 + 9 + 14 + · · ·+ 44.
• (Stewart: p. 369, Q. 4) Estimate the area under f (x) =√
xover [0, 4] using four approximating rectangles.
• (Briggs, et al, p. 377) Given that4∑
k=1
f (1 + k) · 1 is a
Riemann sum of a certain function f over an interval [a, b]with a partition of n subdivisions. Identify the f , [a, b] and n.
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Definite integrals Riemann Sums Integrable functions
Upper/Lower Riemann sumsRiemann originally considered for a partition Pn = {x0, · · · , xn}
Area = limn→∞
Rn = [f (x1)∆x + f (x2)∆x · · ·+ f (xn)∆x ]
Area = limn→∞
Ln = [f (x0)∆x + f (x1)∆x · · ·+ f (xn−1)∆x ]
and more generally, the upper and lower sums defined respectivelyby
Area = limn→∞
Un = limn→∞
[f (x∗1 )∆x + f (x∗2 )∆x · · ·+ f (x∗n )∆x ]
where f (x∗k ) attains the largest value in the k−th interval[xk−1, xk ];
Area = limn→∞
Ln = limn→∞
[f (x1∗)∆x + f (x2∗)∆x · · ·+ f (xn∗)∆x ]
where f (x∗k ) attains the smallest value in the k−th interval[xk−1, xk ]
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Definite integrals Riemann Sums Integrable functions
Upper Riemann sum figure
Figure: (Stewart, Figures 5.1.13)
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Definite integrals Riemann Sums Integrable functions
Upper/lower Riemann figures
Figure: (Stewart, Figures 5.1.14)
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Definite integrals Riemann Sums Integrable functions
Definite Integrals definition
Figure: (Stewart Figure p. 372)
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Definite integrals Riemann Sums Integrable functions
Definite Integral notation
Figure: (Publisher Figure 5.21)
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Definite integrals Riemann Sums Integrable functions
Integrable functions
Theorem (Compare Stewart p. 373) Let f be a continuousfunction except on a finite number of discontinuities over theinterval [a, b]. Then f is integrable on [a, b]. That is,
limδx→0
n∑k=1
f (x∗k ) ∆xk =
∫ ba
f (x) dx ,
exists irrespective to the x∗k and the partition [xk−1, xk ] chosen.So
• Since f (x) = x2 is continuous over [0, 1] so it is integrable
and
∫ 10
x2 dx =1
2according to a previous calculation.
• Since f (x) = x3 is continuous over [0, 1] so it is integrable
and
∫ 10
x3 dx =1
3according to a previous calculation.
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Definite integrals Riemann Sums Integrable functions
Special partition and sample points
Figure: (Stewart p. 374)
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Definite integrals Riemann Sums Integrable functions
Piecewise continuous functionsThe following function has a finite number of discontinuities and sois integrable. However, we note that part 2 of the area is negative:
Figure: (Briggs, et al, Figure 5.23)
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Definite integrals Riemann Sums Integrable functions
Negative area
Figure: (Briggs, et al, Figure 5.18)
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Definite integrals Riemann Sums Integrable functions
Negative area
Figure: (Briggs, et al, Figure 5.17)
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Definite integrals Riemann Sums Integrable functions
Net area
Figure: (Briggs, et al, Figure 5.20)
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Definite integrals Riemann Sums Integrable functions
Recognizing integral
Figure: (Briggs, et al, Figure 5.24)
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Definite integrals Riemann Sums Integrable functions
Computing net area
Figure: (Briggs, et al, Figure 5.31)
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Definite integrals Riemann Sums Integrable functions
Exercises
1. Write down the right Riemann sum for
∫ 20
√4− x2 dx ;
2. Interpret the sum limn→∞
n∑k=1
3
n(1 + 3kn )as a certain Riemann
integral.
3. Let
f (x) =
{2x − 2, if x ≤ 2;−x + 4, if x > 2.
Compute both the net area and actual area of
∫ 50
f (t) dt.
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Definite integrals Riemann Sums Integrable functions
Hints to Exercises
One can analyze the the sum as
n∑k=1
3
n(1 + 3kn )=
n∑k=1
1
(1 + 3kn )× 3
n
=n∑
k=1
1
(1 + k 3n )× 3
n=
n∑k=1
f (k × 3n
)× 3n
which represents a right Riemann sum for f (x) =1
1 + x, and for
the integration range from a = 0 and b = 3. But then
lim∆→0
n∑k=1
3
n(1 + 3kn )=
∫ 30
1
1 + xdx
= ln |1 + x |∣∣∣30
= (ln 4− ln 1) = ln 4.
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Definite integrals Riemann Sums Integrable functions
Hints to Exercises
The net area of the last example is given by∫ 20
(2x − 2) dx +∫ 5
2(−x + 4) dx = (x2 − 2x)
∣∣20
+ (−x2/2 + 4x)∣∣52
= (22 − 2 · 2) + 12
(22 − 52) + (20− 8) = 0 +−12· 21 + 12 = 3
2.
The actual area is given by∫ 10
(2x − 2) dx +∣∣∣∣∫ 2
1(2x − 2) dx
∣∣∣∣+ ∫ 42
(−x + 4) dx +∣∣∣∣∫ 5
4(−x + 4) dx
∣∣∣∣=∣∣∣(x2 − 2x)∣∣10∣∣∣+ (x2 − 2x)∣∣21 + (−x2/2 + 4x)∣∣42 + ∣∣∣(−x2/2 + 4x)∣∣54∣∣∣
= | − 1|+ 1 + 2 + | − 12| = 9
2.
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Definite integrals Riemann Sums Integrable functions
Properties of Definite Integral
Figure: (Briggs, et al, Table 5.4)
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Definite integrals Riemann Sums Integrable functions
Sum of integrals
Figure: (Briggs, et al, Figure 5.29)
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Definite integrals Riemann Sums Integrable functions
Comparison properties of Definite Integral
Figure: (Stewart, p. 381)
Definite integralsRiemann SumsIntegrable functions