Calculus (I)
WEN-CHING LIEN
Department of Mathematics National Cheng Kung University
2008
W -C L Calculus (I)
5.1 The Definite Integral
1. The area problem.
Q: How to estimate the area of the unit disk?
W -C L Calculus (I)
2. Riemann Integral
Definition (Partition)
A partition of the closed interval[a,b]is a finite subset of [a,b]which contains a and b.
Construction of the Riemann integral:
(1) Let P = {x0,x1,· · ·,xn−1,xn}be a partition of[a,b].
W -C L Calculus (I)
2. Riemann Integral
Definition (Partition)
A partition of the closed interval[a,b]is a finite subset of [a,b]which contains a and b.
Construction of the Riemann integral:
(1) Let P = {x0,x1,· · ·,xn−1,xn}be a partition of[a,b].
W -C L Calculus (I)
2. Riemann Integral
Definition (Partition)
A partition of the closed interval[a,b]is a finite subset of [a,b]which contains a and b.
Construction of the Riemann integral:
(1) Let P = {x0,x1,· · ·,xn−1,xn}be a partition of[a,b].
W -C L Calculus (I)
2. Riemann Integral
Definition (Partition)
A partition of the closed interval[a,b]is a finite subset of [a,b]which contains a and b.
Construction of the Riemann integral:
(1) Let P = {x0,x1,· · ·,xn−1,xn}be a partition of[a,b].
W -C L Calculus (I)
Then, on each interval[xi−1,xi], f takes a maximum value Mi and a minimum value mi. We calculate
Uf(P) =M1∆x1+ · · · +Mn∆xn, the P upper sum.
Lf(P) =m1∆x1+ · · · +mn∆xn, the P lower sum.
(2) On each interval[xk−1,xk], choose a point ck ∈ [xk−1,xk].
(3) Evaluate the Riemann Sum:
S(P) = Xn
k=1
f(ck) · ∆xk
W -C L Calculus (I)
Then, on each interval[xi−1,xi], f takes a maximum value Mi and a minimum value mi. We calculate
Uf(P) =M1∆x1+ · · · +Mn∆xn, the P upper sum.
Lf(P) =m1∆x1+ · · · +mn∆xn, the P lower sum.
(2) On each interval[xk−1,xk], choose a point ck ∈ [xk−1,xk].
(3) Evaluate the Riemann Sum:
S(P) = Xn
k=1
f(ck) · ∆xk
W -C L Calculus (I)
Then, on each interval[xi−1,xi], f takes a maximum value Mi and a minimum value mi. We calculate
Uf(P) =M1∆x1+ · · · +Mn∆xn, the P upper sum.
Lf(P) =m1∆x1+ · · · +mn∆xn, the P lower sum.
(2) On each interval[xk−1,xk], choose a point ck ∈ [xk−1,xk].
(3) Evaluate the Riemann Sum:
S(P) = Xn
k=1
f(ck) · ∆xk
W -C L Calculus (I)
Then, on each interval[xi−1,xi], f takes a maximum value Mi and a minimum value mi. We calculate
Uf(P) =M1∆x1+ · · · +Mn∆xn, the P upper sum.
Lf(P) =m1∆x1+ · · · +mn∆xn, the P lower sum.
(2) On each interval[xk−1,xk], choose a point ck ∈ [xk−1,xk].
(3) Evaluate the Riemann Sum:
S(P) = Xn
k=1
f(ck) · ∆xk
W -C L Calculus (I)
Then, on each interval[xi−1,xi], f takes a maximum value Mi and a minimum value mi. We calculate
Uf(P) =M1∆x1+ · · · +Mn∆xn, the P upper sum.
Lf(P) =m1∆x1+ · · · +mn∆xn, the P lower sum.
(2) On each interval[xk−1,xk], choose a point ck ∈ [xk−1,xk].
(3) Evaluate the Riemann Sum:
S(P) = Xn
k=1
f(ck) · ∆xk
W -C L Calculus (I)
Note:
Lf(P) ≤S(P) ≤Uf(P).
(4) Find the limit
kPk→0lim S(P).
Definition (The Definite Integral) Z b
a
f(x)dx = lim
kPk→0S(P)
is called the definite integral of f from a to b, if the limit exists. And f is said to be integrable.
W -C L Calculus (I)
Note:
Lf(P) ≤S(P) ≤Uf(P).
(4) Find the limit
kPk→0lim S(P).
Definition (The Definite Integral) Z b
a
f(x)dx = lim
kPk→0S(P)
is called the definite integral of f from a to b, if the limit exists. And f is said to be integrable.
W -C L Calculus (I)
Note:
Lf(P) ≤S(P) ≤Uf(P).
(4) Find the limit
kPk→0lim S(P).
Definition (The Definite Integral) Z b
a
f(x)dx = lim
kPk→0S(P)
is called the definite integral of f from a to b, if the limit exists. And f is said to be integrable.
W -C L Calculus (I)
Theorem
If f is continuous on[a,b], then Z b
a
f(x)dx = lim
kPk→0S(P).
W -C L Calculus (I)
Thank you.
W -C L Calculus (I)
