Business Mathematics lecture: integral calculus

0.06 MB·English

by  KIT Tyabandha

Department of Mathemati
s, Mahidol University Kit Tyabandha, PhD
Integral
al
ulus
th
10 January 2005
0
De(cid:12)nition 1. Let f(x) be a fun
tion, and let f (x) be its derivative. The reverse pro
ess of
di(cid:11)erentiation is
alled antidi(cid:11)erentiation or integration. It gives us the original fun
tion, whi
h is

alled the antiderivative or integral of f(x).
x
Theorem 1. Let
, n and k be
onstants. Then
a. Z
kdx=kx+
b. Z
dx=x+

. Z
n 1 n+1
x dx= x +
; n6=(cid:0)1
n+1
d. Z
(cid:0)1
x dx=lnx+
; x>0
e. Z
(cid:0)1
x dx=lnjxj+
; 06=x<0
f. Z
kx 1 kx
e dx= e +
k
g. Z Z
kf(x)dx=k f(x)dx
h. Z Z Z
(f(x)(cid:6)g(x))= f(x)dx(cid:6) g(x)dx
i. Z Z
(cid:0)f(x)dx=(cid:0) f(x)dx
x
P
n i
De(cid:12)nition 2. The approximation (i = 1) (f(xi)(cid:1)x ) of the area under a
ontinuous
urve
A is
alled a Riemann sum. That area under the
urve is
Xn
A= lim f(xi)(cid:1)xi
n!1
i=1
x
Theorem 2. Let F(x) be the integral of f(x). We
all the fundamental theorem of
al
ulus the
expression.
Z b
b
f(x)dx= F(x)ja =F(b)(cid:0)F(a)
a
x
Theorem 3.
a.
Z b Z a
f(x)dx=(cid:0) f(x)dx
a b
b. Z a
f(x)dx=F(a)(cid:0)F(a)=0
a
th
Business Mathemati
s, Integral
al
ulus {1{ From 5 Nov 05, as of 10 January, 2006
Department of Mathemati
s, Mahidol University Kit Tyabandha, PhD

.
Z
Z b Z
f(x)dx= f(x)dx+ f(x)dx; a(cid:20)b(cid:20)
a a b
d.
Z b Z b Z b
f(x)dx(cid:6) g(x)dx= (f(x)(cid:6)g(x))dx
a a a
e.
Z b Z b
kf(x)dx=k f(x)dx
a a
x
Property d of Theorem 3 is used to (cid:12)nd the area between two
urves.
Theorem 4. The pro
ess of integration by parts is
Z Z
0 0
(f(x)(cid:1)g (x))dx=f(x)(cid:1)g(x)(cid:0) (g(x)(cid:1)f (x))dx
Proof. From
d 0 0
(f(x)(cid:1)g(x))=f(x)(cid:1)g (x)+g(x)(cid:1)f (x)
dx
we have Z Z
0 0
f(x)(cid:1)g(x)= (f(x)(cid:1)g (x))dx+ (g(x)(cid:1)f (x))dx
{
Bibliography
Edward T Dowling. Mathemati
al methods for business and e
onomi
s. S
haum's outline series,
1993
th
Business Mathemati
s, Integral
al
ulus {2{ From 5 Nov 05, as of 10 January, 2006