Business And Financial Mathematics-integral

Integral Sub Topics I. Definition of Integral II. Properties of integral III Definite Integral IV Application of definite Integral V. Problems I. Definition of Integral Suppose Function F(x) where F’(x)=f(x). F(x) is called an antiderivative of f(x). In general, F(x)+c where c=constant is also an antiderivative of f(x). F(x)+c is called integral indefinite integral (integral tak tentu ) of f(x), denoted

∫ f(x) dx = F(x) + c II Properties of Integral 1. ∫ 1 dx = c

2. . ∫ a f(x) dx = a ∫ f(x) dx where a = contant 3. . ∫ {f(x) + g(x)} dx =

∫ f(x) dx + ∫ g(x) dx

a n +1 x +c n +1 5. . ∫ u dv = uv - ∫ v du is called integral by part (integral parsial) 4. . ∫ a x n dx =

Example: x ∫ 2 x e dx = ..........

Suppose : u =2x and dv=ex , then du=2 dx and v= ex x x x x x ∫ 2 x e dx = ∫ u dv = 2x e - 2∫ e dx = 2x e - 2 e + C.. Problems 8 x3 − 4x 1. ∫ dx = 4 2 x −x 2. ∫ 2 x sin x dx = .......... Note: When F(x)=sin(x) then F’(x)= cos(x) and when F(x)=cos(x)then F’(x)= -sin(x) When F(x)=sin(ax) maka F’(x)= a cos(ax) and when F(x)=cos(ax) then F’(x)= -a sin(ax) 43

So: ∫ cos(ax) dx = 1/a sin(ax) + C and ∫ sin(ax) dx = -1/a cos(ax) + C ax ax When F(x)= e then F’(x)= a e ax ax So : ∫ e dx = 1/a e +C III Definite Integral

Definite Integral f(x) in close interval [a,b] denoted, b

∫ f ( x)dx = F (b) − F (a) a

where F(x) antiderivative of f(x). It is described by an area under a curve y=f(x) , line x=a and x=b and X axis. Example. 3

∫−x

2

+ 9dx = F (3) − F (1) ,where F(x)= -1/3 X3+9X

1

= [-1/3 · 33+9 · 3] - [-1/3 · 13+9 · 1] = 18 – 26/3 = 28/3 Graph 10 8 6 4 Y

2 0 -2

-1

-2 0

1

2

3

4

5

-4 -6 -8 X

IV Application of definite Integral Look at supply curve and demand curve and market equilibrium.

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80 70 60 50 40 P

30 20 10 0 -2

-10 0

2

4

6

8

10

-20 -30 Q

where : S : P=5Q +28 D: P=-Q2+64 Equilibrium point at: 5Q +28 = -Q2+64 Q2+5Q-36=0 (Q+9)(Q-4)=0 Q=-9 or Q=4 So, Equilibrium point at Q0=4 dan P0=48 4.a Consumer Surplus In the interval 0<=Q<= Q0, actually the consumers ready to pay more than P0 . In fact they pay at a lower price P0 So, they have an advantage which is called consumer surplus. The formula is:

CS =

Q0

∫ P dQ − P Q D

0

0

0

where PD is Demand curve Based on the example above,

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4

CS = ∫ − Q 2 + 64dQ − 4 • 48 0

= F(4) –F(0) – 192 , where F(Q)= -1/3 · Q3+64Q = [ -1/3 · 43+64·4]-0-192 =-64/3+256-192= 128/3

4.b Producer Surplus

In the interval 0<=Q<= Q0, actually the producers ready to offer more than P0 . In fact they got a higher price P0 So, they have an advantage which is called producer surplus. The formula is: Q0

PS = P0Q0 − ∫ PS dQ 0

where PS Supply Curve Based on the example above,, 4

PS = 4 • 48 − ∫ 5Q + 28dQ 0

= 192- {F(4) –F(0)} , where F(Q)= 5/2 · Q2+28Q = 192- [ 5/2 · 42+28·4]-0 =192-80/2-112=40 V Problems Problem1. f(x)= x4- 3x2+2,: Find : 4

∫ f ( x)dx = 0

Problem2. Given the supply function and demand function as follows: S : P=3Q2 +5 D: P=-2Q2+50

Find consumer surplus Find producer surplus

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