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positive direction
b F (a)
∫ f(x)dx = F(b)  F(a)
a
b F(b
F(a
F( b) = ∫ f(x)dx + F( a)
a
a If F( b) is known, then from b going RIGHT to LEFT in the negative direction to a :
F(b
negative direction
a F(b
F(a
F(
∫b f(x)dx = F( a )  F( b )
a F(a) = ∫b f(x)dx + F(b)
Using the relations above we may compute F C ( x) . We must choose a
small inter val x . Then {f(x). ∆ x} is an element of ar ea under the
area
.
ar
cur v e o f f(x) over the inter val ∆ x . The smaller the ∆ x the more accurate
the approximation of the ar ea under the cur v e o f f(x) over the interval
area
ar
∆ x . Another reason why we have to choose ∆ x very small is that we do not
skip or miss the EXTREMUM points (maximum and minimum) in plotting
F C ( x). These are the points where f(x) CHANGES SIGN.
next FC (x) = CHANGE in FC (x) over next ∆x + current value of FC (x)
CHANGE in FC (x) over next ∆x = area under the curve of f(x) over next ∆x
area
curv
ar
159 Going RIGHT (in the postive direction ) from x 0 with i = 1 :
p ostive
w ith
F C (x i ) =
(x ∫ x x + ∆x
i −1 f(x) dx + F C (x i − 1 )
f (x)
(x
i−1 +∆
(x
(x
F C (x i ) ≅ f (x i ) .( +∆ x ) + F C (x i − 1 ) for i = 1, 2, 3, . . . The + SIGN in ( + ∆ x) is due to the postive direction i n the integration.
p ostive Going LEFT (in the negative direction ) from x 0 with i = − 1:
n egative
w ith
F C (x i ) =
(x ∫ x x − ∆x
i +1 f(x) dx + F C (x i +1 )
f (x)
(x i +1 (x
(x
F C (x i ) ≅ f(x i ) .( − ∆ x ) + F C (x i + 1 ) for i = −1, −2, −3, . . .
The − SIGN in ( − ∆x) is due to the negative direction in the integration.
negative
This is the r ectangular method o f NUMERICAL INTEGRATION. In the next
chapter we shall see how to compute and interpret the meaning of the
CONSTANT OF INTEGRATION from the nature of the problem. 160 EXERCISES
The exercises are to show how to plot FC ( x) when only f(x), x 0 a nd FC ( x 0)
are known and the constant of integration C is not known. Also, we do not
need to know the expression of F(x). We make use of t...
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 Fall '09
 TAMERDOğAN
 Limit, Δx

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