Unformatted text preview: s due to (negative f(x)) . (positive direction ) .
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Note: f(x) δx is an infinitesimal
element of area. f(x) =  2x δ x  1 2 1
δx 0
 1  2  3 F(x) =  x 2 f(x)dx is an i nstantaneous
element o f area. If we try to
depict it, it will appear as a line.  4 151
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321  3 3 2 3 x F(2)  F(1) =  (2) 2  ( (1) 2 ) =  3
The CHANGE in F(x) is 3 in magnitude.
The  SIGN reflects a DECREASE in F(x) over [1, 2] in the positive direction .
positive continuous
Example 6: Integrate contin uous f (x) over the inter val [1, 2] going
c ontin
RIGHT to LEFT in the negative direction .
negative
T he AREA under f(x) over the inter val [1, 2] going RIGHT to LEFT in
the negative direction i s :
negative
∫ 1
2 f (x)dx = ∫ 1
[ x 2 ]
2 1
2  2x dx = [ x 2 ]
 1
2 =  (1) 2  (  (2) 2 ) =  1 + 4 = + 3 The AREA is 3 in magnitude.
The + SIGN is due to (negative f(x)) . (negative direction ) .
F (1)  F(2) =  (1)2  ( (2) 2 ) = + 3
The CHANGE in F(x) is 3 in magnitude.
The + SIGN reflects an INCREASE in F(x) over [1, 2] in the negative direction.
negative a b b a ∫ f (x)dx =  ∫
 f (x)dx RIGHT to LEFT =  LEFT to RIGHT
over [a,b]
over [a,b] 152 In all the previous examples f(x) did not CHANGE SIGN over the interval of integration.
Now let us see two examples where f(x) CHANGES SIGN over the interval of integration. continuous
1,
c ontin
Example 7: Integrate contin uous f (x) = 2x over the inter val [ 2]
going LEFT to RIGHT in the positive direction . See diagram page 148.
positive
The AREA under f(x) = 2x over the inter val [ 2] going LEFT to RIGHT
1,
in the positive direction i s :
positive
2 0 2 1
 1
 0 ∫ 2x dx = ∫ 2x dx + ∫ 2 x dx
We must brea...
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 Fall '09
 TAMERDOğAN
 Limit, Δx

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