alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

We continuous contin saw the plot of f0x in the 2

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Unformatted text preview: k this up into 2 parts because f(x) CHANGES SIGN at x = 0 . 0 0 ∫ 2x dx = [x 2 ]-- = ( 0) 2 -- (--1) 2 = 0 -- 1 = --1 -- ---1 --1 -- The AREA is 1 in magnitude. The -- SIGN is due to (negative f(x)) . ( positive direction ) . positive 2 2 ∫ 2 x dx = [x 2 ] 0 = ( 2) 2 -- (0) 2 = 4 -- 0 = +4 --0 The AREA is 4 in magnitude. The + SIGN is due to (positive f(x)) . (positive direction ) . So: ∫ 2 --1 -- 2 x dx = -- + 4 = + 3 --1 F(2) -- F(--1) = ∫ - 2 --1 -- 2 2 x dx = [x 2 ] ---1 = (2) 2 -- (--1) 2 = 4 -- 1 = +3 -- -The CHANGE in F(x) is 3 in magnitude. The + SIGN reflects an INCREASE in F(x) over [-- 1, 2] in the positive direction. positive -- 153 continuous --1, Example 8: Integrate contin uous f (x) = 2x over the inter val [-- 2] c ontin -going RIGHT to LEFT in the negative direction . See diagram page 148. negative T he AREA under f(x) = 2x over the inter val [-- 2] going RIGHT to LEFT --1, -in the negative direction i s : negative --1 0 --1 2 2 0 ∫ 2x dx = ∫ 2x dx + ∫ 2x dx We must break this up into 2 parts because f(x) CHANGES SIGN at x = 0 . 0 0 ∫ 2x dx = [x 2 ]-- = ( 0) 2 -- (2) 2 = 0 -- 4 = -- 4 ----1 2 The AREA is 4 in magnitude. The -- SIGN is due to (positive f(x)) . (negative direction ) . --1 2 ∫ 2x dx = [x 2 ] 0 = (-- 2 -- (0) 2 = 1 -- 0 = +1 --1) --0 The AREA is 1 in magnitude. The + SIGN is due to (negative f(x)) . (negative direction ) . So: --1 ∫ 2x dx = -- 4 + 1 = -- 3 2 --1 2 F(--1) -- F(2) = ∫ 2x dx = [x 2 ] ---1 2 = (--1)2 -- (2) 2 = 1 -- 4 = --3 --The CHANGE in F(x) is 3 in magnitude. The -- SIGN reflects a DECREASE in F(x) over [--1, 2] in the negative direction. negative 154 A note on AREA and INTEGRATION INTEGRATION Several attempts have been made to equate AREA under f(x) over [a, b] from a mensuration point of view with the CHANGE in the INTEGRAL {F(b) − F(a)} for all possisble f(x). These attempts have not been successful because they failed to take into account the influence of the direction of integration over [a, b]. direction may continuous We may use the INTEGRAL F(x) to ffind the AREA under continuous ind f(x) over [a, b] when : 1. f(x) is positive (above the x-axis) and the direction of integration is positive. positive direction positive In this case : AREA under f(x) CHANGE in the INTEGRAL F(x) =...
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