alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

In general let fx be a continuous function over the

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Unformatted text preview: ” integrating f(x) with a < b left to right left right + direction b ∫a right to left right left − direction curve f(x) is + above the x axis + − ∫b − curve f(x) is − below the x axis a + This is the usual law of signs under multiplication. So the SIGN of the area under area curv the cur ve , which is also the SIGN of the CHANGE, depends on both SIGN of the function and the direction of integration . direction 140 Integration has five parts : Integ 1. The integration operation to find the integral F(x) integration integral F(x) = the expression of CHANGE = ANTIDERIVATIVE {f(x)} . 2. The inter val, say [a, b] with a < b, over which the operand or integrand f(x) is interv integ inter inte to be integrated. direction o integ 3. The direction off integration over the interval [a, b] with a < b, where the dir integration is being done : b ∫a = positive direction = a to b with a < b positive a ∫b = negative direction = b to a with a < b negative integ 4. Evaluating the integral F(x) over the interval [a, b] in the given direction to get inte something definite = CHANGE . definite b b ∫ f(x)dx = [ F(x)] f(x)dx a a a ∫ f(x)dx = f(x)dx b = F(b) − F(a) F(b) F(a) F( a [ F(x)] = F(a) − F(b) F(a) F(b) F( b 5. Finding the CONSTANT OF INTEGRATION. To get a clear picture and firm grip of the relationship between : direction integ area curv 1. direction of integr ation and SIGN of the area under the cur ve , dir ar area curv 2. SIGN of the area under the cur ve and the SIGN of the CHANGE , ar 3. the correct physical interpretation , let us look at all 4 cases in the table using our main example. 141 1. (positive f(x)) . (positive direction) = positive CHANGE (positive y(t1) h1 4321 4321 4321 4321 4321 4321 4321 t1 (0,0) ∫ t1 = INCREASE in height from t1 to t2 h2 t T vertical speed y ’(t) = u . s i n θ -- g t t2 → t2 + area under y’(t) over [t1, t2] y(t2) t2 y ’(t)dt = [ y(t)] = y (t2) -- y(t1) = INCREASE in height t1 2. (positive f(x)) . (negative direction) = negative CHANGE (negative y(t1) h1 4321 4321 4321 4321 4321 4321 4321 t1 (0,0) t2 → t1 ∫ t2 area − area under y’(t) over [t1, t2] y(t2) = DECREASE in height from t2 to t1 h2 t T vertical speed y ’(t) = u . s i n θ -- g t t1 y ’(t)dt = [ y(t)] = y (t1) -- y(t2) = DECREASE in height t2 142 3. (negative f(x)) . (positive direction) = negat...
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This note was uploaded on 11/29/2012 for the course PHYSICS 105 taught by Professor Tamerdoğan during the Fall '09 term at Middle East Technical University.

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