{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}


In general let fx be a continuous function over the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online