INTEGRATION.

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Presentation transcript:

INTEGRATION

Area under Curve 3. Find the area under the curve y = x2 on [0, 1]. Find the remaining area as well. Method I: vertical elements 0 1 Method II: Horizontal elements

Partial Fractions Equate the coefficients of the two sides of the polynomial and integrate. Use cover-up method. A = -1/6, B= 3/10, C= -2/15

Applications Find the equation of family of curves whose slope at any point is equal the negative of twice the abscissa of the point. Find the curve passing through (1,3). dy/dx = -2x y = x2 + C 3 = 1 + C => C = 2 y = x2 + 2

Applications Find the equation of family of curves whose slope at any point P(x,y) is m = 3x2y and passes through (0,8). dy/dx = 3x2y dy/y = 3x2dx ln y = x3 + C = x3 + ln c y = cex^3 8 = c y = 8ex^3

Applications At every point of a certain curve, y" = x2 – 1. Find the equation of the curve if it passes through (1, 1) and is tangent to the line x + 12y = 13. y' = x3/3 – x + C = -1/12 => C = 7/12 y = x4/12 – x2/2 + 7x/12 + C 1 = 1/12 – 6/12 + 7/12 + C => C = 5/6 y = x4/12 – x2/2 + 7x/12 + 5/6

Proportion A certain quantity q increases at a rate proportional to itself. If q = 25 when t = 0 and q = 75 when t = 2, find q when t = 6. dq/dt = kq or dq/q = kdt ln q = kt + ln c q = cekt 25 = c and 75 = 25e2k => k = (ln 3)/2 = 0.5493 q(6) = 25e3.2958 = 674.975

Acceleration A car is slowing down at the rate of 0.8 ft/sec2. How far will the car travel if its speed is initially 15 mph? a = -0.8 v = -0.8t + 22 where 15mph = 22 ft/sec; 60 mph = 88 ft/sec s = -0.4t2 + 22t When v = 0, car is stopped. Thus t = 22/0.8 = 27.7 sec s(27.5) = -0.4(27.5)2 + 22(27.5) = 302.5 ft

Acceleration What constant acceleration is required to move a particle 50 ft in 5 seconds? A = -a v = -at + v0; at = v0 s = -at2/2 + v0t 50 = -25a/2 + 5v0 = -25a/2 + 25a 100 = -25a + 50a a = 4 ft/sec2

Acceleration What constant acceleration is required to slow a particle from a velocity of 45 ft/sec to a dead stop in 15 ft? a = -a v = -at + 45; when v = 0, t = 45/a S = -at2/2 + 45t 15 = -a(45/a)2/2 + 45(45/a) 15a = -452/2 + 452 a = -67.5ft/sec2

Area Find the area bounded by y = x3 – 6x2 + 8x and the x-axis. 0 2 4