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AREAS BETWEEN CURVES

a b. a b. a b. Y. Y. Y. X. X. X. AREAS BETWEEN CURVES. y = f(x). y = f(x). C. A. y = g(x). y = g(x). B. b. b. Area C. Area A = . Area B = . f(x) dx. g(x) dx. a. a. = AreaA - AreaB. b. b. = . f(x) dx - . g(x) dx. a. a. b. = . [ f(x) – g(x) ] dx. a.

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AREAS BETWEEN CURVES

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  1. a b a b a b Y Y Y X X X AREAS BETWEEN CURVES y = f(x) y = f(x) C A y = g(x) y = g(x) B b b Area C Area A = Area B = f(x) dx g(x) dx a a = AreaA - AreaB b b = f(x) dx -  g(x) dx a a b = [ f(x) – g(x) ] dx a

  2. a b Y X NB: In the interval a < x < b if f(x) > g(x) then the area enclosed between the curves from x = a to x = b is given by b  [ f(x) – g(x) ] dx a ie MUST BE LEARNED !!

  3. Ex 19 Find the finite between area between the curves. y = x2 – 6x Upper curve is y = 12x - 2x2 Limits are 0 and 6. y = 12x - 2x2 Brackets round 2nd formula 6 Area = 12x - 2x2 - (x2 – 6x) dx 0 6 = 12x - 2x2 - x2 + 6x dx 0 6 = 18x - 3x2 dx 0 = [ ] 6 9x2 – x3 0 = (324 – 216) - 0 = 108units2

  4. Ex20 Y y = x4 OK! Who hid my banana? y = 8x X Find shaded area! 2 Limits: Area =  8x - x4 dx 0 < x < 2 0 x4 = 8x take x = 1 = [ ] 2 4x2 – 1/5x5 x4 - 8x = 0 0 8x = 8 x(x3 – 8) = 0 = (16 – 32/5) - 0 x4 = 1 x = 0 or x3 = 8 = 93/5 units2 So 8x > x4 x= 2

  5. Ex21 Find the finite area enclosed by the parabolic curves y = 2x2 + x – 9 and y = x2 + 2x – 3. ********* Roughly Limits 2x2 + x – 9 = x2 + 2x – 3 x2 - x – 6 = 0 (x - 3)(x + 2) = 0 x = -2 or x = 3 For -2 < x < 3 taking x = 0 2x2 + x – 9 = -9 & x2 + 2x – 3 = -3 so y = x2 + 2x – 3 is the upper curve

  6. ctd 3 Area =  x2 + 2x – 3 – (2x2 + x – 9 ) dx -2 3 = x2 + 2x – 3 – 2x2 - x + 9 dx -2 3 =  –x2 + x + 6 dx -2 = [ ] 3 –1/3x3 + 1/2x2 + 6x -2 = (-9 + 41/2 + 18) – (8/3 + 2 - 12) = 205/6 units2

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