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INTEGRATION & AREA

INTEGRATION & AREA. NB: we should think of a line as a very simple curve!. Consider. y = 2x + 3. Y. 8. B. Area(I) = 8 X 4  2= 16. (I). A. Area(II) = 5 X 4 = 20. 4. Total = 36. (II). 5. X. 1 5. A is (1,5) & B is (5,13). Now consider. [ ]. 5. 5. .

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INTEGRATION & AREA

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  1. INTEGRATION & AREA NB: we should think of a line as a very simple curve! Consider y = 2x + 3 Y 8 B Area(I) = 8 X 4  2= 16 (I) A Area(II) = 5 X 4 = 20 4 Total = 36 (II) 5 X 1 5 A is (1,5) & B is (5,13) Now consider [ ] 5 5  (2x + 3) dx = x2 + 3x = (25 + 15) – (1 + 3) = 36 1 1 Comparing answers we should see that the area can also be obtained by integration so we can use this for curves as follows….

  2. Area Under a Curve Y The area under the curve y = f(x) from x = a to x = b is given by y = f(x) b  f(x) dx X a a b MUST BE LEARNED !!! Ex13 Y y = p(x) 7  p(x) dx Shaded area = -2 X -2 7

  3. Ex14 Y y = x2 – 4x + 5 X 2 5 = [ ] 5 5  1/3x3 – 2x2 + 5x Area = (x2 – 4x + 5) dx 2 2 = (125/3 – 50 + 25) – (8/3 – 8 + 10) = 12 units2

  4. Ex15 Y X y = 2x(6 – x) NB: need limits! Curve cuts X-axis when 2x(6 – x) = 0 so x = 0 or x = 6 6 6 Area = = 2x(6 – x) dx (12x – 2x2) dx 0 0 = [ ] 6 6x2 – 2/3x3 0 = (216 – 144) - 0 = 72 units2

  5. Ex16 Y y = (x – 1)(x – 6) X NB: need limits! Curve cuts X-axis when (x – 1)(x – 6) = 0 so x = 1 or x = 6 6 6 = Area = (x2 -7x + 6) dx (x – 1)(x – 6) dx 1 1 = [ ] 6 1/3x3 – 7/2x2 + 6x 1 (**) Area can’t be negative. Negative sign indicates area is below X-axis. Actual area = 205/6 units2 = (72 – 126 + 36) - (1/3 – 7/2 + 6) = -205/6 units2 (**)

  6. Ex17 Y y = x(x – 4) Need to find each section separately ! partB 0 4 6 X partA 6 4 Area B = Area A = x(x – 4) dx x(x – 4) dx 4 0 = [ ] 4 6 1/3x3 – 2x2 = (x2 – 4x) dx 4 0 = [ ] = (72 – 72) - (211/3 – 32) 4 1/3x3 – 2x2 0 = 102/3 = (211/3 – 32) - 0 = -102/3 (really 102/3) Total = 102/3 +102/3 = 211/3 units2

  7. NB: both areas are identical in size however the different signs indicate position above and below the X-axis! If we try to calculate the area in one step then the following happens 6 Area = It is obvious the total area is not zero but the equal magnitude positive and negative parts have cancelled each other out. Hence the need to do each bit separately. x(x – 4) dx 0 = [ ] 6 1/3x3 – 2x2 0 = (72 – 72) – 0 = 0

  8. Ex18 Y y = x3 – 4x2 – x + 4 X To find limits must solvex3 – 4x2 – x + 4 = 0 using polynomial method. Start with x = 1. 1 1 -4 -1 4 1 -3 -4 1 -3 -4 0 f(1) = 0 so (x – 1) a factor Other factor is x2 – 3x – 4 = (x – 4 )(x + 1)

  9. Solving (x + 1)(x – 1)(x – 4) = 0 gives x = -1 or x = 1 or x = 4 so 1 = [ ] 1 1st area =  1/4x4 - 4/3x3 – 1/2x2 + 4x (x3 – 4x2 – x + 4) dx -1 -1 = (1/4 - 4/3 – 1/2 + 4) – (1/4 + 4/3 – 1/2 - 4) = 51/3 = [ ] 4 2nd area =  1/4x4 - 4/3x3 – 1/2x2 + 4x 4 (x3 – 4x2 – x + 4) dx 1 1 = (64 – 256/3 – 8 + 16) - (1/4 - 4/3 – 1/2 + 4) = -153/4 (Really 153/4) So total area = 51/3 + 153/4 = 211/12 units2

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