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Published byXavier Cuthbertson Modified over 2 years ago

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Page 253 Example 1 1.Red line = ”over function” = 4 – 3x. 2.Blue line = ”under function” = 3x 2 – 3x + 1. 3.Work out 1 – 2 = 4 – 3x – (3x 2 – 3x + 1) = 3 – 3x 2 4.Work out primitive function for 3 – 3x 2. This is F(x) = 3x – x 3 5.Work out F(x) for x=1. This is 3.1 – 1 3 = 2. Call this value1. 6.Work out F(x) for X=-1. This is 3.-1 – (-1) 3. = -2 Call this value2. 7.Value1 – value2 = 2 – (-2) = 4. Answer = 4. Area

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Beräkna arean av det område som begränsas av kurvorna Y=X 2 -4 och Y=2X-X 2. Page 254. ”Over function” is Y=2X – X 2. ”Under function” is Y=X 2 – 4. Work out Y=2X – X 2 – (X 2 – 4). This gives Y=2X-2X 2 +4. Primitive function F(X) = x 2 – 2x 3 /3 + 4X This time we have to work out the upper and lower limits as we are not given them IN THE EXAMPLE. Therefore solve X 2 -4=2X-X 2. 2X 2 -2x-4 = 0 X 2 -x-2 = 0 (x-2)(x+1)=0 X 1 = 2 and x 2 = -1 Area

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Primitive function F(X) = x 2 – 2x 3 /3 + 4X X 1 = 2 and x 2 = -1 Work out F(X) for X=2. This gives 4 – 2.8/3 + 8 = 20/3. Work out F(X) for X=-1. This gives 1 – (2.-1 3 )/3 + 4.-1 = 1+(2/3)-4. This gives 2/3 – 3 = -7/3. Answer= F(2) – F(-1) = 9 Note: If there is no ”over function” and ”under function” please see page 258 Area

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3.7 Absolute value DAY 2. Solve for x----no notes on this slide (just watch). |x| = 5 |x + 2| = 5 x = 5 or x = -5 x + 2 = 5 or x + 2 = -5 x = 3 - 2 -2.

3.7 Absolute value DAY 2. Solve for x----no notes on this slide (just watch). |x| = 5 |x + 2| = 5 x = 5 or x = -5 x + 2 = 5 or x + 2 = -5 x = 3 - 2 -2.

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