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**Higher Maths 2 2 Integration**

1 UNIT OUTCOME SLIDE

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**UNIT OUTCOME SLIDE NOTE**

Higher Maths Integration 2 UNIT OUTCOME SLIDE NOTE average speed Speed Time Graphs D ÷ ÷ Calculate the distance travelled in each journey. S × T 40 40 40 Speed (mph) 20 Speed (mph) 20 Speed (mph) 20 2 4 2 4 6 2 4 6 Time (hours) Time (hours) Time (hours) D = 4 × 30 = 120 miles D = 5 × 20 = 100 miles 30 miles 90 miles + 15 miles ! In Speed Time graphs, the distance travelled is the same as the area under the graph. 135 miles

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**UNIT OUTCOME SLIDE NOTE ! REMEMBER y x f (x)**

Higher Maths Integration 3 UNIT OUTCOME SLIDE NOTE Reverse Differentiation ! REMEMBER D ‘rate of change of distance with respect to time’ speed = T If we know how the speed changes, and want to find distance, we need to ‘undo’ finding the rate of change with respect to time. In other words we need to reverse differentiate. y f (x) Differentiating backwards is used to find the area under a function. x

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**S S UNIT OUTCOME SLIDE NOTE y ( ) x y x S f (x) x f (x) x f (x) x**

Higher Maths Integration 4 UNIT OUTCOME SLIDE NOTE S means ‘the sum of...’ Estimating Area Under Curves To estimate area under a function, split the area into vertical strips. The area of each strip is the height, multiplied by : y f (x) x f (x) × x S ( Total Area f (x) × x ) = x x As the strips get narrower, the estimate becomes more accurate. y f (x) Area under the function f (x) S f (x) x as x x = x

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**∫ ∫ S UNIT OUTCOME SLIDE NOTE y x f (x) f (x) d x d x f (x) x x x = =**

Higher Maths Integration 5 UNIT OUTCOME SLIDE NOTE Integration ‘Integrate’ means ‘join together all the pieces’ The algebraic method for finding area under a function is called Integration. Integration uses reverse differentiation to ‘undo’ finding the rate of change. y x Area under the function f (x) x ∫ = f (x) d x ∫ d x S f (x) x as x Expression or function to be integrated. =

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**UNIT OUTCOME SLIDE NOTE**

Higher Maths Integration 6 UNIT OUTCOME SLIDE NOTE Differentiating Backwards Integration involves differentiating in reverse. f (x) multiply by the power reduce the power by one f ′(x) f (x) divide by the power increase the power by one f ′(x) How to differentiate backwards: • increase the power of every x-term by one • divide every x-term by the new power

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**∫ ∫ UNIT OUTCOME SLIDE NOTE ! y y dx d x dy 8 x 3 + x 2 – 6 x**

Higher Maths Integration 7 UNIT OUTCOME SLIDE NOTE Finding the Anti-Derivative ! How to Reverse Differentiate The result of a differentiation is called a derivative. • increase each power by one • divide by each new power The result of differentiating backwards is called the anti-derivative. ∫ d x Example expression or function Find the anti-derivative for y ∫ dy 8 x 3 + x 2 – 6 x y 8 x x 2 – 6 x dx = = dx 2 x x 3 – 3 x2 1 = 3

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**UNIT OUTCOME SLIDE NOTE**

Higher Maths Integration 8 UNIT OUTCOME SLIDE NOTE The Constant of Integration When differentiating, part of an expression is often lost. Example f (x) = x 4 All three functions have derivative f (x) = x 4 + 2 f ′(x) = 4 x 3 f (x) = x 4 – 7 f ′(x) The anti-derivative of is When differentiating in reverse, it is essential to remember to add back on the unknown number. f (x) = x 4 + c unknown constant This is called the constant of integration.

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**constant of integration**

Higher Maths Integration 9 UNIT OUTCOME SLIDE NOTE Basic Integration ! How to Integrate LEARN THIS The result of integration is called an integral. • increase each power by one • divide by each new power Example 1 • add the constant of integration ∫ f (x) = x 6 f (x) d x Example 2 ∫ ∫ ∫ 4 1 x 6 = d x Find d x 4 x = 2 d x x 1 = x c 1 8 x + c = 2 7 constant of integration 8 x + c =

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**∫ ∫ ∫ UNIT OUTCOME SLIDE NOTE y x – f (x) x2 f (x) d x x1 x2 x1**

Higher Maths Integration 10 UNIT OUTCOME SLIDE NOTE Integration and Area Integration can be used to find the area ‘under’ a function between two different values of x. Area under f (x) between x1 and x2 y f (x) x2 ∫ = f (x) d x x This is called a Definite Integral x1 x2 x1 ‘Lower Limit’ ‘Upper Limit’ ∫ ∫ f ( x2) d x – = f ( x1) d x

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**∫ ∫ ∫ UNIT OUTCOME SLIDE NOTE y y y x x x g (x) h (x) f (x) f (x) d x**

Higher Maths Integration 11 UNIT OUTCOME SLIDE NOTE Area Between a Function and the x-axis The area ‘under’ a function can be described more mathematically as the area between the function and the x-axis. g (x) h (x) y y y f (x) -7 x x -6 3 8 x -2 5 5 3 8 ∫ ∫ ∫ f (x) d x g(x) d x h(x) d x -2 -6 -7

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**Write integral inside square brackets**

Higher Maths Integration 12 UNIT OUTCOME SLIDE NOTE Definite Integral Evaluating Definite Integrals Write integral inside square brackets x2 Example ∫ 3 d x ∫ [ ] 3 2 x 3 dx 1 x c x1 = 2 ( ) ( ) 1 × ( 3 ) c 1 × ( 0 ) c = – 2 2 1 = 40 units 2 2 ! Definite Integrals do not require the constant of integration. NOTICE The constants of integration cancel each other out.

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**[ ] ∫ ( ) ( ) ( ) ( ) UNIT OUTCOME SLIDE NOTE – – y = 9 x 2 – 2 x d x**

Higher Maths Integration 13 UNIT OUTCOME SLIDE NOTE Evaluating Definite Integrals (continued) y = 9 x 2 – 2 x Example 2 Find the area below the curve between x = 1 and x = 4. 4 Write integral inside square brackets (no constant required) [ ] 4 ∫ 9 x 2 – 2 x d x 3 x 3 – x 2 = 1 1 ( ) ( ) – 3 × 43 – 42 3 × 13 – 12 = ...then evaluate for each limit and subtract. ( ) ( ) – = 192 – 16 3 – 1 = 174 units 2 Remember units!

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**∫ ∫ UNIT OUTCOME SLIDE NOTE y x dx f (x) a x-axis : x- f (x) c d a dx**

Higher Maths Integration 14 UNIT OUTCOME SLIDE NOTE Areas Above and Below the x-axis When calculating areas by integration, areas above the x-axis are positive and areas below the x-axis are negative. ∫ dx f (x) a b > 0 How to calculate area between a curve and the • draw a sketch • calculate the areas above and below the axis separately • add the positive value of each area (ignore negative signs) x-axis : x- f (x) y c d x a b ∫ dx f (x) c d < 0

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**∫ ∫ ∫ ( ) UNIT OUTCOME SLIDE NOTE y x ! CAREFUL! – g (x) f (x) f (x)**

Higher Maths Integration 15 UNIT OUTCOME SLIDE NOTE Area Between Functions Integration can also be used to find the area between two graphs, by subtracting integrals. Area enclosed by f (x) and g (x) g (x) between intersection points a and b y f (x) b b ∫ ∫ – = f (x) d x g (x) d x x a a a b b ! CAREFUL! ∫ ( ) f (x) g (x) f (x) is above g (x) = – d x a

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