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Higher Maths 2 2 Integration1. Speed Time Graphs D ST × ÷÷ 20 40 0 024 Time (hours) Speed (mph) Calculate the distance travelled in each journey. 20 40.

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Presentation on theme: "Higher Maths 2 2 Integration1. Speed Time Graphs D ST × ÷÷ 20 40 0 024 Time (hours) Speed (mph) Calculate the distance travelled in each journey. 20 40."— Presentation transcript:

1 Higher Maths 2 2 Integration1

2 Speed Time Graphs D ST × ÷÷ Time (hours) Speed (mph) Calculate the distance travelled in each journey Time (hours) Speed (mph) Time (hours) Speed (mph) D=4 × 30 = 120 miles average speed D=5 × 20 = 100 miles 30 miles 90 miles + 15 miles 135 miles In Speed Time graphs, the distance travelled is the same as the area under the graph. Higher Maths 2 2 Integration2

3 Reverse Differentiation D T speed = rate of change of distance with respect to time 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. Differentiating backwards is used to find the area under a function. f (x)f (x) Higher Maths 2 2 Integration3

4 Estimating Area Under Curves To estimate area under a function, split the area into vertical strips. f (x)f (x) x f (x)f (x) x The area of each strip is the height, multiplied by : x Total Area f (x)f (x) x × f (x)f (x) x × () = As the strips get narrower, the estimate becomes more accurate. Area under the function = f (x)f (x) f (x)f (x) x as 0 means the sum of... Higher Maths 2 2 Integration4 x

5 5 Integration The algebraic method for finding area under a function is called Integration. Integrate means join together all the pieces Integration uses reverse differentiation to undo finding the rate of change. Area under the function = f (x)f (x) f (x)f (x) x d xd x x as 0 f (x)f (x) = x d xd x Expression or function to be integrated.

6 Higher Maths 2 2 Integration6 Differentiating Backwards Integration involves differentiating in reverse. multiply by the power reduce the power by one f (x)f (x) f ( x ) divide by the power increase the power by one f (x)f (x) f ( x ) divide every x -term by the new power How to differentiate backwards: increase the power of every x -term by one

7 Higher Maths 2 2 Integration7 Finding the Anti-Derivative The result of a differentiation is called a derivative. divide by each new power How to Reverse Differentiate increase each power by one The result of differentiating backwards is called the anti-derivative. Example dy dx = 8 x 3 + x 2 – 6 x Find the anti-derivative for y y = dx 8 x 3 + x 2 – 6 x = 2 x 4 + x 3 – 3 x d xd x expression or function

8 Higher Maths 2 2 Integration8 The Constant of Integration When differentiating, part of an expression is often lost. Example f (x)f (x) = x f (x)f (x) = x 4= x 4 f (x)f (x) = x 4 – 7 f ( x ) = 4 x 3 All three functions have derivative The anti-derivative of f ( x ) is f (x)f (x) = x 4 + c unknown constant When differentiating in reverse, it is essential to remember to add back on the unknown number. This is called the constant of integration.

9 Higher Maths 2 2 Integration9 Basic Integration The result of integration is called an integral. f (x)f (x) Example 1 d xd x f (x)f (x) = x 6 divide by each new power How to Integrate increase each power by one = 1 7 = x 7 + c d xd x x 6x 6 add the constant of integration constant of integration Example 2 d xd x 4 x Find = d xd x 4 x 1 2 = 8 x + c = 8 x c+ c

10 Higher Maths 2 2 Integration10 Integration and Area Integration can be used to find the area under a function between two different values of x. f (x)f (x) x1x1 x2x2 d xd x f (x)f (x) x1x1 x2x2 = d xd x f ( x2)f ( x2) – d xd x f ( x1)f ( x1) Area under f ( x ) between x 1 and x 2 = Upper Limit Lower Limit This is called a Definite Integral

11 The area under a function can be described more mathematically as the area between the function and the x -axis. Area Between a Function and the x - axis Higher Maths 2 2 Integration11 f (x)f (x) d xd x f (x)f (x) g (x)g (x) d xd x g(x)g(x) h (x)h (x) d xd x h(x)h(x)

12 Higher Maths 2 2 Integration12 Evaluating Definite Integrals = dx 2 x 32 x [ x 4 + c 1 2 ] 0 3 = ( × ( 3 ) 4 + c 1 2 )( × ( 0 ) 4 + c 1 2 ) – = The constants of integration cancel each other out. Definite Integrals do not require the constant of integration. d xd x x1x1 x2x2 Definite Integral Write integral inside square brackets units 2 Example

13 [ ] Higher Maths 2 2 Integration13 Evaluating Definite Integrals (continued) Example d xd x x 2 – 2 x 3 x 3 – x 2 = ( ) 3 × 4 3 – 4 2 = – ( ) 3 × 1 3 – 1 2 ( ) 192 – 16 = – ( ) 3 – 1 = 174units 2 Find the area below the curve between x = 1 and x = 4. y = 9 x 2 – 2 x Write integral inside square brackets... (no constant required)...then evaluate for each limit and subtract. Remember units!

14 When calculating areas by integration, areas above the x - axis are positive and areas below the x - axis are negative. Areas Above and Below the x - axis Higher Maths 2 2 Integration14 b a cd dx f (x)f (x) a b > 0 dx f (x)f (x) c d < 0 f (x)f (x) 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-x-

15 Area Between Functions Integration can also be used to find the area between two graphs, by subtracting integrals. Higher Maths 2 2 Integration15 f (x)f (x) g (x)g (x) Area enclosed by f ( x ) and g ( x ) d xd x f (x)f (x) = = – d xd x g (x)g (x) f (x)f (x) d xd x g (x)g (x) ( – ) b a b a b a b a between intersection points a and b is above f (x)f (x) g (x)g (x)


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