2UNIT OUTCOME SLIDE NOTE Higher Maths Integration2UNITOUTCOMESLIDENOTEaverage speedSpeed Time GraphsD÷÷Calculate the distance travelled in each journey.S×T404040Speed (mph)20Speed (mph)20Speed (mph)2024246246Time (hours)Time (hours)Time (hours)D = 4 × 30 = 120 milesD = 5 × 20 = 100 miles30 miles90 miles+ 15 miles!In Speed Time graphs, the distance travelled is the same as the area under the graph.135 miles
3UNIT OUTCOME SLIDE NOTE ! REMEMBER y x f (x) Higher Maths Integration3UNITOUTCOMESLIDENOTEReverse Differentiation!REMEMBERD‘rate of change of distance with respect to time’speed =TIf 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.yf (x)Differentiating backwards is used to find the area under a function.x
4S S UNIT OUTCOME SLIDE NOTE y ( ) x y x S f (x) x f (x) x f (x) x Higher Maths Integration4UNITOUTCOMESLIDENOTESmeans ‘the sum of...’Estimating Area Under CurvesTo estimate area under a function, split the area into vertical strips.The area of each strip is the height, multiplied by :yf (x)xf (x)×xS(Total Areaf (x)×x)=xxAs the strips get narrower, the estimate becomes more accurate.yf (x)Area under the functionf (x)Sf (x)xasxx=x
5∫ ∫ S UNIT OUTCOME SLIDE NOTE y x f (x) f (x) d x d x f (x) x x x = = Higher Maths Integration5UNITOUTCOMESLIDENOTEIntegration‘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.yxArea under the functionf (x)x∫=f (x)d x∫d xSf (x)xasxExpression or function to be integrated.=
6UNIT OUTCOME SLIDE NOTE Higher Maths Integration6UNITOUTCOMESLIDENOTEDifferentiating BackwardsIntegration involves differentiating in reverse.f (x)multiply by the powerreduce the power by onef ′(x)f (x)divide by the powerincrease the power by onef ′(x)How to differentiate backwards:• increase the power of every x-term by one• divide every x-term by the new power
7∫ ∫ UNIT OUTCOME SLIDE NOTE ! y y dx d x dy 8 x 3 + x 2 – 6 x Higher Maths Integration7UNITOUTCOMESLIDENOTEFinding the Anti-Derivative!How to Reverse DifferentiateThe result of a differentiation is called a derivative.• increase each power by one• divide by each new powerThe result of differentiating backwards is called the anti-derivative.∫d xExampleexpression or functionFind the anti-derivative fory∫dy8 x 3 + x 2 – 6 xy8 x x 2 – 6 xdx==dx2 x x 3 – 3 x21=3
8UNIT OUTCOME SLIDE NOTE Higher Maths Integration8UNITOUTCOMESLIDENOTEThe Constant of IntegrationWhen differentiating, part of an expression is often lost.Examplef (x)= x 4All three functions have derivativef (x)= x 4 + 2f ′(x)= 4 x 3f (x)= x 4 – 7f ′(x)The anti-derivative ofisWhen differentiating in reverse, it is essential to remember to add back on the unknown number.f (x)= x 4 + cunknown constantThis is called the constant of integration.
9constant of integration Higher Maths Integration9UNITOUTCOMESLIDENOTEBasic Integration!How to IntegrateLEARN THISThe result of integration is called an integral.• increase each power by one• divide by each new powerExample 1• add the constant of integration∫f (x)= x 6f (x)d xExample 2∫∫∫41x 6=d xFindd x4 x=2d xx1= x c18 x+ c=27constant of integration8 x + c=
10∫ ∫ ∫ UNIT OUTCOME SLIDE NOTE y x – f (x) x2 f (x) d x x1 x2 x1 Higher Maths Integration10UNITOUTCOMESLIDENOTEIntegration and AreaIntegration can be used to find the area ‘under’ a function between two different values of x.Area under f (x) between x1 and x2yf (x)x2∫=f (x)d xxThis is called a Definite Integralx1x2x1‘Lower Limit’‘Upper Limit’∫∫f ( x2)d x–=f ( x1)d x
11∫ ∫ ∫ UNIT OUTCOME SLIDE NOTE y y y x x x g (x) h (x) f (x) f (x) d x Higher Maths Integration11UNITOUTCOMESLIDENOTEArea Between a Function and the x-axisThe area ‘under’ a function can be described more mathematically as the area between the function and the x-axis.g (x)h (x)yyyf (x)-7xx-638x-25538∫∫∫f (x)d xg(x)d xh(x)d x-2-6-7
12Write integral inside square brackets Higher Maths Integration12UNITOUTCOMESLIDENOTEDefinite IntegralEvaluating Definite IntegralsWrite integral inside square bracketsx2Example∫3d x∫32 x 3dx1x cx1=2()()1× ( 3 ) c1× ( 0 ) c=–221=40units 22!Definite Integrals do not require the constant of integration.NOTICEThe constants of integration cancel each other out.
13[ ] ∫ ( ) ( ) ( ) ( ) UNIT OUTCOME SLIDE NOTE – – y = 9 x 2 – 2 x d x Higher Maths Integration13UNITOUTCOMESLIDENOTEEvaluating Definite Integrals (continued)y = 9 x 2 – 2 xExample 2Find the area below the curve between x = 1 and x = 4.4Write integral inside square brackets (no constant required)[ ]4∫9 x 2 – 2 xd x3 x 3 – x 2=11( )( )–3 × 43 – 423 × 13 – 12=...then evaluate for each limit and subtract.( )( )–=192 – 163 – 1=174units 2Remember units!
14∫ ∫ UNIT OUTCOME SLIDE NOTE y x dx f (x) a x-axis : x- f (x) c d a dx Higher Maths Integration14UNITOUTCOMESLIDENOTEAreas Above and Below the x-axisWhen calculating areas by integration, areas above the x-axis are positive and areas below the x-axis are negative.∫dxf (x)ab> 0How 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)ycdxab∫dxf (x)cd< 0
15∫ ∫ ∫ ( ) UNIT OUTCOME SLIDE NOTE y x ! CAREFUL! – g (x) f (x) f (x) Higher Maths Integration15UNITOUTCOMESLIDENOTEArea Between FunctionsIntegration 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 byf (x)bb∫∫–=f (x)d xg (x)d xxaaabb!CAREFUL!∫()f (x)g (x)f (x)is aboveg (x)=–d xa