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Unformatted text preview: Higher Unit 2
Higher Outcome 2 What is Integration www.mathsrevision.com The Process of Integration ( Type 1 ) Area under a curve ( Type 2 ) Area under a curve above and below xaxis ( Type 3) Area between to curves ( Type 4 ) Working backwards to find function ( Type 5 )
www.mathsrevision.com You have 1 minute to come up with the rule.
Higher Integration
Outcome 2 Integration can be thought of as the opposite of differentiation www.mathsrevision.com (just as subtraction is the opposite of addition). 3x dx =
4 we get 3x5 +c 5 x n +1 x n dx = +c (n + 1) 3x dx =
4 3 x 4+1 +c = (4 + 1) 3x5 +c 5 Integration
Higher Outcome 2 Differentiation www.mathsrevision.com x n multiply by power divide by new power decrease power by 1 increase power by 1 Integration x n +1 x n dx = +C n +1
Where does this + C come from? Integration
Higher Outcome 2 Integrating is the opposite of differentiating, so: differentiate integrate
2 www.mathsrevision.com f ( x)
But: f ( x)
= f ( x) = g ( x) = h( x) f ( x) = 3x + 1 f ( x) = 3 x 2 + 4 f ( x) = 3 x 2  10 differentiate 6x
integrate Integrating 6x..........which function do we get back to? Integration
Higher Outcome 2 Solution: When you integrate a function remember to add the www.mathsrevision.com Constant of Integration...............+ C Integration
Higher Notation Outcome 2 means "integrate 6x with respect to x" means "integrate f(x) with respect to x" 6x dx
www.mathsrevision.com f ( x) dx This notation was "invented" by Gottfried Wilhelm von Leibniz Integration
Higher Outcome 2 Examples: www.mathsrevision.com x dx 7 3x  2 x + 1 dx 2 x +C 8 8 x x 3 2 + x 3 2
x  x + x+C
3 2 3 2 Integration
Higher Just like differentiation, we must arrange the function as a series of powers of x before Outcome 2 we integrate. www.mathsrevision.com 1 4 + x+ dx 5 3 2x x
1 3  x 5 ( + x 2 + 4 x 2 )dx 2 x 4 x x + 3 +4 1 +C 8 2 2
1 2 x 8 4+  1 +C 8x 3 x2
3 2 3 2  1 2 1 2 x3 8 4+  +C 8x 3 x 4x dx
4x +c 4 x4 + c
4 3 3 x dx
3x dx
3x
3 2 3 2 1 2 + c 2x + c
i g ra t In te on = A re a rv e u e r c und 3 2 2 dx 2 3x 2 x 2 dx 3 2 x 1 +c 3(1) 2  +c 3x
In te g ra tio n te c h n iq u e s 1 2 3 x2
 2 3 dx x dx 2 3x +c 2
ti g ra In te on
1 3 x 2 + 3 dx 3 x x + 2 x 3 dx 3 x 2 2 x 2 + +c 6 2 2 x 1  2 +c 6 x
=
de a u n Are e c u rv r 1 + 5 4 x dx x x
 1 2 ( x  2)(3x + 1) dx 2x  3 dx x
1 2  1 2
3 2 1 2 x 1 2 + 5 x dx 1 4 3 x 2  5 x  2 dx
3x3 5 x 2   2x + c 3 2 5x2 x3   2x + c 2 1 2 1 2 + 5x
5 4 5 4 2 x  3 x dx
2x
3 2 +c  3x
1 2 +c 2x + 4x + c
Name : 5 4 4x 3 3 2 6 x +c 1 2 Real Application of Integration
Find area between the function and the xaxis between x = 0 and x = 5 A = bh = x5x5 = 12.5 x A = x dx = 2 0 0 2 5 5 2 2 25 5 0 =  =  0 = 12.5 2 2 2 Real Application of Integration
Find area between the function and the xaxis between x = 0 and x = 4 A = bh = x4x4 = 8 A = lb = 4 x 4 = 16 AT = 8 + 16 = 24 x A = x + 4 dx = + 4 x 2 0 0
2 4 4 2 4 = + 4 4 [ 0] = 8 + 16 = 24  2 Real Application of Integration
Find area between the function and the xaxis between x = 0 and x = 2
2 A = x dx
2 0 x = 3 0 3 2 3 3 8 2 0 8 = +0=  = 3 3 3 3 Real Application of Integration
Find area between the function and the xaxis between x = 3 and x = 3
3 x A = x dx = 2 3 3 2 3 2  3) 2 9 9 3 ( =  =  =0 2 2 2 2 ? Houston we have a problem ! Areas under the xaxis ALWAYS give negative values By convention we simply take the Real Application of Integration positive value since we cannot get a negative area. We need to do separate integrations for above and below the xaxis. x A = x dx = 2 0 0 2 3 3 2 3 9 9 = [ 0] =  0 =  2 2 2 x  3) 2 9 ( A = x dx = = [ 0]  =  2 3 2 2 3 2 0 0 9 9 AT = + = 9 2 2 Real Application of Integration
Integrate the function g(x) = x(x 4) between x = 0 to x = 5 We need to sketch the function and find the roots before we can integrate Real Application of Integration
We need to do separate integrations for above and below the xaxis.
Since under xaxis take positive value x 2 A = x( x  4) dx =  2 x 0 3 0
3 4 4 3 4 64 32 2 =  2 4 [ 0] =  32 =   3 3 3 32 3 5 3 3 5 4 25 32 7 2 2 A = x( x  4) dx =  2 5  2 4  +  = = 3 3 3 3 3 4 Real Application of Integration 32 7 AT = x( x  4) dx + x( x  4) dx = + = 13 3 3 0 4 4 5 Area between Two Functions
Find upper and lower limits. y=x x2 = x 2 x x=0 y=x
2 x( x  1) = 0 x=0 x =1 then integrate top curve bottom curve. x 2 13 1 1 x 2  A = ( x  x ) dx =  =  [ 0] = 2 6 2 3 3 0 0
2 1 3 1 1 6 Area between Two Functions
Find upper and lower limits. x2  1 =  x2 + 1 2x2  2 = 0 2( x 2  1) = 0 2( x  1)( x + 1) = 0 x = 1
then integrate top curve bottom curve.
Take out common factor
1 1 1 x =1 A = ( x 2 + 1)  ( x 2 1) dx = (2 x 2 + 2) dx = 2 ( x 2 + 1) dx
1 1 1 Area between Two Functions
1 = 2 ( x 2 + 1) dx
1 x  = 2 + x 3 1 
3 1  (1)3 ( 1)3  = 2 + 1  + ( 1) 3 3 1 2 8 1 = 2 + 1  1) = 2 + 2 =   3 3 3 3 Integration
Higher Outcome 2 If f '( x) = 3 x 2 + 4 x  1 and f (2) = 11, find f ( x). www.mathsrevision.com To get the function f(x) from the derivative f'(x) we do the opposite, i.e. we integrate. f ( x) = (3 x 2 + 4 x  1)dx
3 2 f (2) = 11 23 + 2.22  2 + C = 11 8 + 8  2 + C = 11 x x =3 +4  x+C 3 2 C = 3
Hence: = x + 2x  x + C
3 2 f ( x) = x + 2 x  x  3
3 2 Integration
Higher Outcome 2 Example : www.mathsrevision.com 5 1 x 2 2 5 52 )  ( 12 ) = 24 2x dx = = = ( x 2 1 1
2 5 x 4 4 2 4x dx = = x 1 4 2 2
2 3
4 2 24 )  ( (  2) 4 ) = =( 6 Calculus
Integrate x  4 x + 3 dx
2 Revision Integrate term by term x 4x  + 3x + c 3 2 3 2 simplif y 1 3 2 x  2 x + 3x + c 3 Back Quit Next Calculus Revision
Integrate 3 x + 4 x dx
3 Integrate term by term 3x 4x + +c 4 2
3 4 4 2 x + 2x + c
4 2 Back Quit Next Calculus
Evaluate 3 2 4 Revision
x dx
4 4 1 Straight line form
2 ( x)3 3 1
4 x
1 1 2 dx 2 x 3 1 2 3 ( ) 3 2 4  3 ( ) 3 1 16 2  3 3 14 3 4
Back Quit Next 2 3 Calculus Revision
2
Evaluate 1
1 2 dx 2 x 2 Straight line form x 2 dx
1 x  1 ( 2 )  ( 1 )
1 1 1  + 1 2
Back 1 2
Next Quit Calculus
Integrate 2 x  3 dx x
Straight line form Revision x  2 x dx 1 2 3 2 3 2x x  +c 2 3 2 2 2 3 x + x 2 + c 3 2 Back Quit Next Calculus Revision
Integrate 1 x + dx x
3 Straight line form
1 2 x +x
3  1 2 dx
1 2 x x + 1 +c 4 2
Back Quit 4 1 4 x4 + 2x + c Next Calculus Integrate x  5x dx x
3 Revision Straight line form x x 3
1 2  5x x
1 2 dx
2 7
7 2 x 5 2  5 x dx
3 2 1 2 x 10  x 3 +c Back Quit Next Calculus Revision
Integrate 4x  x dx 2 x 3 2 Split into separate fractions 4x 2x
1 2  x 3 2 1 2 2x dx 2x  x dx 1 2 1 2 4 3 x  x2 + c 3 2 1 4 Back Quit Next Calculus
Integrate x 5 dx x x
2 Revision Straight line form x2  5 x
 3 2
3 2 dx x2 x x
3 2
3 2  5 x

3 2 dx x  5x 1 2 dx 3 2  5x 1 2 1  2 +c 2 3 x + 10 x 3 2  1 2 +c Back Quit Next Calculus
p Revision Find p, given
1
p x dx = 42 2 x = 42 3 1
3 2 p x
1 1 2 dx = 42
2 3 p3 2 3 p  3 2 2 (1) 3 3 2 = 42  2 3 = 42 2 p3  2 = 126 p= (2
1 12 3 p3 = 64 p 3 = 642 = 212 ) = 16 Back Quit Next Calculus
Integrate (3x  1)( x + 5) dx
3 x + 14 x  5 dx
2 Revision Multiply out brackets Integrate term by term simplify 3x 3 14 x 2 +  5x + c 3 2 x + 7 x  5x + c
3 2 Back Quit Next Calculus Revision
Integrate (5 + 3 x) 2 dx (5 + 3x) +c 3 3
3 Standard Integral (from Chain Rule) 1 3 (5 + 3 x) 9 +c Back Quit Next Calculus
Integrate (x 2  2) ( x + 2)
2 Revision
dx, x 0 x4  4 dx 2 x x4 4  2 dx 2 x x x 2  4 x 2 dx x 2 Multiply out brackets Split into separate fractions x 3 4 x 1  +c 3 1 1 3 4 x + +c x
3 Back Quit Next Calculus
Evaluate 2 1 2 1 x + dx x 2 Revision 2 1 2 ( x 2 + x 1 ) 2 dx Cannot use standard integral So multiply out
1 x5 + x 2  x 1 5 1
2 x 4 + 2 x + x 2 dx
1 (
64 10 1 5 25 + 2 2  1 2 ) (
 1 5 1 5 + 12  1 1 )
8
1 5 ( 32 5 + 4 1 2 ) ()

1 5 + 40 10  20 10  2 10 82 10 Back Quit Next Calculus Revision The graph of y = g ( x ) passes through the point (1, 2). If dy 1 1 3 =x + 2 dx x 4 express y in terms of x. dy 1 3 2 =x +x  dx 4 x 4 x 1 1 y= +  x+c 4 1 4
Use the point simplify x 1 1 y =   x+c 4 x 4
Evaluate c Back 4 ( 1) 2= 1 1   ( 1) + c 4 ( 1) 4
1 4
4 4 c=3 1 1 y = x   x+3 x 4
Next Quit Calculus Revision dy = 6 x 2  2 x passes through the point (1, 2). A curve for which dx Express y in terms of x. 6 x3 2 x 2 y=  +c 3 2
Use the point y = 2 x3  x 2 + c 2 = 2(1)3  (1) 2 + c c=5 y = 2 x3  x 2 + 5 Back Quit Next Integration
Higher Outcome 2 www.mathsrevision.com Further examples of integration Exam Standard Area under a Curve
Higher Outcome 2 The integral of a function can be used to determine the area between the x axis and the graph of the function. www.mathsrevision.com y y = f ( x) Area =
x b a f ( x) dx a b If f ( x) dx = F ( x) then b a f ( x) dx = F (b)  F (a) NB: this is a definite integral. It has lower limit a and an upper limit b. Area under a Curve
Higher Outcome 2 Examples: www.mathsrevision.com 5 1 x 25 1 3 (3 + x) dx = x + = 15 + 2  3 + 2 = 2 1
2 5 24 2 0 (3 x  1) dx x
2  0
3 = ( 23  2 )  ( 0 3 + 0 ) x 2 =6 Area under a Curve
Higher Outcome 2
From the definition of f ( x) dx it follows that:
a b www.mathsrevision.com b a f ( x) dx =  x) dx f( b a F (b)  F (a ) =  ( F (a )  F (b) )
Conventionally, the lower limit of a definite integral is always less then its upper limit. Area under a Curve
Higher Outcome 2 y=f(x) Very Important Note: When calculating integrals:
areas above the xaxis are positive areas below the xaxis are negative www.mathsrevision.com a b c
d d a b f ( x) dx > 0 c f ( x) dx < 0 When calculating the area between a curve and the xaxis: make a sketch calculate areas above and below the xaxis separately ignore the negative signs and add Area under a Curve
Higher Outcome 2 The Area Between Two Curves www.mathsrevision.com To find the area between two curves we evaluate:
y = f ( x) Area = (top curve  bottom curve)
b a y = g ( x) b Area = ( f ( x )  g ( x) dx
a Area under a Curve
Higher Example: Outcome 2 Calculate the area enclosed by the lines x = 2, x = 4 and the curves y = x 2 and y = 4  x 2 www.mathsrevision.com y y=x 2 Area = [ x 2  (4  x 2 )] dx = (2 x 2  4) dx
2 2 4 4 x=4 x=2 y = 4  x2 x 4 x 2 (2 x  4) dx =  4 x 3 2 2
2 3 4 128 16 =(  16)  (  8) 3 3 112 = 8 3 88 = 3 Area under a Curve
Higher Complicated Example: Outcome 2 The cargo space of a small bulk carrier is 60m long. The shaded part of the diagram represents the uniform crosssection of this space. www.mathsrevision.com x2 It is shaped like a parabola with y = ,  6 4 between lines y = 1 and y = 9. x 6, Find the area of this crosssection and hence find the volume of cargo that this ship can carry. 9 1 y y= Higher9 x 4
2 Area under a Curve
The shape is symmetrical about the yaxis. So we calculate the area of one of the light shaded rectangles and one of the dark shaded wings. The area is then double their sum. y= www.mathsrevision.com y = 1 t s s t
x The rectangle: let its width be s The wing: extends from x = s to x = t The area of a wing (W ) is given by: s2 y =1= 4 t2 y=9= 4
t s=2 t =6 x2 Aw = (9  ) dx 4 s Area under a Curve
Higher Outcome 2 www.mathsrevision.com x x Aw = (9  ) dx = x  ) (9 12 4 2 2
3 6 2 6 63 23 2 (9 6  (9 2  12 )  12 ) = = 18 3
The area of a rectangle is given by: R = (9  1) s = 16
A = 2( R + W ) The area of the complete shaded area is given by:
2 208 A = 2(16 + 18 ) = 3 3 The cargo volume is: 208 60 A = 60. = 20.208 = 4160m3 3 Exam Type Questions
Higher Outcome 2 www.mathsrevision.com At this stage in the course we can only do Polynomial integration questions. In Unit 3 we will tackle trigonometry integration ...
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This note was uploaded on 02/13/2012 for the course MAT 205 math 205 taught by Professor Google during the Spring '10 term at University of Phoenix.
 Spring '10

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