june 21, ‘99
DESCRIPTION
June 21, ‘99. EVALUATING DEFINITE INTEGRALS. MATLAB can evaluate definite integrals like This is provided that the integrand f(x) be available as a function , not an array of numbers. HOW DOES MATLAB DO IT?. The primary function for evaluating definite integrals is quad8 - PowerPoint PPT PresentationTRANSCRIPT
©1999 BG Mobasseri 2 04/20/23
EVALUATING DEFINITE INTEGRALS
MATLAB can evaluate definite integrals like
This is provided that the integrand f(x) be
available as a function, not an array of numbers
f (x)dxa
b
∫
©1999 BG Mobasseri 3 04/20/23
HOW DOES MATLAB DO IT?
The primary function for evaluating definite
integrals is quad8
quad8 has the following syntax– q=quad8(‘function’,a,b)
This is equivalent to the expression
( function)dxa
b
∫
©1999 BG Mobasseri 4 04/20/23
BUILT-IN MATLAB FUNCTIONS
Evaluate the following
Since cosine is a built-in MATLAB function;
y=quad8(‘cos’,0,3*pi/2)
cos(x)0
3π 2
∫ dx
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USER-DEFINED FUNCTION(we haven’t covered function writing)
You can integrate functions that are not part of
MATLAB library.
For example, you can write a function of your
own such as gauss,
y =1
2 πe
−x2
2
-3 -2 -1 0 1 2 30
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
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Try it! area under humps
humps is a pre-defined function in MATLAB. Let’s
first plot it usingx=0:0.01:2;
plot(x,humps(x))
Now let’s find the area over a number of intervals– area=quad8(‘humps’,0,0.5)
– area=quad8(‘humps’,0,1)
– area=quad8(‘humps’,0,2)
©1999 BG Mobasseri 7 04/20/23
APPROXIMATION TO INTEGRALS - trapz
Function trapz uses areas of trapezoids to
approximate the area under the curve
-3 -2 -1 0 1 2 30
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
x=0:0.01:2;y=humps(x)area=trapz(x,y)
You might be surprisedhow large the answeris?answer next
©1999 BG Mobasseri 8 04/20/23
Sample Spacing
trapz assumes unit spacing between samples
If that is not true, the output of trapz must be
scaled by the actual spacing, e.g. 0.1
So what is the right answer in the previous slide?
1
©1999 BG Mobasseri 10 04/20/23
Try it !
Energy of a signal
Using trapz, find the energy of a gaussian pulse
(slide 5) in the range (-1,1)
E = s2 t( )T1
T2
∫ dt
©1999 BG Mobasseri 11 04/20/23
EXTENSION OF 1D INTEGRALS
1-D integral can geometrically be interpreted as
an area.
It is possible to evaluate volumes, not by
multidimensional integrals as is generally done ,
but as 1-D integrals.
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DEFINING VOLUMES
There are a number of ways a 3D shape can be
generated– Sweeping a Cross Section
– The Disc Method
– The Washer Method
– The Shell Method
©1999 BG Mobasseri 13 04/20/23
CROSS-SECTIONAL METHOD
Imagine sweeping a 1D shape, of varying cross
sections A(x), along a path. This action will
generate a swept volume.
x
V = A x( )a
b
∫ dx
= (a×b)0
c
∫ dx=abcb
a
c
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VOLUME OF A PYRAMID
In problems like this you must first do two things– write a function for the cross section as a function
of x
– determine the lower and upper limit of the sweep
x
h
b
A x( ) =bh
h−x( ) ⎡ ⎣ ⎢
⎤ ⎦ ⎥2
©1999 BG Mobasseri 15 04/20/23
THE DISC METHOD
Take a 1D curve f(x) and revolve it around the x-
axis. This is a volume of revolution:– semi-circle---> sphere
– triangle --> cone
Every cross section is a circle. The radius of the
circle at xo is f(xo).
V = π f x( )[ ]2a
b
∫ dx
©1999 BG Mobasseri 16 04/20/23
Try it! REVOLVING A SINUSOID
Take one period of and revolve it
around the x-axis. Plot the shape then find the
volume of the revolution
1+cos2πx( )
-1-0.5
00.5
1
-1-0.5
00.5
10
0.2
0.4
0.6
0.8
1
©1999 BG Mobasseri 17 04/20/23
THE SHELL METHOD
Define a function f(x) in a<x<b. Revolve R
around the y-axis
Examples:– revolve a rectangle --> cylinder with a thickness
– revolve a circle --> torus/donut
yf(x)
x
RV= 2πxfx( )dxa
b∫
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Try it!
Let f(x)=1-(x-2)2 for 1<x<3. Revolve this around
the y-axis and find its volume
31
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ARC LENGTH
Another important application of integrals is
finding arc lengths
xa b
f(x)
L= 1+ f' x( )[ ]2a
b∫ dx
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PARAMETRIC CURVES
It is frequently easier to work with a parametric
representation of a curve,i.e.
x=f(t)
y=g(t)
For example, a circle
x(t)=rcos(t)
y(t)=rsin(t)
r
t
©1999 BG Mobasseri 21 04/20/23
LENGTH OF PARAMTERIC CURVES
Using derivatives of f(t) and g(t)
L = f' t( )[ ]2 + g' t( )[ ]2a
b
∫ dt
©1999 BG Mobasseri 22 04/20/23
CYCLOID
P
P
P
xfull perimeter=2.pi.r
Path length traversed by a point on a wheel is of
interest
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LENGTH OF A CYCLOID
The parametric equation of a cycloid with r=1 is
given by
x=t-sin(2.pi.t)
y=1-cos(2.pi.t)
First, plot the cycloid for 0< t <1.
Then find its length for one cycle and compare it
with the horizontal distance
©1999 BG Mobasseri 24 04/20/23
INTEGRALS IN POLAR COORDINATES
A curve can be represented in polar coordinates
by
Equivalently
r = f θ( )
x = f θ( )cosθ( )y= f θ( )sinθ( ) ⎧ ⎨ ⎩
r = f θ( )
θ
θ
x
y
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CURVE LENGTH
The length of a curve represented in polar
coordinates is given by
L = ′ f θ( )[ ]2 + f θ( )[ ]2α
β
∫ dθ
©1999 BG Mobasseri 26 04/20/23
PERIMETER OF AN ELLIPSE
Find the perimeter of an ellipse given by
r = 63+2cosθ
′ f θ( ) =−12sinθ
(3+ 2cosθ)2
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cardioid, 3-leaved rose
cardioid is defined by r=1+cosθ. 3-leaved rose is given by r=cos3θ.
0.5
1
1.5
2
30
210
60
240
90
270
120
300
150
330
180 0
0.2 0.4 0.6 0.8 1
30
210
60
240
90
270
120
300
150
330
180 0
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LENGTH OF A cardioid
We need the derivative of f(θ– f’(θ -sin(θ
Then,
L = sin2θ + (1+ (cosθ))2
−π2
π2
∫ dθ = 2+ 2cosθ−π2
π2
∫ dθ
©1999 BG Mobasseri 30 04/20/23
AREA OF AN ELLIPSE
For the ellipse given by
find its area and verify
r = 63+2cosθ
©1999 BG Mobasseri 31 04/20/23
AREA OF A cardioid
Here we have
Then
f θ( ) =1+ cosθ
A=120
2π
∫ f θ( )[ ]2dθ =120
2π
∫ 1+cosθ[ ]2dθ
©1999 BG Mobasseri 32 04/20/23
HOMEWORK-1
Find the length and area of a cardioid. Use the relevant equations for length and area given previously
0.5
1
1.5
2
30
210
60
240
90
270
120
300
150
330
180 0