[PDF] Lecture 17 : Double Integrals Using Partial Integration to Evaluate

View - Download Lecture 17 : Double Integrals

Previous PDF

Next PDF

Lecture 17 : Double Integrals0/ 15

1/ 15Some of you have not learned how to do double integrals. In this course you will

need to do double integrals over rectangles and I will now explain how to do such calculations.

Partial (Indefinite) Integration

In one variable calculus you learned about the indefinite integralRf(x)dx. The point of the indefinite integral was that it was an inverse of the derivative ddx Z f(x)dx! =f(x) (In fact this is the definition of the) indefinite integral.

Lecture 17 : Double Integrals

2/ 15So

Z x

3dx=x44

and ddx x 44
=x3 The indefinite integral is defined only up to an arbitrary constant, "the constant of integration". The fundamental theorem of calculus then says that to evaluate thedefinite integral bR

0f(x)dxyou takeany indefiniteintegral, evaluate it at the upper limitb

and at the lower limitaand subtract the latter from the former.Lecture 17 : Double Integrals

3/ 15Now in two variable calculus you havetwo"partial" derivatives@f@x(x;y)and

@f@y(x;y)so you have two partial indefinite integralsRf(x;y)dxandRf(x;y)dy.

So (by definition)

@@x Z f(x;y)dx! =f(x;y) @@y Z f(x;y)dy! =f(x;y):

Let"s compute

Z (x2+y2)dx: The idea is totract y as a constant.Lecture 17 : Double Integrals

4/ 15The partial indefinite integral inxisdefined only up to a function of y.(because@@xg(y) =0).

Let"s do a harder one

Z sinxy dy=1x cosxy:(checkchain ruleLecture 17 : Double Integrals

5/ 15Partial (Definite) Integrals

Once you have the partial indefinite integral you have the partial definite integral 2 Z 1 (x2+y2)dx= x33 +y3x! x=2 x=1 83
+2y2! 13 +y2! =y2+73

The Golden Rule

Treatyas a constant throughout and do the one variable integral with respect to x. Notethe output is a function of y.Lecture 17 : Double Integrals

6/ 15Using Partial Integration to Evaluate Double Integrals over Rectangles

Just as we use the indefinite integral in one variable to evaluate definite integrals in one variable we will use the partial integrals to evaluate (definite) double integrals(*) The notation is very confusing because of tradition. Here thex-limits areaandb and they-limits arecandd so a and b go with dx and c and d go with dy.

Watch out for this later.

Lecture 17 : Double Integrals

7/ 15So we have a rectangleR[a;b][c;d]in thexy-planeWhen you learn integration theory correctly you will write this integral as

Z Z R f(x;y)dx dy(**) However putting in the limitsa,bandc,dis helpful for computations. Think of rewriting (**) as x=bZ x=ay=dZ y=cf(x;y)dx dy:Lecture 17 : Double Integrals

8/ 15So how do you compute

2Z 11 Z 0 (x2+y2)dx dy(**)

So the rectangleRis a square01

1 2

Here is how you compute (**).

There aretwoways. It is a famous theorem of Fubini that they lead to the same result. We will see this in our example.Pick the way that leads to the easiest computations.

Lecture 17 : Double Integrals

9/ 15First way (do thex-integration first)1Write (**) asZRemember 1 and 2 go withdxand 0 and 1 go withdy(see page 6)2Do the inside partial definite integral. The output will be the function ofy

g(y) =y2+73 (from page 5).Lecture 17 : Double Integrals

10/ 153Do a one variable definite integral ofg(y)with respect toyfrom 0 to 1.

1 Z 0 y 2+73 dy= y33 +73
y! y=0 y=0 13 +73
=83

That"s it.

The above method is said to evaluate the double integral by iterated one variable integrals. However the first step is new it is a partial integral with respect to x.

Lecture 17 : Double Integrals

11/ 15Second way (do they-integration first)1This time we write (**) as2Now we do the partial integration with respect toysox is a constant.

1 Z 0 (x2+y2)dy= x

2y+y33

y=1 y=0 =x2+13 The output is a function ofx.Lecture 17 : Double Integrals

12/ 153Perform the one variable integration of the output functionh(x) =x2+13

with respect tox. 2 Z 1 x 2+13 dx= x33 +x3 x=2 x=1 83
+23
13 +13 83
So as predicted we got the same answer no matter which order we chose to perform the iterated integrals.

Lecture 17 : Double Integrals

13/ 15Double Integrals of Product Functions over Rectangles

There is one case in which double integrals one particularly easy to compute.Definition Let f(x;y)be a function of two variables x and y. The f(x;y)is a product function if there exist g(x)and h(g)such that f(x;y) =g(x)h(y)Lecture 17 : Double Integrals

14/ 15Examples

f(x;y) =exsiny YES (it is a product) f(x;y) =ex+siny NO (it is not a product) f(x;y) =xy YES f(x;y) =x+y NOTheorem b Z ad Z c (g(x)h(y))dx dy 0

BBBBBBBB@b

Z a g(x)dx1

CCCCCCCCA0

BBBBBBBB@d

Z c h(y)dy1

CCCCCCCCA

ZMost functions of x and y are NOT product functions.Lecture 17 : Double Integrals

15/ 15Theorem (Cont.)

So 1 Z 01 Z 0 (xy2)dx dy=0

BBBBBBBB@1

Z 0 x dx1

CCCCCCCCA0

BBBBBBBB@1

Z 0 g 2dy1

CCCCCCCCA

12 13 =16 So a double integral of a product function over a rectangle is the product of two one variable integrals (one in x, the other in y).

Lecture 17 : Double Integrals

quotesdbs_dbs17.pdfusesText_23

[PDF] double split complementary colors examples

[PDF] double split complementary colors list

[PDF] double taxation agreement italy

[PDF] double taxation agreement us france

[PDF] double taxation spain usa

[PDF] double torus euler characteristic

[PDF] double waler forming system

[PDF] douglas county oregon election results 2016

[PDF] douglas county school calendar 2019 2020

[PDF] dover nh assessor database

[PDF] dowel basket assembly with expansion joint

[PDF] dowel basket stakes

[PDF] dowel baskets manufacturers

[PDF] down south zip codes

[PDF] download 3 tier architecture project in asp.net