Using Partial Integration to Evaluate Double Integrals over Rectangles integrals However the first step is new it is a partial integral with respect to x Lecture
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[PDF] Lecture 17 : Double Integrals - UMD MATH
Using Partial Integration to Evaluate Double Integrals over Rectangles integrals However the first step is new it is a partial integral with respect to x Lecture
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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 x3dx=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 Integrals3/ 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 Integrals4/ 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 Integrals5/ 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 Integrals6/ 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 Integrals8/ 15So how do you compute
2Z 11 Z 0 (x2+y2)dx dy(**)So the rectangleRis a square01
1 2Here 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 Integrals10/ 153Do a one variable definite integral ofg(y)with respect toyfrom 0 to 1.
1 Z 0 y 2+73 dy= y33 +73y! 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= x2y+y33
y=1 y=0 =x2+13 The output is a function ofx.Lecture 17 : Double Integrals12/ 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.