第4.4节 换元积分法

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4.4INTEGRATION BY CHANGE OF VARIABLES

Wehaveseen that the sum, constant, and power rules for differentiation canbe turned around to give the sum, constant, and power ruler forintegration. In this section we shall show how to make use of theChain Rule for differentiation in problems of integration. The ChainRule will lead to the important method of integration by change ofvariables. The basic idea is to try to simplify the function to beintegrated by changing from one independent variable to another.

IfF is an antiderivative of f and we take u as the independentvariable, then ∫f(u)du is a family of functions of u,

∫ f(u)du = F(u) + C.

Butif we take x as the independent variable and introduce u as adependent variable u=g(x), then du and ∫ f(u) du mean thefollowing:

du= g' (x) dx, ∫ f(u)du = ∫ f (g(x)) g'(x)dx = H(x) +C.

Thenotation ∫ f(u)du always stands for a family of functions of theindependent variable, which in some cases is another variable such asx. The next theorem can be used as follows. To integrate a givenfunction of x, properly choose a new variable u= g(x) and integrate anew function with respect to u.

DEFINITION

LetI and J be intervals. We say that a function g maps J into I if forevery

Pointxin J, g(x) is defined and belongs to I (Figure 4.4.1).

Figure4.4.1

THEOREM1 ( Indefinite Integration by Change of Variables)

SupposeI and J are open intervals, f has domain I, g maps J into I, and g isdifferentiable on J.Assume that when we take u as the independentvariable,

Thenwhen x is the independent variable and u = g(x),

PROOFlet H(x) = F(g(x)). For any x in J, the derivatives g' (x) and F'(g(x))= f(g(x))

exist.Therefore by the Chain Rule,

Itfollows that

Sowhen u = g(x), we have

Theorem1 gives another proof of the general power rule

Whereu is given as a function of the independent variable x, from thesimpler power rule

Wherex is the independent variable.

EXAMPLE1 Find

EXAMPLE2 Find

Letu= 1 + 1/x, Then du =－1/x²dx and thus

So

Ina simple problem such as this example, we can save writing by usingthe term 1+1/x instead of introducing a new letter u,

Inexamples such as the above one, the trick is to find a new variable usuch that the expression becomes simpler when change variables. Thisusually must be done by an “educated” trial and error process.

Onemust be careful to express dx in terms of du before integrating withrespect to u.

EXAMPLE3 Find ∫(1+5x)² dx, Let u = 1+5x. For emphasis weshall do it correctly and incorrectly.

Correct:du =5dx, dx = ___ du,

Incorrect:

Incorrect:

EXAMPLE4 Find

Wetryto express the integral in terms of u.

Sinceu=2－x²,x²=2－u.Therefore

Wenext describe the method of definite integration by change ofvariables. In a definite integral

Itis always understood that x is the independent variable and we areintegrating between the limits x= a and x=b. Thus when change to anew independent variable u, we must also change the limits ofintegration. The theorem below will show that if u = c when x=a andu=d when x=b, then c and d will be the new limits of integration.

THEOREM2 (Definite Integration by Change of Variables)

SupposeI and J are open intervals, f is continuous and has an antiderivativeon 1, g has a continuous derivative on J, and g maps J into I. Thenfor any two points a and b in J,

PROOFLet F be an antiderivative of f. Then by Theorem 1, H(x) = F(g(x)) isan antiderivative of h(x)= f(g(x))g'(x). Since f,g, and g' arecontinuous, h is continuous on J. Then by the Fundamental Theorem ofCalculus,

EXAMPLE5 Find the area under the line y= 1+3x from x = 0 to x = 1.This can be done either with or without a change of variables.

Withoutchange of variable: ∫ (1+3x) dx = x +3x²/2 + C, so

Withchange of variable: Let u = 1+ 3x. Then du=3dx, dx= ___ du.

Whenx = 0, u= 1+3·0 = 1. When x =1, u =1+3·1 =4.

EXAMPLE5 shows us that ____ (1+3x) dx = ___ (u/3) du; that is, theareas shown in Figure 4.4.2 are the same.

Figure4.4.2

Figure4.4.3

EXAMPLE6 Find the area under the curve y=2x/ (x²-3)² from x=2 to x=3

(Figure4.3.3).

Letu= x² -3. Then du = 2x dx. At x=2, u= 2² -3 =1. At x=3,

u= 3² -3 =6. Then

EXAMPLE7 Find ________ xdx, the function _____ x as given is onlydefined on the closed interval [-1,1]. In order to use Theorem2, weextend it to the open interval J=(－∞,∞)by

Letu=1－x².Then du= －2xdx, dx=－du/2x.At x=0, u=1. At x=1,u=0. Therefore

Wesee in Figure 4.4.4 that as x increases from 0 to 1, u decreases from1 to 0, so the limits become reversed. The areas shown in Figure4.4.5 are equal.

Figure4.4.4

Figure4.4.5

Wecanuse integration by change of variables to derive the formula for thearea of a circle, A = πr², where r is the radius. It is easier towork with a semicircle because the semicircle of radius r is just theregion under the curve

Tostart with we need to give a rigorous definition of π. Bydefinition, π is the area of a unit circle. Thus π is twice thearea of the unit semicircle, which means:

DEFINITION

Thearea of a semicircle of radius r is the definite integral

Toevaluate this integral we let x=ru. Then dx=r du. When x= ____ r, u =___ 1.

Thus

Thereforethe semicircle has area π r²/2 and the circle area π r² (Figure4.4.6).

EXAMPLE8 Find

Letu= x -x3. Then du =(1 -3x²) dx. When x=0, u=0 -03 =0.

Whenx =1, u =1 -13 =0. Then

Asx goes from 0 to 1, u starts at 0, increases for a time, then dropsback to 0

(Figure4.4.7).

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Figure4.4.6

Figure4.4.7

Wedo not know how to find the indefinite integrals in this example.Nevertheless the answer is 0 because on changing variables bothlimits of integration become the same. Using the Addition Property,we can also see that, for instance,

PROBLEMSFOR SECTION 4.4

InProblems 1-90, evaluate the integral.

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InProblems 91-108, evaluate the definite integral.

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105______106______

107_______108______

109Find the area of the region below the curve y=1/(10-3x)fromx=1and x=2.

110Find the area of the region under one arch of the curve y=sin xcosx.

111Find the area of the region under one arch of the curve y=cos(3x).

112Find the area of the region below the curve y=4x______betweenx=0and x=2

113Find the area below the curve y=(1+7x)²/3betweenx=0and x=1

114Find the area below the curve y=x/(x²+1)between x=0and x=3.

□115Evaluate: ________dx

□116Evaluate: ___2x_______dx

□117let f and g have continuous derivatives and evaluate ∫f ' (g(x))g'(x)dx.

□118a real function fissaid to be even if f(x) =f(-x)forall x.Show that if fisa continuous even function, then __f(x)dx=__f(x) dx.

□119an odd function is a real function g such that g(-x)= -g(x) forall x.Prove that for a continuous odd function g,__g(x)dx=0.