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Change of Variable Method for Integral
Motivation: Let’s start with the following exercise:
f(x) = x2 = =
For a non-negative function f(x) , we know that the integral stands for the area under its curve. To measure the area of a region, the area itself will not change; but the unit we use to measure the area can be different. In other words, if we use different coordinate system with different unit length to measure the area, we get the result in a different unit. And we would like to find out its relationship between original coordinate system and the new coordinate system. This is the basic idea behind “change of variable” method for integral.
Back to this problem, if u = 2x , then x= u f(x) = u2 .
And we can represent it as a function of u : g(u) = u2
When x=0, u=0; when x=1, u=2 . The following integral is = =
We are not surprised that it is not as above. But the whole process is very confusing. Actually, this is the common mistake made by the people. If you think about the geometrical meaning for u=2x, that means the new “unit length” is twice the original unit length. And we are measuring the area from x=0 and x=1. Under the “new measure”, it should be the area between u=0 and u=1/2 . But =
There are many mistakes that need to be clarified. At first, the way we obtain g(u) is by using x= u and substitute x into f(x) . So, g(u) will be equal to f(x) if you always keep the relation x= u. Thus, we still need to use u=0 and u=2 as the end points for the integral. Second, remember the integral is the limit of “Riemann Sum”. So, for that tiny interval we use to multiply the function in Riemann Sum, we just replace it with x= u . The correct way to find the value of the original integral after change of the variable is
= =
But we should not go through those thinking process while we handle the change of variable for integral. The following is a systematic approach to do it.
Theorem ( Change of Variable ) Let f be a function x that f is integrable on [a,b]. And x=T(u) where T is a differentiable monotonic function with range containing [a,b]. Then
=
Proof:
f is integrable on [a,b] . Thus, it does not matter to use the limit by choosing left end points in each small interval if we equally partition the interval into n pieces:
a = x0 < x1 < x2 < … < xn-1 < xn = b , where xi = i , i = 1,2, .., n = =
And there exits ui such that xi = T(ui) , i = 1, 2, 3, … n .
Since T is a differentiable function, there exists ci between ui-1 and ui such that xi – xi-1= T’(ci)( ui – ui-1) ( Mean Value Theorem )
So,
=
T is a monotonic function u0 , u1 , u2, … , un is a partition of the interval with end points T-1(a) and T-1(b) .
So, when n , we have =
This is the most basic type for change of variable in integral. When the transform involves multiple variables, a determinant known as “Jacobian” will be introduced in the integral.
We might think a step further: originally, the symbol is just used to represent the limit of Riemann sum. With the theorem here, we might consider it as : f(x) is multiplied by an infinitesimal amount dx and we take summation from x=a to x=b . For each infinitesimal strip, its area is f(x)dx .
Hence, when x=T(u), naturally we will have
dx = T’(u)du and f(x)dx =
And you can just express in the following way:
f(x)dx = f(T)dT where dT=T’du
This notation is more intuitively while applying on the problems of physics or other fields.
Example:
Sol: We do not like to have x appear in square root. So, set
x=sin(t) In this case, the interval of t is [0, ] . And 1-x2 = 1- sin2(t) = cos2(t) = cos(t) , t [0, ]
= cos(t) dx = cos(t)dt
= = = =
Example:
Sol: We try to avoid the variable showing in the denominator.
Set x=tan(t) . The range of tan(t) can cover any interval in real values. Thus, we do not have to worry about if it is applicable.
= cos2(t) dx = sec2(t) dt
Thus, = = = t + C where C is a constant .
And t = tan-1(x) . So, = tan-1(x) + C
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