Hi, megahog-ga:
The definition and theory of Riemann integrals is commonly presented
first for the case of a continuous function integrated over a finite
(bounded) interval.
An improper integral in this context is then some sort of case where
either the integrand (function) is not continuous over the interval of
integration, the interval is not bounded, or both.
In your problem the interval of integration [-1,8] is certainly
bounded, so the only issue that arises is the one point at with the
function to be integrated is discontinous, i.e. 1/x^(1/3) is
discontinuous at x = 0.
For simplicity we will write the function to be integrated as:
f(x) = x^(-1/3)
since one over the cube root of x is also expressed as the minus
one-third power of x. The circumflex ^ is often used in math
"texting" to mean a superscript, esp. as in this case an exponent.
Many examples treat a discontinuity occuring at an endpoint of the
interval of integration, but here the discontinuity lies strictly
inside that interval. The approach to be taken as far as finding out
whether the improper integral converges or diverges is to break the
interval of integration into two parts, so that the discontinuity lies
at the upper limit of [-1,0] and at the lower limit of [0,8]. If both
"parts" of the integral converge, then we say the original improper
integral also converges. Otherwise we conclude that it diverges:
INTEGRAL f(x) dx OVER [-1,8] = INTEGRAL f(x) dx OVER [-1,0]
+ INTEGRAL f(x) dx OVER [0,8]
Then we use a standard approach to each piece, taking a limit of
integrals whose upper (resp. lower) limits of integration approach 0,
thus reducing the question of the convergence of the improper integral
to a limit of "proper" integrals:
INTEGRAL f(x) dx OVER [-1,0] = LIMIT ( INTEGRAL f(x) dx OVER [-1,u] )
u -> 0-
(i.e. the limit as u tends to zero from below)
INTEGRAL f(x) dx OVER [0,8] = LIMIT ( INTEGRAL f(x) dx OVER [v,8] )
v -> 0+
(i.e. the limit as v tends to zero from above)
To evaluate the integrals inside the LIMIT's, we can use the fact that
f(x) is continuous on intervals [-1,u] and [v,8] and has an
"anti-derivative" there:
f(x) = dF(x)/dx where F(x) = (3/2) * x^(2/3)
(Check my differentiation of F(x) by using the Power Rule for
derivatives. One can obtain other anti-derivatives of f(x) by adding
any constant to F(x).)
The Fundamental Theorem of Calculus then gives us the definite
integrals of f(x) in terms of this anti-derivative:
INTEGRAL f(x) dx OVER [-1,u] = F(u)-F(-1)
= (3/2)*(u^(2/3)) - (3/2)*((-1)^(2/3))
= (3/2)*(u^(2/3)) - (3/2)*1
INTEGRAL f(x) dx OVER [v,8] = F(8)-F(v)
= (3/2)*(8^(2/3)) - (3/2)*(v^(2/3))
= (3/2)*4 - (3/2)*(v^(2/3))
Due to the positive exponent 2/3rds in these expressions, the limits
as u,v tend to zero both exist, and both are obtained by
"substituting" 0 for u and v since the "2/3rds power" function is
continuous there:
LIMIT ( INTEGRAL f(x) dx OVER [-1,u] ) = 0 - (3/2) = -3/2
u -> 0-
LIMIT ( INTEGRAL f(x) dx OVER [v,8] ) = (3/2)*4 - 0 = 6
v -> 0+
That is, both "pieces" of the original improper integral converge, and
so it converges as well.
The value of the original improper integral is furthermore the sum of
these two values above, namely -3/2 + 6 = 9/2.
regards, mathtalk-ga |