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1 1 This last limit is + We therefore conclude that the improper integral diverges
YOU TRY IT Evaluate the improper integral
1 2 dx dx ( x + 1) 4/5 Improper integrals with integrand which is infinite at the left endpoint of integration are handled in a manner similar to the right endpoint case: EXAMPLE
1/2 0 Evaluate the integral 1 x ln2 x
SOLUTION
This integral is improper with infinite integrand at 0 The value of the integral is defined to be
1 x ln2 x
dx , provided that this limit exists Since 1/( x ln2 x) is continuous on the interval [ , 1/2] for > 0, this last integral can be evaluated directly and will have a finite real value For clarity, write ( x) = ln x, ( x) = 1/x Then the (indefinite) integral CALCULUS DeMYSTiFieD
becomes ( x) dx 2 ( x) Clearly the antiderivative is 1/ ( x) Thus we see that
1 x ln x
dx = lim
1 ln x
= lim
1 ln( 1/2) 1 ln
Now as 0+ we have ln hence 1/ ln 0 We conclude that the improper integral converges to 1/ ln 2 YOU TRY IT Evaluate the improper integral
0 2 1/( x + 2) 1/2 dx
Many times the integrand has a singularity in the middle of the interval of integration In these circumstances we divide the integral into two pieces for each of which the integrand is infinite at one endpoint, and evaluate each piece separately EXAMPLE
4 4 Evaluate the improper integral
4( x + 1) 1/5 dx
SOLUTION
The integrand is unbounded as x tends to 1 Therefore we evaluate separately the two improper integrals 1 4 4 1 4( x + 1) 1/5 dx and
4( x + 1) 1/5 dx
5 I N D E T E R M I N A T E F O R M S
The first of these has the value
0+ 4 4( x + 1) 1/5 dx = lim 5( x + 1) 4/5 1 4 = lim 5 ( ) 4/5 ( 3) 4/5 = 5 34/5 The second integral has the value
0+ 1+
4( x + 1) 1/5 dx = lim 5( x + 1) 4/5 4 1+ = lim 5 54/5 = 59/5 We conclude that the original integral converges and
4 4 1 4( x + 1) 1/5 dx 4( x + 1) 1/5 dx +
4 1 4( x + 1) 1/5 dx
= 5 34/5 + 59/5 YOU TRY IT Evaluate the improper integral
3 4 x 1 dx
It is dangerous to try to save work by not dividing the integral at the singularity The next example illustrates what can go wrong EXAMPLE
2 2 Evaluate the improper integral x 4 dx
CALCULUS DeMYSTiFieD
SOLUTION
What we should do is divide this problem into the two integrals
0 2 x 4 dx and
x 4 dx
( ) Suppose that instead we try to save work and just antidifferentiate: 2 2 x 4 dx =
1 3 x 3 2 2 1 12 A glance at Figure 54 shows that something is wrong The function x 4 is positive, hence its integral should be positive too However, since we used an incorrect method, we got a negative answer In fact each of the integrals in line ( ) diverges, so by definition the improper integral 2 2 x 4 dx
diverges
FIGURE 54
5 I N D E T E R M I N A T E F O R M S
EXAMPLE
Analyze the integral 1 dx x( 1 x) 1/2 SOLUTION
The key idea is that we can only handle one singularity at a time This integrand is singular at both endpoints 0 and 1 Therefore we divide the domain of integration somewhere in the middleat 1/2 say (it does not really matter where we divide)and treat the two singularities separately First we treat the integral 1/2 0 1 dx x( 1 x) 1/2 Since the integrand has a singularity at 0, we consider
1 dx x( 1 x) 1/2 This is a tricky integral to evaluate directly But notice that 1 1 x( 1 x) 1/2 x ( 1) 1/2 when 0 < x 1/2 Thus 1 dx x( 1 x) 1/2 1 dx = x ( 1) 1/2 1 dx x
We evaluate the integral: it equals ln Finally, lim ln = +
The first of our integrals therefore diverges But the full integral
1 dx x( 1 x) 1/2 CALCULUS DeMYSTiFieD
converges if and only if each of the component integrals
1/2 0 1 dx x( 1 x) 1/2 1 1/2 1 dx x( 1 x) 1/2 converges Since the first integral diverges, we conclude that the original integral diverges as well
YOU TRY IT Calculate
3 1/3 2 ( 2x) dx as an improper integral
533 An Application to Area
Suppose that f is a nonnegative, continuous function on the interval (a, b] which is unbounded as x a+ Look at Figure 55 Let us consider the area under the graph of f and above the xaxis over the interval (a, b] The area of FIGURE 55
5 I N D E T E R M I N A T E F O R M S
the part of the region over the interval [a + , b], > 0, is
f (x) dx
Therefore it is natural to consider the area of the entire region, over the interval (a, b], to be
0+ a+
f (x) dx
This is just the improper integral
Area =
f (x) dx
EXAMPLE
1 x ln4/3 x
Calculate the area above the xaxis and under the curve
0 < x 1/2 SOLUTION
According to the preceding discussion, this area is equal to the value of the improper integral
1/2 0 1 x ln
dx = lim
1 x ln4/3 x
For clarity we let ( x) = ln x, ( x) = 1/x Then the (indefinite) integral becomes ( x) 3 dx = 1/3 4/3 ( x) ( x)

