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The limit of this expression as a —> 0+ is 6 — 0 = 6. Thus the given integral is convergent (to 6).

The situation where the integrand becomes infinite at the upper limit of integration is perfectly similar. It is an altogether different proposition, however, when an infinite value of the integrand occurs in the open interval (a, b) rather than at a or b. In this eventuality, it is necessary to take advantage of the additivity of definite integrals and first decompose the given integral into subintegrals. Assume that /(*)—> oo asx —> p, where p is a point in the interval (a, 6); then, by the additivity property, we have fhf(x)dx= [Pf(x)dx+ (hf(x)dx

The given integral on the left can be considered as convergent if and only if each subintegral has a limit.

Example 5 Evaluate / —dx. The integrand tends to infinity when x ap-

1 x proaches zero; thus we must write the given integral as the sum "1 , . ro

( x 3 dx = f x 3 dx + f x 3 dx (say, = /, + I2) J - 1 J — i Jn

The integral /, is divergent, because rb lim / x dx = lim

0 0

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