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This point is equidistant from all sides and is called the mc^rt/^r (the center of the inclrcle of the triangle). Concerning the radius r of the incircle. see Paragraph 3.1.1-3. The angle bisector through a vertex cuts the opposite side in ratio proportional to the adjacent sides of the triangle. The length of the angle bisector la drawn to the side a is given by the formulas | 304 Integrals Example 3. Let us show that the improper integral smX dx is convergent for a 0 and A 0. J xX Set f x sin x and g x x x and verify conditions i and ii of Theorem 8. We have i sinxdx cosa -cosH 2 d a ii since A 0 the function x x is monotonically decreasing and goes to zero as x to. So both conditions of Theorem 8 are met and therefore the given improper integral is convergent. 7.2.7-3. Some remarks. 1 . If an improper integral is convergent and the integrand function tends to a limit as x to then this limit can only be zero it is such situations that were dealt with in Examples 1-3 . However the property lim f x 0 is not a necessary condition for x convergence of the integral 7.2.7.1 . An integral can also be convergent if the integrand function does not have a limit as x to. For example this is the case for Fresnel s integrals y sin x2 dx y cos x2 dx 2 2. Furthermore it can be shown that the integral 6x dx is convergent regard- less of the fact that the integrand function being everywhere positive is not even bounded f nk nk k 1 2 . . The graph of this function has infinitely many spikes with heights increasing indefinitely and base widths vanishing. At the points lying outside the spike bases the function rapidly goes to zero. 2 . If f x is a monotonic function for x 0 and the improper integral J f x dx is convergent then the following limiting relation holds f x dx lim f en . Jo 0 n 1 7.2.8. General Reduction Formulas for the Calculation of Improper Integrals Below are some general formulas involving arbitrary functions and parameters that may facilitate the calculation of improper integrals. 7.2.8-1. Improper integrals involving power functions. f- fif f f x dx 0 1 x J 1 x 2 b - a Ja f ax f bx dx y 0 - f to ln if a 0 b 0 f x is continuous on 0 to and f to lim f x is a finite quantity x 7.2. Definite Integral 305 f axxf x dx f 0 In if a 0 b 0 f x is continuous on 0 to 0x a and the integral dx exists c 0 Jc x ro 0 fro ro f x dx if a 0 b 0 0 fro ro x2