241 - 5 Because x^3 is increasing, the minimum on (k/n, (k+1)/n) is (k^3)/(n^3) and the maximum is (k+1)^3/(n^3). The base of each interval is (1/n). hence the upper sum is SUM[i = 1 to n] (1/n)(i^3 / n^3) =.25(1+2/n+1/n^2). Similaly, the lower sum is SUM[i = 0 to n-1] (1/n)(i^3 / n^3) = .25(1-2/n+1/n^2). Since both approach .25 as n appraches infinity, the Riemann integral exists and equals .25 6 For delta arbitrarily small, we may surround the c_i with neighborhoods radius delta, and use the endpoints of those intervals to define the partition (a and b must also be members of the partition; it may be necessary to make some of the neighborhoods even smaller so they will not intersect, but will only strengthen the result). If B is the bound on the function, for that partition the upper sum will be less than n times 2 times delta times B, and the lower sum greater than -n times 2 times delta times B. Since delta can be arbitrarily small (recall that n is fixed), the upper and lower sums are both arbitrarily close to 0, hence the integral is 0. 13 If f1 = 0 on the rationals and 1 on the irrationals, and f2 = 1 on the rationals and 0 on the irrationals; L(f1)=L(f2)=0, but L(f1 + f2) = 1 249 - 1 For a fixed partition, the heighs of all the rectangles are multiplied by k, hence L(kf, P) = kL(f,P) For a sequence of partitions which provide L(f), there will also be convergence of kL(f, P_i)=L(kf,P_i) (the same argument holds for U). It follows from above that if L(f)=U(f), L(kf)=U(kf), k passes across the integral. 11 Let h(x) = x for x not equal to 0, h(0)=1. i/h is unbounded, hence not integrable. 19 This follows from the previous result that sup(f,g)=.5(f+g+|f-g|), and that the sum and absolute value of integrable functions are integrable. 259 - 5 For x=0, use the definition of the derivative to get lim[x goes to 0+](x^2 sin(pi/x^2))/(x-0) = 0. For x not equal 0, take the derivative, which is 2xsin(pi/x^2) + x^2cos(pi/x^2)(-2/x^3); this is not bounde near 0 because of the factor (1/x) whixh is not always killed by the cosine function. For a greater than 0, int[a,1]g(x)dx is just G(1)-G(a), and since G(a) goes to zero as a goes to 0+, the limit as a goes to 0+ is just G(1)= 0. 9 If f is integrable, then F' is integrable, because it differs from f at only a finite number of points, Therefore by theorem 7.3.1 Int[a,b]F' = F(b)-F(a). Theorem 7.2.4 essentially tells us we can ignore a finite number of points, hence Intf = IntF'. 17 Assume f(x) is positive somewhere, hence on an interval, call it [c,e]. Then int[0,c] = int[c,1] =int[c,d] + int[d,1] similarly, int[0,c] + int[c,d] = int[0,d] = int[d,1]. subtracting these equations provides int[c,d] = -int[c,d], hence int[c,d]=0. This contradicts that f is positive and continuous on [c,d].

Sun Dec 07 1997