EE 518 Homework 6 Due on Monday 2 00pm October 10 2016 Problem 1 Suppose f g a b R is differentiable on its domain and g 0 6 0 Show that c a b such that f 0 c f a f c 0 g c g b g c Hint Construct proper function to apply mean value theorem Define h x f a g x g b f x f x g x then we have h x is differentiable on a b and h a h b f a g b Therefore by mean value theorem there exists a c a b such that h0 c h a h b 0 a b h0 x f a g 0 x g b f 0 x g x f 0 x f x g 0 x therefore f a g 0 c g b f 0 c g c f 0 c f c g 0 c 0 f a f c f 0 c 0 g c g b g c Problem 2 If f x g x on a b and f g are Riemann integrable on a b show that Z b Z b f x dx g x dx a a Hint Use definition of Riemann integral Since both f x and g x are integrable on a b by definition of Riemann integral we have Z b b Z f x dx sup L P f a P Assume that b Z g x dx sup L P g P a Z b f x dx a g x dx a i e sup L P f sup L P g P P Then there exists a partition Pf of a b such that for any partition P of a b L Pf f L P g However since for any x a b f x g x L Pf f L Pf g This is a contradiction therefore Z b Z f x dx a b g x dx a Page 1 of 5 EE 518 Homework 6 Due on Monday 2 00pm October 10 2016 Problem 3 Let f 0 denote the Gamma Function i e Z f x 1 e x dx 0 i Show that f is well defined for any 0 i e show that 0 both 1 Z x 1 e x dx lim 0 and Z 1 Z t 0 x 1 e x dx lim t 1 x 1 e x dx t Z t x 1 e x dx 1 exists ii Show that f 1 f 1 1 and consequently f n n 1 n Z Page 2 of 5 EE 518 Homework 6 Due on Monday 2 00pm October 10 2016 i Let 1 Z x 1 e x dx h1 t t Then we need to show that limt 0 h1 t exists Since 0 and e x 1 for x 0 1 therefore R1 x 1 e x x 1 for x 0 1 Hence h1 t t x 1 dx for t 0 1 Since x 1 0 for any t0 0 1 thus Z 1 Z 1 1 1 x 1 h1 t0 x lim t 0 t t0 i e h1 t is bounded And h1 t is monotonically increasing as t 0 Therefore the limit limt 0 h1 t exists bounded and monotonic Let Z t x 1 e x dx h2 t 1 Then we need to show that limt h2 t exists For any 0 x 1 e x 0 x x 2 lim hence there exist a M 0 such that for all x M x 1 e x x 2 For t M Z h t M x 1 e x dx 1 Z x 1 e x dx M The first term is integrable via continuity The second term is monotonically increasing with t and R bounded from above by M x 2 dx 1 M Therefore the limit exists To conclude above Gamma function is well defined for all 0 ii For all 1 using integration by parts we have Z Z 1 x x 1 f x e dx e x 1 x 2 e x dx 0 0 0 Z 1 1 x 2 e x dx 1 f 1 0 For Z when 1 f 1 1 1 1 Assume that when n f n n 1 then we have f n 1 nf n n therefore for any Z f n n 1 Problem 4 Calculate the following integrals i Z e x ln x xex 1 Page 3 of 5 EE 518 Homework 6 Due on Monday 2 00pm October 10 2016 ii Z 2 sinn xdx n 0 1 2 0 hint consider both n is even and odd number i Z e x ln x xex 1 Z e Z 1 e xdex ln xdx2 2 1 1 Z e Z e 1 e 2 x e x x ln x 1 xdx xe 1 e dx 2 1 1 1 1 e2 ee 1 ee 4 4 ii Let In R 2 0 sinn xdx I1 1 2 I0 For n 2 Z In 2 sinn xdx 2 Z 0 sinn 1 xd cos x 0 2 0 sinn 1 x cos x Z 2 0 2 Z n 2 n 1 sin cos2 x sinn 2 xdx n 1 Z xdx n 1 0 2 sinn xdx 0 n 1 In 2 n 1 In Therefore In When n is odd In When n is even In n 1 In 2 n n 1 n 3 2 n 1 n 3 2 I1 n n 2 3 n n 2 3 n 1 n 3 1 n 1 n 3 1 I0 n n 2 2 n n 2 2 2 Problem 5 Suppose a function f is continuous on a c and differentiable on a c some b a c Show that there exists a a c such that f 0 0 Rt Let h t a f x dx for t a c Then we have Rb a f x dx Rc b f x dx 0 for h a h b h c 0 Apply mean value theorem we know there exist a ta a b such that h0 ta 0 i e f ta 0 Similarly there exists a tb b c such that f tb 0 Use MVT again there exist a ta tb a b such that f 0 0 Page 4 of 5 EE 518 Homework 6 Due on Monday 2 00pm October 10 2016 Problem 6 For the following parts submit your code and your results R 1 Calculate sin x x cos x dx R 10 2 Calculate x dx R 10 3 Calculate 0 x2x 5 dx using some numerical method The results are the following R 1 sin x x cos x dx x sin x 2 R 10 3 R 10 0 x dx x x2 5 dx 2 10 a 1 5 3 1 5221 and they are generated by the following code Page 5 of 5