EE 518 : Homework #6 Due on Monday 2:00pm, October 10, 2016Problem 1Suppose f, g : [a, b] → R is differentiable on its domain, and g06= 0. Show that ∃c ∈ (a, b), such thatf(a) − f(c)g(c) − g(b)=f0(c)g0(c)Hint: Construct proper function to apply mean value theoremDefine h(x) = f(a)g(x) + g(b)f(x) − f(x)g(x), then we have h(x) is differentiable on [a, b], andh(a) = h(b) = f(a)g(b)Therefore, by mean value theorem, there exists a c ∈ [a, b], such thath0(c) =h(a) − h(b)a − b= 0h0(x) = f(a)g0(x) + g(b)f0(x) − g(x)f0(x) − f(x)g0(x)therefore,f(a)g0(c) + g(b)f0(c) − g(c)f0(c) − f(c)g0(c) = 0f(a) − f(c)g(c) − g(b)=f0(c)g0(c)Problem 2If f(x) > g(x) on [a, b], and f, g are Riemann integrable on [a, b] show thatZbaf(x)dx >Zbag(x)dxHint: Use definition of Riemann integral.Since both f(x) and g(x) are integrable on [a, b], by definition of Riemann integral, we haveZbaf(x)dx = supPL(P, f ),Zbag(x)dx = supPL(P, g)Assume thatZbaf(x)dx ≤Zbag(x)dxi.e.supPL(P, f ) ≤ supPL(P, g)Then there exists a partition Pfof [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, thereforeZbaf(x)dx >Zbag(x)dxPage 1 of 5EE 518 : Homework #6 Due on Monday 2:00pm, October 10, 2016Problem 3Let f : (0, ∞) → < denote the Gamma Function, i.e.f(α) =Z∞0xα−1e−xdx(i) Show that f(α) is well defined for any α > 0, i.e., show that ∀α > 0, bothZ10xα−1e−xdx = limt→0Z1txα−1e−xdxandZ∞1xα−1e−xdx = limt→∞Zt1xα−1e−xdxexists.(ii) Show thatf(α) = (α − 1)f(α − 1), ∀α > 1and consequently,f(n) = (n − 1)!, ∀n ∈ Z+Page 2 of 5EE 518 : Homework #6 Due on Monday 2:00pm, October 10, 2016(i) Leth1(t) =Z1txα−1e−xdx.Then we need to show that limt→0+h1(t) exists. Since α > 0 and e−x< 1 for x ∈ (0, 1), thereforexα−1e−x< xα−1for x ∈ (0, 1). Hence h1(t) <R1txα−1dx for t ∈ (0, 1). Since xα−1> 0 for anyt0∈ (0, 1), thush1(t0) <Z1t0xα−1< limt→0+Z1txα−1=1αi.e. h1(t) is bounded. And h1(t) is monotonically increasing as t → 0+. Therefore the limitlimt→0+h1(t) exists (bounded and monotonic).Leth2(t) =Zt1xα−1e−xdx.Then we need to show that limt→∞h2(t) exists. For any α > 0,limx→∞xα−1e−xx−2= 0hence there exist a M > 0 such that for all x > M, xα−1e−x< x−2. For t > M,h(t) =ZM1xα−1e−xdx +Z∞Mxα−1e−xdxThe first term is integrable via continuity. The second term is monotonically increasing with t andbounded from above byR∞Mx−2dx = 1/M. Therefore the limit exists. To conclude above, Gammafunction is well-defined for all α > 0.(ii) For all α > 1, using integration by parts, we havef(α) =Z∞0xα−1e−xdx = −e−xxα−1∞0−Z∞0(α − 1)xα−2(−e−x)dx=Z∞0(α − 1)(α − 1)xα−2(−e−x)dx = (α − 1)f(α − 1)For α ∈ Z+, when α = 1f(1) = 1 = (1 − 1)!Assume that when α = n, f(n) = (n − 1)!, then we havef(n + 1) = nf(n) = n!therefore, for any α ∈ Z+, f(n) = (n − 1)!.Problem 4Calculate the following integrals(i)Ze1(x ln x + xex)Page 3 of 5EE 518 : Homework #6 Due on Monday 2:00pm, October 10, 2016(ii)Zπ/20sinnxdx, n = 0, 1, 2, . . .(hint: consider both n is even and odd number)(i)Ze1(x ln x + xex)=12Ze1ln xdx2+Ze1xdex=12x2ln xe1−Ze1xdx+xex|e1−Ze1exdx=14e2+14+ ee+1− ee(ii) Let In=Rπ/20sinnxdx.I0=π2, I1= 1For n ≥ 2,In=Zπ/20sinnxdx = −Zπ/20sinn−1xd cos x= −sinn−1x cos xπ/20+ (n − 1)Zπ/20cos2x sinn−2xdx= (n − 1)Zπ/20sinn−2xdx − (n − 1)Zπ/20sinnxdx= (n − 1)In−2− (n − 1)InTherefore,In=n − 1nIn−2.When n is odd,In=n − 1n·n − 3n − 2· ··· ·23I1=n − 1n·n − 3n − 2· ··· ·23When n is even,In=n − 1n·n − 3n − 2· ··· ·12I0=n − 1n·n − 3n − 2· ··· ·12·π2Problem 5Suppose a function f is continuous on [a, c], and differentiable on (a, c).Rbaf(x)dx =Rcbf(x)dx = 0, forsome b ∈ (a, c). Show that there exists a ξ ∈ (a, c), such that: f0(ξ) = 0Let h(t) =Rtaf(x)dx, for t ∈ [a, c]. Then we haveh(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 ξ ∈ (ta, tb) ⊂ (a, b), such that f0(ξ) = 0.Page 4 of 5EE 518 : Homework #6 Due on Monday 2:00pm, October 10, 2016Problem 6For the following parts, submit your code and your results:1. CalculateRsin(x) + x cos(x)dx2. CalculateR10α√x − αdx3. CalculateR100xx2+5dx using some numerical methodThe results are the following:1.Rsin(x) + x cos(x)dx = x sin(x)2.R10α√x − αdx =2∗(10−a)1.533.R100xx2+5dx∼=1.5221and they are generated by the following code:Page 5 of