MA 212 Further Mathematical Method (Calculus) - London School of Economics. Solutions Section A and B.

MA 212 Further Mathematical Method (Calculus) - London School of Economics. Solutions Section A and B.The Laplace transform is de…ned by Lff(t)g = Zt=0 1 estf(t)dt. We have Lff0(t)g = sLff(t)g f(0 ). Indeed, integrating by parts, we get Lff0(t)g = Zt=0 1 estf0stf(t)]1 0 Zt=0 1(s)estf(t)dt = sLff(t)g f(0): We used that f is of exponential growth at most to deduce that estf(t) ! 0 as t ! 1. For g(t) = Zu=0 t h(u) du, we have g(0) = 0 and g0(t) = h(t) (by the Fundamental Theorem of Calculus). Hence h~(s) = Lfg0(t)g = sLfg(t)g g(0) = sLfg(t)g and g~(s) = 1sh~(s) follows. (b) Set H = H(s) = Lfh(t)g. By applying the Laplace Transform to both sides of the equation and using (a), we obtain that sH h(0) + 41 s H = Lf1g: We use the fact that Lf1g = 1s and h(0) = 1 to obtain: sH 1 + 41 s H = 1 s : Hence, H s + 4s = 1s + 1 and, consequently, H = s + 1 s2 + 4 = 1 2 2 s2 + 22 + s s2 + 22 : We deduce that h(t) = 1 2 sin(2t) + cos(2t). 1(c) We have problems at x = 1 and at 1. Hence we split the integral into two and deal with each separately: We know that Z1 2 (x 1 1)p dx converges for p 1. Also, xlim !1 (x3 x x)p =(x 1 1)p = lim x!1 x (x2 + x)p = 1=2p. Hence, Z1 2 (x3 x x)p dx = Z1 2 (x 1 1)p (x2 +x x)p dx converges i¤ p 1 by the limit comparison test. Similarly, Z21 (x3 x x)p dx = Z21 xx3p (1 +1x2)p dx = Z21 x31p1 (1 +1x2)p dx: Also, lim x!1 x (x3 x)p =x31p1 = lim x!1 (1 +1x2)p = 1. By the limit comparison test, this integral converges i¤ 3p 1 1. Consequently, the original integral converges i¤ 2=3 p 1. 2Question 2 (a)(i) lim x!1 2x2 + cos x sin x (2x + 1)2 = lim x!1 2 + cos x x2 sin x x2 (2 + 1=x)2 = 2 + 0 + 0 (2 + 0)2 = 1 2 : Here we used the fact that sin x and cos x are bounded. (ii) The next limit is 0=0-type and we use L’Hospital Rule: lim x!1 x ln(cos(1=x)) = lim x!1 ln(cos(1=x)) 1=x = lim x!1 sin(1=x) cos(1=x) 1 x2 1=x2 = lim x!1 cos(1 sin(1=x =x)) = 0: (iii) The next limit is 0=0-type and we use L’Hospital Rule: lim x!0 ex2 1 x sin x = lim x!0 ex22x sin x + x cos x = lim x!0 ex22 sin x x + cos x = 1: (b)(i) we use the substitution t = eu …rst: Z1 2 ln1t dt = Z0 ln 2 u1eudu. Now, lim u!0 1u eu 1u = lim u!0 eu = 1; the integral R0ln 2 u1du diverges, hence the original integral also diverges by LCT. (ii) we have a problem at 0 and at 1, so we divide into two integrals: Z11 sin t7=32t dt converges absolutely because sin t7=32t  1=t7=2 and R11 t7dt =2 converges. Z0 1 sin t7=32t dt also converges. We use that limt!0 sin t=t = 1 and obtain that lim t!0 sin3 t t7=2 1p t = lim t!0 sin3 t t3 = 1: Since Z0 1 t11=2 dt converges, by LCT, Z0 1 sin t7=32t dt converges as well. 3Question 3 (a)(i) We have Z0 1 r1 x x dx = Z0=2 s1 sinsin 2 t2 t2 sin t cos t dt = Z0=2 cos sin tt2 sin t cos t dt = Z0=2 2 sin2 t dt = Z0=2(1 cos(2t)) dt = t sin(2 2 t)= 0 2 = 2 : The region of integration D is bounded by lines y = 0, y = 1 and x = y, and the curve x = y2. (Picture to follow.)