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Homework answers / question archive / Examiners’ commentaries 2020 Examiners’ commentaries 2020 MT2176 Further calculus Important note This commentary reflects the examination and assessment arrangements for this course in the academic year 2019–20

Examiners’ commentaries 2020 Examiners’ commentaries 2020 MT2176 Further calculus Important note This commentary reflects the examination and assessment arrangements for this course in the academic year 2019–20

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Examiners’ commentaries 2020 Examiners’ commentaries 2020 MT2176 Further calculus Important note This commentary reflects the examination and assessment arrangements for this course in the academic year 2019–20. The format and structure of the examination may change in future years, and any such changes will be publicised on the virtual learning environment (VLE). Information about the subject guide and the Essential reading references Unless otherwise stated, all cross-references will be to the latest version of the subject guide (2012). You should always attempt to use the most recent edition of any Essential reading textbook, even if the commentary and/or online reading list and/or subject guide refer to an earlier edition. If different editions of Essential reading are listed, please check the VLE for reading supplements – if none are available, please use the contents list and index of the new edition to find the relevant section. Comments on specific questions Candidates should answer all FOUR questions. All questions carry equal marks. Question 1 (a) Use Taylor’s theorem to show that, for x > 0: x − x2 ≤ tan−1 x ≤ x. Hence find the limit: lim tan−1 x x→0+ x . (b) For t > 0, the function J is given by: Z 1/(2t) J (t) = sin−1 (tx) dx. 0 Use the Leibniz rule to show that: t2 J 0 (t) = k for some constant k that you should find. (c) Determine whether each of the integrals: Z ∞√ Z 1√ 1 + x2 − 1 1 + x2 − 1 i. dx, ii. dx x2 x2 0 1 Z and ∞ iii. 0 √ 1 + x2 − 1 x2 dx is convergent or divergent. 5 MT2176 Further calculus Reading for this question For (a), Section 2.2.3 of the subject guide. For (b), Section 6.1.3 of the subject guide. For (c), Chapter 4 of the subject guide. Approaching the question (a) Taylor’s theorem states that for x in an interval around c, an (n + 1) times differentiable function f can be written as: f (x) = f (c) + (x − c)f 0 (c) + (x − c)n (n) (x − c)n+1 (n+1) (x − c)2 00 f (c) + · · · + f (c) + f (d) 2! n! (n + 1)! for some d lying strictly between c and x. For x ≥ 0, we can get the given inequality by taking the function tan−1 x and applying Taylor’s theorem around the point x = 0. For instance, we can take: f (x) = tan−1 x, f 0 (x) = 1 1 + x2 and f 00 (x) = −2x (1 + x2 )2 which gives us: f (0) = 0, f 0 (0) = 1 f 00 (d) = and −2d (1 + d2 )2 so that, applying Taylor’s theorem, we have: tan−1 x = 0 + (x − 0)(1) + (x − 0)2 2! −2d (1 + d2 )2 =x− d x2 (1 + d2 )2 for some d ∈ (0, x). (Observe that the question explicitly asks you to use Taylor’s theorem. Using a Taylor series with no explicit remainder term is not sufficient.) Now, as: d>0 ln(1 + x) = x − we have d x2 ≤ x (1 + d2 )2 whereas, we can see that: d 0, we have d < (1 + d2 )2 .) That is, for x ≥ 0, we have: x − x2 ≤ tan−1 x ≤ x as required. Lastly, for x > 0, this inequality gives us: 1−x≤ tan−1 x ≤1 x and so, as: lim (1 − x) = 1 x→0 we can see that: 6 lim 1 = 1 x→0 tan−1 x =1 x→0 x lim by the Sandwich theorem. and Examiners’ commentaries 2020 (b) For t > 0, use the Leibniz rule to differentiate the function: Z 1/(2t) J(t) = sin−1 (tx) dx 0 to get: J 0 (t) = − 1 2t2 sin−1 t 2t Z 1/(2t) −0+ 0 x p dx 1 − (tx)2 provided that its use is justified. To see that it is, observe that both 0 and 1/(2t) are differentiable for t > 0 and, furthermore, the functions: F (t, x) = sin−1 (tx) and ∂F x =p ∂t 1 − (tx)2 are jointly continuous for t > 0 and 0 ≤ x ≤ 1/(2t). Now, to simplify our answer, we notice that: Z 1/(2t) i1/(2t) x 1 1 h p p =− 2 dx = − 2 2 1 − (tx)2 2 2t t 0 1 − (tx) 0 √ 3 −1 2 ! and so we see that: 1 π J (t) = − 2 − 2 12t t √ 0 ! 3 1 −1 = 2 2 t which gives us t2 J 0 (t) = k with: √ 3 π − 1− 2 12 ! √ k =1− π 3 − 2 12 as the sought-after value of k. (c) The three integrals in this question all have the same integrand and it is convenient to start by noting that, if we rationalise this, we get: √ √ √ 1 + x2 − 1 1 + x2 − 1 1 + x2 + 1 (1 + x2 ) − 1 1 √ √ f (x) = = = =√ 2 2 2 2 x x 1+x +1 x ( 1 + x2 + 1) 1 + x2 + 1 for x > 0. i. We note that, even though this looks like an improper integral of the second kind since the integrand is undefined at x = 0, we have: √ 1 + x2 − 1 1 1 lim = lim √ = 2 + + 2 x 2 x→0 x→0 1+x +1 and so we can ‘repair’ the integrand by taking f (0) = 1/2 and: √ 1 + x2 − 1 f (x) = x2 if 0 < x ≤ 1. With this done, the integrand is now continuous on the interval [0, 1] and so we can say that the given integral is proper, i.e. it must be convergent. ii. This is an improper integral of the first kind. For large x, given what we know about f (x), it makes sense to take g(x) = 1/x so that: lim x→∞ f (x) x 1 = lim √ = 1. = lim p g(x) x→∞ 1 + x2 + 1 x→∞ (1/x2 ) + 1 + (1/x) With this, we can then use the Limit Comparison Test (LCT1) to see that as: Z ∞ Z ∞ dx g(x) dx = x 1 1 is divergent, so is the original integral. 7 MT2176 Further calculus iii. This is an improper integral of the third kind with problems at x = 0 and as x → ∞. As such, we write it as: Z 0 ∞ √ 1 + x2 − 1 dx = x2 √ 1 Z 0 1 + x2 − 1 dx + x2 Z ∞ √ 1 1 + x2 − 1 dx x2 and note that, as we saw in part (ii), the second of these integrals is divergent and so the given integral is divergent too. Question 2 (a) Consider the repeated integral: Z ? √ ? Z 9−y2 3/2 ? xe(9−x) dx? dy. 3 y=−3 x=0 Sketch the region of integration and, by changing the order of integration, evaluate this repeated integral. (b) i. Show that: π/2 Z sin2p−1 θ cos2q−1 θ dθ. B(p, q) = 2 0 Deduce that B(1/2, 1/2) = π and Γ(1/2) = √ π. ii. Hence show that: Z 1 (− ln x)5/2 dx = 15 √ 0 8 π. iii. Using your answer to part (i) and the integral: Z t 0 du √ √ u t−u √ find the Laplace transform of 1/ t. Reading for this question For (a), Section 5.1 of the subject guide. For (b), Sections 7.1.2 and 7.2.2 of the subject guide. Approaching the question (a) We are given the repeated integral: Z 3 Z √9−y2 I= ! xe y=−3 (9−x)3/2 dx dy x=0 and, looking at its limits, we see that the region ofp integration takes a fixed −3 ≤ y ≤ 3 so that the corresponding values of x satisfy 0 ≤ x ≤ 9 − y 2 . This means that a sketch of this region should look like the one illustrated in Figure 1. 8 Examiners’ commentaries 2020 Figure 1: The sketch for Question 2 (a). Using this sketch, we√can see that when √ we change the order of integration, each fixed 0 ≤ x ≤ 3 gives us − 9 − x2 ≤ y ≤ 9 − x2 . As such, the repeated integral we are asked to evaluate can be written as: ! Z √9−x2 Z 3 (9−x)3/2 I= xe dy dx √ y=− 9−x2 x=0 and we can evaluate the y-integral by noting that: Z √9−x2 h i√9−x2 (9−x)3/2 (9−x2 )3/2 xe dy = xye √ √ y=− 9−x2 y=− 9−x2 p 2 3/2 = 2x 9 − x2 e(9−x ) . This means that our repeated integral becomes: Z 3 i p 2 3/2 2 3/2 3 2h 2 I= 2x 9 − x2 e(9−x ) dx = − e(9−x ) = − (1 − e27 ) 3 3 0 x=0 which is the sought-after answer. (b) i. Using the definition of the Beta function and the substitution u = sin2 θ, we have: Z 1 B(p, q) = up−1 (1 − u)q−1 du 0 π/2 Z (sin2p−2 θ)(1 − sin2 θ)q−1 (2 sin θ cos θ) dθ = 0 π/2 Z sin2p−2 θ cos2q−2 θ(2 sin θ cos θ) dθ = 0 Z π/2 =2 sin2p−1 θ cos2q−1 θ dθ 0 as required. With this, we can easily see that: Z π/2 π 1 1 , =2 dθ = 2 =π B 2 2 2 0 and, using the formula: B(p, q) = we also have: 1 1 (Γ(1/2))2 B , = 2 2 Γ(1) ⇒ Γ(p)Γ(q) Γ(p + q) 2 1 Γ = πΓ(1) 2 ⇒ √ 1 Γ = π 2 as Γ(1) = 1 and Γ(1/2) > 0. 9 MT2176 Further calculus ii. Using the substitution u = − ln x, we have: Z 1 Z 0 Z (− ln x)5/2 dx = u5/2 (−e−u ) du = ∞ 0 ∞ u5/2 e−u du = Γ 0 7 2 if we use the √ definition of the Gamma function. Then, using Γ(α + 1) = αΓ(α) and Γ(1/2) = π from part (i), this gives us: Z 1 5 3 1 1 15 √ (− ln x)5/2 dx = × × × Γ = π 2 2 2 2 8 0 as required. iii. Using the substitution u = vt, we have: Z 1 Z 1 Z 1 Z t dv 1 1 t dv du 1/2−1 1/2−1 √ √ √ v (1−v) dv = B = = = , =π √ √ √ 2 2 u t−u v 1−v vt t − vt 0 0 0 0 √ using part (i). We also know that the given integral is the convolution of 1/ t with itself which means that, by the convolution theorem, we have: Z t 2 du 1 π =L L √ = L{π} = . √ √ s u t−u t 0 √ Therefore, noting that the Laplace transform of 1/ t must be positive, we see that: r 1 π L √ = s t as expected. Question 3 (a) Determine the limits: ln t i. lim √ t→∞ t and ii. lim and ii. 2ln(1+t) − 1 t t→0 . (b) Determine whether the integrals: Z ∞ i. 1 ln t t2 Z dt 1 2ln(1+t) − 1 t3 0 dt are convergent or divergent. (c) A transformation from (x, y) to (r, θ) is defined by: x = r cos θ i. Find the Jacobian, ∂(x, y) ∂(r, θ) and y = r sin θ. , of this transformation. ii. The region D in the positive quadrant is given by the inequalities: 1 ≤ x2 + y 2 ≤ 9 and √ y 1 ≤ 3. √ ≤ x 3 Sketch the region D and its image, ?, under this transformation. ZZ y −1 iii. Hence evaluate the double integral tan dx dy. x D 10 Examiners’ commentaries 2020 Reading for this question For (a), Section 2.1.2 of the subject guide. For (b), Chapter 4 of the subject guide. For (c), Section 5.2 of the subject guide. Approaching the question (a) i. As the numerator and denominator both tend to infinity as t → ∞, we can use L’Ho?pital’s rule to see that: 1/t 2 ln t √ = lim √ = 0 lim √ = lim t→∞ t→∞ t 1/(2 t) t t→∞ is the sought-after limit. ii. As the numerator and denominator both tend to zero as t → 0, we can use L’Ho?pital’s rule to see that: 2ln(1+t) − 1 (ln 2)2ln(1+t) /(1 + t) (ln 2)2ln(1+t) = lim = lim = ln 2 t→0 t→0 t→0 t 1 1+t lim is the sought-after limit. (b) i. This is an improper integral of the first kind and, as the integrand is: f (t) = ln t t2 it makes sense to let g(t) = 1/t3/2 so that: lim t→∞ f (t) ln t = lim √ = 0 g(t) t→∞ t using what we saw in (i) from part (a). Then, as the integral: Z ∞ Z ∞ dt g(t) dt = t3/2 1 1 is convergent, by the Limit Comparison Test (LCT2), the given integral must be convergent too. ii. This is an improper integral of the second kind as the integrand: f (t) = 2ln(1+t) − 1 t3 is undefined at t = 0. Looking at what we saw in (ii) from part (a), it makes sense to let g(t) = 1/t2 so that: f (t) 2ln(1+t) − 1 lim = lim = ln 2 t t→0+ g(t) t→0+ and then, as the integral: Z 1 Z 1 dt g(t) dt = 2 0 0 t is divergent, by the Limit Comparison Test (LCT1), the given integral must be divergent too. (c) i. We are given the transformation defined by: x = r cos θ and y = r sin θ and so the Jacobian of this transformation is: ∂(x, y) x = r yr ∂(r, θ) xθ cos θ = yθ sin θ −r sin θ = r cos2 θ + r sin2 θ = r(cos2 θ + sin2 θ) = r r cos θ as cos2 θ + sin2 θ = 1. 11 MT2176 Further calculus ii. The region D should look like the one illustrated in Figure 2(A). Figure 2: The sketches for Question 3 (c) part ii. Under the transformation this region becomes 1 ≤ r ≤ 3 as r2 = x2 + y 2 and: √ 1 √ ≤ tan−1 θ ≤ 3 3 π π ≤θ≤ 6 3 ⇒ as tan θ = y/x. A sketch of its image, ?, should look like the one illustrated in Figure 2(B). iii. Using what we have seen in parts (i) and (ii) we can evaluate the integral as follows: ! ZZ Z 3 Z π/3 −1 y −1 tan dx dy = tan (tan θ)r dθ dr x D r=1 θ=π/6 ! Z Z 3 π/3 r=1 θ=π/6 = rθ dθ dr 3 θ2 = r 2 r=1 Z π2 = 24 Z π/3 dr π/6 3 r dr r=1 = 3 π2 r2 24 2 1 = π2 (4) 24 = π2 . 6 Question 4 (a) You may assume that, under the customary assumptions, L{f 0 (t)} = sL{f (t)} − f (0). Deduce that L{f 00 (t)} = s2 L{f (t)} − sf (0) − f 0 (0). Hence solve the differential equation: f 00 (t) − 3f 0 (t) + 2f (t) = t when f (0) = 1 and f 0 (0) = 1. 12 Examiners’ commentaries 2020 (b) For t > 0, the function F is given by: Z F (t) = ∞ 0 e−tu 1+u du. i. Show that, for t > 0, |F (t)| ≤ 1/t and deduce that F (t) → 0 as t → ∞. ii. Show that: dF = F (t) − 1 dt justifying any manipulations you perform. iii. Hence show that: −t e ∞ Z F (t) = t e−x x t dx. Reading for this question For (a), Section 7.2.1 of the subject guide. For (b), Section 6.2 of the subject guide. Approaching the question (a) As f 00 (t) is the derivative of f 0 (t) we can use the given fact twice to see that: L{f 00 (t)} = sL{f 0 (t)} − f 0 (0) = s (sL{f (t)} − f (0)) − f 0 (0) = s2 L{f (t)} − sf (0) − f 0 (0) as required. Taking the Laplace transform of both sides of the differential equation: f 00 (t) − 3f 0 (t) + 2f (t) = t we get: 1 s2 f˜(s) − sf (0) − f 0 (0) − 3 sf˜(s) − f (0) + 2f˜(s) = 2 . s It is, of course, essential that you get this correct if you want to get any significant credit for what follows! Now, using the initial conditions f (0) = 1 and f 0 (0) = 1 as well, this gives us: 1 1 s2 f˜(s) − s − 1 − 3 sf˜(s) − 1 + 2f˜(s) = 2 ⇒ (s2 − 3s + 2)f˜(s) = 2 + s − 2 s s so that we have: s3 − 2s2 + 1 (s − 1)(s − 2)f˜(s) = s2 ⇒ f˜(s) = s3 − 2s2 + 1 A B C D = + 2+ + − 1)(s − 2) s s s−1 s−2 s2 (s for some constants A, B, C and D. (Alternatively, notice that s − 1 is a factor of the numerator (as 13 − 2(12 ) + 1 = 0) and so, on factorising, we have: s3 − 2s2 + 1 = (s − 1)(s2 − s − 1) so that: s2 − s − 1 A B C f˜(s) = 2 = + 2+ s (s − 2) s s s−2 which is a slightly simpler partial fractions decomposition to work with.) To find these constants, we can (say) observe that we need: s3 − 2s2 + 1 = As(s − 1)(s − 2) + B(s − 1)(s − 2) + Cs2 (s − 2) + Ds2 (s − 1) so that setting s = 0 gives us B = 1/2, setting s = 1 gives us C = 0, setting s = 2 gives us D = 1/4 and, looking at the s3 coefficient (say), we also get 1 = A + C + D so that A = 3/4 if we use our values for C and D. That is, we find that: 3/4 1/2 1/4 f˜(s) = + 2 + s s s−2 13 MT2176 Further calculus which means that: f (t) = 3 t 1 + + e2t 4 2 4 is the required solution. (b) i. For t > 0, we have: Z F (t) = 0 ∞ e−tu du and 1+u 0≤ e−tu ≤ e−tu 1+u for u ≥ 0. Therefore, as F (t) > 0 for t > 0, we have: −tu ∞ Z ∞ Z ∞ −tu e 1 e −tu e du = − du ≤ = |F (t)| = 1 + u t t 0 0 0 as required. This means that: 1 1 ≤ F (t) ≤ t t and so, as both −1/t and 1/t tend to zero as t → ∞, we can use the Sandwich theorem to conclude that F (t) → 0 as t → ∞. − ii. We want to argue that: −tu Z Z ∞ Z ∞ dF d ∞ e−tu ∂ e −u −tu = du = du = e du dt dt 0 1 + u ∂t 1 + u 1+u 0 0 and, in order to manipulate the integral in this way, we need to be sure of the following. • The functions: e−tu −u −tu and e 1+u 1+u are clearly jointly continuous for t > 0 and u ≥ 0. • The function: −u −tu e 1+u has dominated convergence for t > 0 and u ≥ 0 since, for any t > 0, there is a c such that 0 < c < t which gives us: u −tu −u −tu = e e ≤ e−tu ≤ e−cu 1+u 1+u and the integral: Z ∞ e−cu du 0 is convergent as c > 0. As such, we can now go on and see that: −tu ∞ Z ∞ Z ∞ Z ∞ −tu −u −tu 1 e e 1 dF −tu = e dt = −1 e dt = du− − = F (t)− dt 1 + u 1 + u 1 + u t t 0 0 0 0 as required. iii. Using part (ii) and the dummy variable x, we can see that: 1 dF − F (x) = − dx x is a linear first-order differential equation. In this case, the integrating factor is e−x and so, following the standard method, we have: Z ∞ −x h i∞ e d −x e−x (e F (x)) = − ⇒ e−x F (x) =− dx dx x x t t 14 Examiners’ commentaries 2020 if we integrate both sides with respect to x from x = t to infinity. That is, using part (i), we have: Z ∞ −x Z ∞ −x e e −x lim (e F (x)) − F (t) = − dx ⇒ F (t) = dx x→∞ x x t t as required. 15 1. (a) For t > 0, let √ Z I(t) = 1 t ln(x2 + t) dx. x2 Find I 0 (t) and hence find I 0 (3) and lim tI 0 (t). t→∞ (b) Use Laplace transforms to find the functions f (t) and g(t) that satisfy the differential equations f 0 (t) = g(t) + 2f (t) and g 0 (t) = 3f (t) + et with f (0) = 0 and g(0) = 0. 2. (a) Find the limits (i) x , lim x x→0 e − cos(x) (ii) x2 lim x→0 ln(1 + x2 ) and (iii) x lim e ln x→∞ ex + 1 ex (b) Use the limit comparison test to determine whether the integrals Z 1 Z 1 dx x dx (i) and (ii) x 2 0 e − cos(x) 0 ln(1 + x ) are convergent or divergent. (c) Use the limit comparison test to find the values of p ∈ R for which the integral Z ∞ p x tan−1 (x) dx 1 + x2 0 is convergent. UL21/0154 Page 3 of 4 . 3. (a) Consider the repeated integral Z 1 Z e xy dy dx. x=0 y=ex ln(y) Sketch the region of integration and, by changing the order of integration, evaluate this repeated integral. (b) Consider the repeated integral Z 1 Z x x3 I= dy dx. 2 2 3/2 x=0 y=0 (x + y ) (i) Sketch the region of integration, D, in the xy-plane. (ii) Sketch the image of the region of integration, ?, in the rθ-plane where (r, θ) are the polar coordinates of the point (x, y). (iii) Hence evaluate the repeated integral I. 4. (a) Show that the function given by h(u) = eu − e−u , u for u 6= 0 and h(0) = 2 is continuous for all u ∈ R. (b) Consider the function given by L(t, x) = ext − e−xt , t for t 6= 0 and L(0, x) = 2x. Show that L(t, x) = xh(tx) using the function in part (a). Deduce that L(t, x) is jointly continuous for all (t, x) ∈ R2 . (c) Let d be any number such that 0 < d < 1. Show that the function K(t, x) = e−t (ext + e−xt ) has dominated convergence for t ≥ 0 and −d ≤ x ≤ d. (d) For |x| < 1, find the integral Z ∞ I(x) = e−t (ext + e−xt ) dt. 0 (e) Hence, for |x| < 1, find Z ∞ ext − e−xt e−t dt. t 0 You should clearly explain what you are doing. END OF PAPER UL21/0154 Page 4 of 4 4a) hu) = sel - uto u U=0 u uto hu) 2 u=0 2u uto h (u) = ueu u=0 2 clearly, the definition for u to i es continuous for all values where we touto Now, L= lim e zu lo form) ue u uto lim 2e²u [Applying Islapital uso elut u So L= lim uto 2e uti 2011-2 So lim hu)=2=hlo So, hu) is continuous for all UER 2b) (i) As shown in ala) part (1) 6 X (1 1. lim xto eo- cos(x) g(x)= t a he So let let (a)= T e-wsx DC So f(x) () T g(x) e - Cosa T= lim , ?? ) As lim fin) where limm f/1.79 918) > so using linit comparison test So f(x) dx converges if and only if So gla) dx converges so ging da- s't da= [en 181}, 0-(-0)=diverges Daiwego Só (1) dx = So' ad also divedges e-cosx So, using limit comparison test solat (1.797 ( let f(x) = ln (1+x2) ga) = 2 x ${ gia) = so lullta) lim fa) ) So using limit comparison test, using 82001) xo glu S flwax converges if 4 only if So g(x) dx converges 1 1/2 dx diverges S x so s ftude = . ? da also diverges en 11+8²) So glador de = 78 ? w -u 46) h(U)= e-e u xt -ext hltx) = e xt xt ? e -xt -e tto . zhltx)= t t=0 X(2)=2x Xhltx) = by the given definition: Llt, x) = xhita) As, hlu) is continuous for all uer So xhita) is continuous for (60) ER? so, ult, &) is jointly continuous for all (t, x) ER²

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