2_calculus_tutorial_3_solutions.pdf - School of Computing Electronics and Mathematics 101MS Engineering Mathematics I Week 6  tutorial solutions

2_calculus_tutorial_3_solutions.pdf - School of Computing...

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Unformatted text preview: School of Computing, Electronics and Mathematics 101MS Engineering Mathematics I Week 6  tutorial solutions Questions 1) 8 marks ˆ true ˆ false ˆ false ˆ true ˆ false ˆ true ˆ false ˆ true 2) 14 marks ˆ Z 2 x x −1/2 x +e Z dx = x 2 Z dx + ex dx = x3 + ex + c. 3 ˆ Z 2 √ − 3 sinh x dx = 2 x Z Z dx − 3 sinh x dx = 2 √ x1/2 − 3 cosh x + c = 4 x − 3 cosh x + c. (1/2) ˆ Z (3x) −3/2 −3/2 Z −3/2 + 4 tanh(2x) dx = 3 x Z dx + 4 tanh(2x) dx = 3−3/2 x−1/2 + 2 ln[cosh(2x)] + c −(1/2) 2 = − √ + 2 ln[cosh(2x)] + c. 3 3x ˆ Let u = 2 + 4x du then dx Z (2 + 4x) ˆ Let u = 2x + 3 1/4 du then dx dx = Let du then dx Z e ˆ Let u = 2x dx Z u = 2(x + 1) then sin u = 2 =⇒ 2(x+1) v 0 = sin x. = du 2 dx Z dx = e Z 2x sin x dx = −2x cos x + 4 . Therefore Z u1/4 du = 1 u5/4 (2 + 4x)5/4 = + c. 4 (5/4) 5 du 2 . Therefore 1 2 = = Z sin u du = − cos u cos(2x + 3) =− + c. 2 2 du 2 . Therefore u du 1 = 2 2 u0 = 2 Z eu du = eu e2(x+1) = + c. 2 2 v = − cos x. Hence Z 2 cos x dx = −2x cos x + 2 cos x dx = −2x cos x + 2 sin x + c. Therefore Z du 1 = 4 4 u = 2 =⇒ sin(2x + 3) dx = = 1/4 du Z Z ˆ dx = 4 =⇒ and ˆ Let u = ln x v 0 = x2 . Therefore u0 = x−1 and v = x3 /3. Hence   Z Z Z x3 ln x 1 x3 x3 ln x 1 x3 ln x x3 x3 1 2 2 x ln x dx = − dx = − x dx = − +c = − + ln x +c 3 x 3 3 3 3 9 3 3 ˆ Let u = sin x then v 0 = cos x. Therefore u0 = cos x and v = sin x. Hence Z Z Z 2 sin x cos x dx = sin x − sin x cos x dx =⇒ 2 sin x cos x dx = sin2 x Z sin2 x =⇒ sin x cos x dx = + c. 2 then 3) 1 mark Let u = x3 − 3x2 + 1 then du/dx = 3x2 − 6x =⇒ dx = du/(3x2 − 6x), therefore Z Z Z 3x2 − 6x 3x2 − 6x du 1 3 2 dx = = du = ln u = ln(x − 3x + 1) + c. 3 2 2 x − 3x + 1 u 3x − 6x u 4) 1 mark Let v0 = 1 then u = ln x. Therefore v = x and u0 = x−1 . Hence Z Z x ln x dx = x ln x − dx = x ln x − x + c = x(−1 + ln x) + c. x 5) 6 marks ˆ Z 1 2 x3 + ex x + e dx = 3 2 x  2 = 1 8 1 7 + e2 − − e = + e(e − 1) = 7.0041. 3 3 3 ˆ √ √ 2 √ − 3 sinh x dx = [4 x − 3 cosh x]21 = 4 2 − 3 cosh 2 − 4 + 3 cosh 1 x √ 3(1 − e)(e − e−2 ) = 4( 2 − 1) + = −5.0005. 2 2 Z 1 ˆ 2 Z (3x) −3/2 1  2 2 + 4 tanh(2x) dx = − √ + 2 ln[cosh(2x)] 3 3x 1   1 1 = 2 − √ + ln(cosh 4) + √ − ln(cosh 2) 3 6 3 3 " # √ 2−1 = 4.0771. = 2 ln(cosh 4 sech 2) + √ 3 6 ˆ Z 0 ˆ π/2   cos(2x + 3) π/2 cos(π + 3) cos 3 sin(2x + 3) dx = − =− + 2 2 2 0 cos π cos 3 − sin π sin 3 cos 3 =− + = cos 3 = −0.9900. 2 2 Z 0 π/2 π/2 2x sin x dx = [−2x cos x + 2 sin x]0 ˆ Z 0 π/2 sin2 x sin x cos x dx = 2  = −4 cos π/2 = 0 π π + 2 sin − 2 sin 0 = 2.0000. 2 2  1 2π sin − sin2 0 = 0.5000. 2 2 6) 1 mark 2 Z Z −t 2 e−t dt = [20t]20 − [−20e−t ]20 = [20t + 20e−t ]20 0 0 0 2 20 dt − 20 20(1 − e ) dt = s= Z = 20(1 + e−2 ) = 22.7067. 7) 1 mark Z b s= a 2mx3 2m dx = 2 2 2 b −a b − a2 Z a b  4 b 2m x m(b4 − a4 ) x dx = 2 = 2 b −a 4 a 2(b2 − a2 ) 3 = m(b2 + a2 )(b2 − a2 ) m(a2 + b2 ) = . 2(b2 − a2 ) 2 8) 4 marks −1  4 x2 1 4 2 12 16 32 8 24 25 x − 3x =− − + + + + − − = , −x3 + 3x2 + x − 3 dx = − + x3 + 4 2 4 4 4 4 4 4 4 4 4 −2 −2  4  Z 1 1 x x2 1 4 2 12 1 4 2 12 16 −x3 + 3x2 + x − 3 dx = − + x3 + − 3x =− + + − + + − − =− , 4 2 4 4 4 4 4 4 4 4 4 −1 −1  4  Z 3 3 x x2 81 108 18 36 1 4 2 12 16 −x3 + 3x2 + x − 3 dx = − + x3 + − 3x = − + + − + − − + = , 4 2 4 4 4 4 4 4 4 4 4 1 1 Z −1 Therefore the area encompassed by the plot is (25 + 16 + 16)/4 = 57/4. 9) 2 marks Z Since v(t = 0) = 0, then c1 Z a dt = −2 v= = 0. cos(2πt) dt = − Therefore v=− Hence Since sin(2πt) + c1 . π sin(2πt) . π Z x(t = 0) = 0, then Z 1 cos(2πt) x = v dt = − sin(2πt) dt = + c2 . π 2π 2 c2 = −1/2π 2 . Therefore x= cos(2πt) − 1 . 2π 2 10) 2 marks Z ln 2 1 2 3 2 2 cosh t dt = [sinh t]ln = = 2.1640, 0 ln 2 0 ln 2 2 ln 2 s r r Z ln 2 4 2 15 2 ln 2 = cosh t dt = [t + sinh t cosh t]0 = 2 + ln 2 0 ln 2 ln 256 ymean = yrms = 2.1691. Total: 40 marks ...
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