NATIONAL UNIVERSITY OF SINGAPORE Deprtment of Mthemtics Semester, /3 MA55 Mth I Suggested Solutions to T. 3. Using the substitution method, or otherwise, find the following integrls. Solution. ) b) x sin(x 3 +)dx = csc t cot tdt= sin(x 3 +) 3 d(x 3 +)= 3 cos(x 3 +)+C. cot td(cot t) = 4 cot t + C. c) θ sin θ cos θ dθ = sin θ cos θ ( ) d ( θ ) = sin ( θ d sin ) = θ sin θ + C. d) 8 tn x sec x ( + tn 3 x) 8 tn xd(tn x) 6 d(tn 3 x +) dx = ( + tn 3 = x) ( + tn 3 =6ln tn 3 x + + C. x) e) sin θ θ cos 3 θ dθ = (cos θ) 3 d(cos θ)=(cos θ) + C =sec θ + C.. Applying the method of integrtion by prts, or otherwise, find the following integrls. Solution. () ) x sin dx = = = ) ) xd cos = x cos ) x cos ) x cos sin ) cos ) + C. ) d + C ) cos dx + C (b) t e 4t dt = 4 = 4 = 4 t d(e 4t )= 4 t e 4t t e 4t t e 4t ( te 4t te 4t dt ) e 4t dt + C + C = 4 t e 4t ) (te 4t e4t + C (continue to simplify). 4 td(e 4t ) + C
(c) e y cos ydy= e y d(sin y) =e y sin y + e y sin ydy+ C e y d(cos y)+c = e y sin y e y cos y = e y sin y e y cos ydy= e y (sin y cos y)+c. e y cos ydy (There is no hrm to renme C/ sc.) (d) θ sin(θ) dθ = = = = θ dcos(θ) = θ cos(θ) θ cos(θ) dθ + C θ cos(θ) θdsin(θ) + C θ cos(θ) θ sin(θ)+ sin(θ) dθ + C θ cos(θ) θ sin(θ) cos(θ) + C. (e) z(ln z) dz = (ln z) d(z )= z (ln z) = z (ln z) (ln z) d(z ) + C = z (ln z) z (ln z)+ zdz + C = z (ln z) z (ln z)+ z + C. z(ln z) dz + C 3. ) Wht vlues of nd b with <b mximize the vlue of b (x x ) dx? b) Wht vlues of nd b ( <b) minimize the vlue of b (x 4 x ) dx? Solution. ) The integrnd x x = x(x ) is positive for x in (, ), nd is non-positive otherwise. With integrl b of f(x) from x = to x = b : f(x) dx interpreted s lgebric re of the region under the grph integrl = re of subregion bove x-xis re of subregion below x-xis,
the integrl gets its lrgest vlue if it is tken over (, ), i.e. with = nd b =. 3
b) Likewise the integrnd x 4 x = x (x ) = x (x + )(x ) is negtive or when x is in,, but is positive otherwise. By the sme line of resoning s in ), the integrl is the lest (lgebriclly) if = ndb =. 4. Evlute the following integrls: (see the solution). Solution. ) b) 4 4 s + s s ds = x dx = 4 ( + s 3/ ) ds =( ) s / =( ) 4 +=+ 3/4. xdx+ ( x) dx = 4 4 + 4 =6. c) π π/ (cos x + cos x ) dx = = π/ π (cos x + cos x ) dx + π/ cos xdx+=sinx π/ =. (cos x + cos x ) dx 4
d) π sin ( + θ ) π dθ = cos( + θ) dθ = π sin( + θ) π = π sin( + π) sin = π +sin. 5. The Fundmentl Theorem of Clculus (I) sys tht d du u f(t) dt = f(u) for continuous function f.here is fixed number. It is sort of chin rule to find To see this, let It follows tht Furthermore, By the chin rule, we hve F (u) = d dx u df du = d du g(x) f(t) dt. f(t) dt nd u = g(x). u F g(x) =F (g(x)) = f(t) dt = f(u). g(x) f(t) dt. df (g(x)) dx = df dg(x) du dx = f(u) g (x) =f(g(x)) g (x). ) y = b) y = c) y = x x sin x cos tdt; Solution. cos x d cos x x = dx x. cos tdt; Solution. cos x x =x cos x =x cos x. dt, x < π t. Solution. sin x d dx sin x = cos x =. cos x 6. For x>, the error function E(x) := x e t dt, π importnt in the theories of het flow, signl trnsmission nd probbility, is required to be evluted numericlly becuse no elementry expression for the ntiderivtive of e t hs been found. Apply Simpson s Rule with n = to estimte E(). (/ π.837967.) 5
Solution. n =, h =. so t =,t =.,t =.,...,t 9 =.9,t =, i.e. t i = i/. Let y i = y(t i )= π e i. y =.837967, y =.7566, y =.8434787, y 3 =.3699, y 4 =.9654988, y 5 =.878785789, y 6 =.78743437, y 7 =.6974864, y 8 =.594985786, y 9 =.5968574, y =.4574974. S = 3. ( y +4y +y +4y 3 +y 4 +4y 5 +y 6 +4y 7 +y 8 +4y 9 + y ).84773. Remrk The computer shows tht the bove nswer is correct up to 5 deciml plces. 7. Find the re of the following region: ) The region bounded between y = sec x, y = 4sin x, x = π 3 nd x = π 3. b) The region in the first qudrnt bounded by y = x, y = 4 x nd y =. c) The region bounded by y =4 x, y = x, x = nd x =3. Ans. ) Solution. 4 3 π, b) 5 6, c) 49 6. ) Observe tht sec x>nd 4sin x on π/3, π/3. 6
π/3 Are = π/3 = tn x + sec x ( 4sin x) dx π/3 ( cosx) dx π/3 =tn π 3 +(x sin x) π/3 π/3 = 3+ 4 3 π sinπ 3 = 4 3 π. b) (There is some mbiguity in the question.) The points of intersection: x = x /4 implies x =orx = 4. Hence the points of intersection re (, ) nd (4, 4). We compute both the res bove nd below the line y =. Notethty = x /4 x = y. Are bove the line = Are below the line = 4 4 y (y) dy = 3 y3/ 4 y = 8 3 5 6 = 6. 4 y (y) dy = 3 y3/ y = 4 3 = 5 6. 7
c) We hve tht ( x) (4 x )=x x =(x +)(x ) is negtive if nd only if x (, ). Hence Are = 3 = + ( x) (4 x ) dx 3 = 3 = 3 x3 3 x x = 3 (x x ) dx + (x x ) dx (x x ) dx 3 x3 x x (7 + 8) (8 + ) (9 4) (4 ) 5 (3) = 3 7 + += 49 6. 8
8. ) Find the volume of the solid generted by revolving the region between the prbol x = y + nd the line x = 3 bout the line x =3. b) The region bounded by the prbol y = x nd the line y =x in the first qudrnt is revolved bout the y-xis to generte solid. Find the volume of the solid. Ans. ) 64 8 π, b) 5 3 π. Solution. ) The prbol nd the line meet t (x, y) with 3=y +, i.e. t (3, ± ). By formul, Volume = π (y +) 3 dy = π y 4 4y +4 dy = π 5 y5 4 3 y3 +4y = π 5 4 4 3 +4 = 64 π. 5 9
b) The prbol nd the line meet t (x, y) with x =x, i.e. t (, ) nd (, 4). Now y =x x = y/ nd y = x x = y, while y (y/) = y ( y/) is positive for y (, 4). So x = y is the outer curve nd x = y/ is the inner curve. Hence, volume = volume of spce enclosed by outer shell volume of hole enclosed by inner shell = 4 π y dy 4 ( y ) π dy = π 4 π 4 4 3 3 = 8 3 3 π.