Unit 3. Integration. 3A. Differentials, indefinite integration. y x. c) Method 1 (slow way) Substitute: u = 8 + 9x, du = 9dx.

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1 Unit 3. Integration 3A. Differentials, indefinite integration 3A- a) 7 6 d. (d(sin ) = because sin is a constant.) b) (/) / d c) ( 9 8)d d) (3e 3 sin + e 3 cos)d e) (/ )d + (/ y)dy = implies dy = / d / y 3A- a) (/5) / c b) (/3) 3/ + / + c y = d = ( d = ) d c) Method (slow way) Substitute: u = 8 + 9, du = 9d. Therefore 8 + 9d = u / (/9)du = (/9)(/3)u 3/ + c = (/7)(8 + 9) 3/ + c Method (guess and check): It s often faster to guess the form of the antiderivative and work out the constant factor afterwards: Guess (8 + 9) 3/ ; d d (8 + 9)3/ = (3/)(9)(8 + 9) / = 7 (8 + 9)/. So multiply the guess by to make the derivative come out right; the answer is then 7 7 (8 + 9)3/ + c d) Method (slow way) Use the substitution: u = 4, du = 48 3 d. 3 ( 4 ) /8 d = u /8 ( /48)du = 48 (8/9)u9/8 + c = 54 ( 4 ) 9/8 + c Method (guess and check): guess ( 4 ) 9/8 ; d d ( 4 ) 9/8 = 9 8 ( 483 )( 4 ) /8 = 54( 4 ) /8. So multiply the guess by to make the derivative come out right, getting the previous 54 answer. e) Method (slow way): Use substitution: u = 8, du = 4d. d = u / ( /4)du = u3/ + c = 6 (8 ) 3/ + c

2 S. SOLUTIONS TO 8. EXERCISES Method (guess and check): guess (8 ) 3/ ; differentiating it: d d (8 ) 3/ = 3 ( 4 )(8 ) / = 6(8 ) / ; so multiply the guess by 6 to make the derivative come out right. The net four questions you should try to do (by Method ) in your head. Write down the correct form of the solution and correct the factor in front. f) (/7)e 7 + c g) (7/5)e 5 + c h) e + c i) (/3) ln(3 + ) + c. For comparison, let s see how much slower substitution is: u = 3 +, du = 3d, so d (/3)du 3 + = = (/3)lnu + c = (/3)ln(3 + ) + c u j) k) ( + 5 d = + 5 ) d = + 5 ln + c ( + 5 d = 5 ) d = 5 ln( + 5) + c + 5 In Unit 5 this sort of algebraic trick will be eplained in detail as part of a general method. What underlies the algebra in both (j) and (k) is the algorithm of long division for polynomials. l) u = ln, du = d/, so ln d = udu = (/)u + c = (/)(ln) + c m) u = ln, du = d/. 3A-3 a) (/5) cos(5) + c d du ln = = lnu + c = ln(ln ) + c u b) (/)sin +c, coming from the substitution u = sin or (/)cos +c, coming from the substitution u = cos. The two functions (/)sin and (/)cos are not the same. Nevertheless the two answers given are the same. Why? (See J-(m).) c) (/3)cos 3 + c d) (/)(sin) + c = (/)csc + c

3 3. INTEGRATION e) 5 tan(/5) + c f) (/7)tan 7 + c. g) u = sec, du = sectan d, sec 9 tan d (sec) 8 sec tan d = (/9)sec 9 + c 3B. Definite Integrals 3B- a) = 3 b) = 6 3B- a) c) = 5 d) + / + /3 + /4 = 5/ 6 ( ) n+ (n + ) b) n= n /k c) k= n sin(k/n) 3B-3 a) upper sum = right sum = (/4)[(/4) 3 + (/4) 3 + (3/4) 3 + (4/4) 3 ] = 5/8 lower sum = left sum = (/4)[ 3 + (/4) 3 + (/4) 3 + (3/4) 3 ] = 7/8 b) left sum = ( ) = 6; right sum = = 4; upper sum = ( ) = 5; lower sum = = 5. c) left sum = (π/)[sin + sin(π/) + sin(π) + sin(3π/)] = (π/)[ + + ] = ; right sum = (π/)[sin(π/) + sin(π) + sin(3π/) + sin(π)] = (π/)[ + + ] = ; upper sum = (π/)[sin(π/) + sin(π/) + sin(π) + sin(π)] = (π/)[ ] = π; lower sum = (π/)[sin() + sin(π) + sin(3π/) + sin(3π/)] = (π/)[ + ] = π. 3B-4 Both and 3 are increasing functions on b, so the upper sum is the right sum and the lower sum is the left sum. The difference between the right and left Riemann sums is (b/n)[f( + + f( n )] (b/n)[f( + + f( n )] = (b/n)[f( n ) f( )] In both cases n = b and =, so the formula is k= (b/n)(f(b) f()) a) (b/n)(b ) = b 3 /n. Yes, this tends to zero as n. b) (b/n)(b 3 ) = b 4 /n. Yes, this tends to zero as n. 3B-5 The epression is the right Riemann sum for the integral so this is the limit. sin(b)d = (/b)cos(b) = ( cosb)/b

4 S. SOLUTIONS TO 8. EXERCISES 3C. Fundamental theorem of calculus 3C- 6 ( ) / d = ( ) / 6 3C- a) (/3)(/3)(3 + 5) 3/ b) If n, then 3 = (/9)(3/ 5 3/ ) 3 = [(4)/ / ] = (/(n + ))(/3)(3 + 5) n+ = (/3(n + ))((n+ 5 n+ ) If n =, then the answer is (/3)ln(/5). c) (/)(cos) π 3π/4 = (/)[( ) ( / ) ] = / 3C-3 a) (/)ln( + ) = (/)[ln5 ln ] = (/)ln(5/) b) (/)ln( + b ) b b 3C-4 As b, b = (/)[ln(5b ) ln(b )] = (/)ln(5/) d = (/9) 9 b = (/9)(b 9 ) (/9)( ) = /9. This integral is the area of the infinite region between the curve y = and the -ais for >. 3C-5 a) b) π π/a sin d = cos π = (cosπ cos) = sin(a)d = (/a)cos(a) π/a = (/a)(cosπ cos) = /a 3C-6 a) 4 = implies = ±. So the area is ( 4)d = ( 4)d = = = 6/3 (We changed to the interval (, ) and doubled the integral because 4 is even.) Notice that the integral gave the wrong answer! It s negative. This is because the graph y = 4 is concave up and is below the -ais in the interval < <. So the correct answer is 6/3. b) Following part (a), a = implies = ± a. The area is a a (a )d = a (a )d = a 3 3 a = ( a 3/ a3/ ) 4 = 3 3 a3/

5 3D- Differentiate both sides; left side L(): L () = d d 3. INTEGRATION 3D. Second fundamental theorem dt a + =, by FT; a + right side R(): R () = d d (ln( + a + ) lna) = + a + + a + = a + Since L () = R (), we have L() = R() + C for some constant C = L() R(). The constant C may be evaluated by assigning a value to ; the most convenient choice is =, which gives L() = = ; R() = ln( + + a ) lna = ; therefore C = and L() = R(). b) Put = c; the equation becomes = ln(c + c + a ); solve this for c by first eponentiating both sides: = c + c + a ; then subtract c and square both sides; after some algebra one gets c = ( a ). y = - t + t y 3D-3 Sketch y = t + t first, as shown at the right. - 3D-4 a) sin(t 3 )dt, by the FT. b) sin(t 3 )dt + c) 3D-5 This problem reviews the idea of the proof of the FT. a) f(t) = t + t 4 / - sin(t 3 )dt f(t) t lim + + f(t)dt = f(t)dt = shaded area width lim b) By definition of derivative, F () = lim by FT, F () = f() =. height. shaded area width F( + ) F() = height = f() =. = lim + f(t)dt; + t 3D-6 a) If F () = dt and F () = dt, then F () = a and F () = a. a a Thus F () F () = a a, a constant. b) By the FT, F () = f() and F () = f(); therefore F = F + C, for some constant C.

6 S. SOLUTIONS TO 8. EXERCISES 3D-7 a) Using the FT and the chain rule, as in the Notes, d d b) = u sinudu = sin( ) d( ) d = sin( ) sin sin cos =. (So dt t = ) c) d tanudu = tan( ) tan d 3D-8 a) Differentiate both sides using FT, and substitute = π/: f(π/) = 4. b) Substitute = u and follow the method of part (a); put u = π, get finally f(π/) = 4 4π. 3E- L( a ) = /a 3E. Change of Variables; Estimating Integrals dt t. Put t = u, dt = du. Then u dt t = u /a u du = L( a ) = dt a t = du u = L(a) 3E- a) We want t = u /, so u = t, du = dt. π e u / du = / π / e t dt = π = E() = π F(/ ) and lim E() = π b) The integrand is even, so e t dt π = N π N e u / du = π N e u / du = E(N) as N lim E() = / because E() is odd. b e u / du = E(b) E(a) by FT or by interval addition Notes PI (3). π a Commentary: The answer is consistent with the limit, N π N e u / du = E(N) E( N) = E(N) as N

7 3E-3 a) Using u = ln, du = d, e π b) Using u = cos, du = sin, sin ( + cos) 3 d = c) Using = sin u, d = cosudu, 3. INTEGRATION ln d = du ( + u) 3 = ( + u) d udu = 3 u3/ = ( 3 ) = = π = 3. cosu cosu du = u π/ = π. 3E-4 Substitute = t/a; then = ± t = ±a. We then have π = a d = t dt a a a = a a a t dt. a Multiplying by a gives the value πa / for the integral, which checks, since the integral represents the area of the semicircle. a -a a 3E-5 One can use informal reasoning based on areas (as in E. 5, Notes FT), but it is better to use change of variable. a) Goal: F( ) = F(). Let t = u, dt = du, then F( ) = Since f is even (f( u) = f(u)), F( ) = f(t)dt = f( u)( du) f(u)du = F(). b) Goal: F( ) = F(). Let t = u, dt = du, then F( ) = Since f is odd ((f( u) = f(u)), F( ) = f(t)dt = f( u)( du) f(u)du = F(). 3E-6 a) 3 < on (,) + 3 > on (,); therefore + 3E-7 d + 3 > d + = ln( + ) = ln =.69 b) < sin < on (, π) sin < sin on (, π); therefore π π sin d < sind = cos = ( ) =. c) N + d > sin d N d = sin d N π = (4 ) = 5 d = N = N + <.

8 S. SOLUTIONS TO 8. EXERCISES 3F. Differential Equations: Separation of Variables. Applications 3F- a) y = (/)( + 5) 5 + c b) (y +)dy = d = (y +)dy = d = (/)(y +) = +c. You can leave this in implicit form or solve for y: y = ± + a for any constant a (a = c) c) y / dy = 3d = (/3)y 3/ = 3 + c = y = (9/ + a) /3, with a = (3/)c. d) y dy = d = y = / + c = y = /( / + c) 3F- a) Answer: 3e 6. y dy = 4d = lny = + c y() = 3 = ln 3 = + c = c = ln 3. Therefore lny = + (ln 3 ) At = 3, y = e 8+ln3 = 3e 6 b) Answer: y = / + 3. (y + ) / dy = d = (y + ) / = + c y() = = ( + ) / = c = c = At = 3, (y + ) / = 3 + = y + = (3/ + ) = 3/ + 3 Thus, y = / + 3. c) Answer: y = 55/3 ydy = d = y / = (/3) 3 + c y() = = c = / = 5 Therefore, at = 5, y / = (/3) = y = 55/3 d) Answer: y = (/3)(e 4 ) (3y + ) dy = d = (/3)ln(3y + ) = + c y() = = (/3)ln = c Therefore, at = 8, (/3)ln(3y + ) = 8 + (/3)ln = ln(3y + ) = 4 + ln = (3y + ) = e 4 Therefore, y = (e 4 )/3

9 Therefore, 3. INTEGRATION e) Answer: y = ln4 at =. Defined for < < 4. The solution y is defined only if < 4. e y dy = d = e y = + c y(3) = = e = 3 + c = c = 4 y = ln(4 ), y() = ln 4 3F-3 a) Answers: y(/) =, y( ) = /, y() is undefined. Therefore, /y = and y dy = d = y = + c y() = = = + c = c = y = The values are y(/) =, y( ) = / and y is undefined at =. b) Although the formula for y makes sense at = 3/, (y(3/) = /( 3/) = ), it is not consistent with the rate of change interpretation of the differential equation. The function is defined, continuous and differentiable for < <. But at =, y and dy/d are undefined. Since y = /( ) is the only solution to the differential equation in the interval (, ) that satisfies the initial condition y() =, it is impossible to define a function that has the initial condition y() = and also satisfies the differential equation in any longer interval containing =. To ask what happens to y after =, say at = 3/, is something like asking what happened to a rocket ship after it fell into a black hole. There is no obvious reason why one has to choose the formula y = /( ) after the eplosion. For eample, one could define y = /( ) for <. In fact, any formula y = /(c ) for c satisfies the differential equation at every point >. 3F-4 a) If the surrounding air is cooler (T e T < ), then the object will cool, so dt/dt <. Thus k >. b) Separate variables and integrate. Eponentiating, (T T e ) dt = kdt = ln T T e = kt + c T T e = ±e c e kt = Ae kt The initial condition T() = T implies A = T T e. Thus T = T e + (T T e )e kt c) Since k >, e kt as t. Therefore, T = T e + (T T e )e kt T e as t

10 S. SOLUTIONS TO 8. EXERCISES d) T T e = (T T e )e kt The data are T = 68, T e = 4 and T(8) =. Therefore, 4 = (68 4)e 8k = e 8k = 6/64 = /4 = 8k = ln4. The number of hours t that it takes to cool to 5 satisfies the equation 5 4 = (64)e kt = e kt = /64 = kt = 3 ln4. To solve the two equations on the right above simultaneously for t, it is easiest just to divide the bottom equation by the top equation, which gives t = 3, t = 4. 8 e) T T e = (T T e )e kt The data at t = and t = are 8 T e = ( T e )e k and 7 T e = ( T e )e k Eliminating e k from these two equations gives ( ) 7 T e 8 Te = T e T e (8 T e ) = ( T e )(7 T e ) 8 6T e + T e = ()(7) 7T e + T e f) To confirm the differential equation: The formula for y is T e = ()(7) 8 T e = 7 64 = 6 y (t) = T (t t ) = k(t e T(t t )) = k(t e y(t)) y(t) = T(t t ) = T e + (T T e )e k(t t) = a + (y(t ) a)e c(t t) with k = c, T e = a and T = T() = y(t ). 3F-6 y = cos 3 u 3 cosu, = sin 4 u dy = (3 cos u ( sinu) + 3 sinu)du, d = 4 sin 3 u cosudu dy d = 3 sinu( cos u) 4 sin 3 = 3 u cosu 4 cosu

11 3F-7 a) y = y; y() = dy y = d = lny = + c 3. INTEGRATION To find c, put =, y = : ln = + c = c =. = lny = = y = e / b) cossin ydy = sind; y() = sinydy = sin d = cosy = ln(cos) + c cos Find c: put =, y = : cos = ln(cos ) + c = c = = cosy = ln(cos) + P 3F-8 a) From the triangle, y = slope tangent = y = dy y = d = lny = + c = y = e +c = Ae (A = e c ) b) If P bisects tangent, then P bisects OQ (by euclidean geometry) So P Q = ( since OP = ). Slope tangent = y = y = dy y = d P = lny = ln + c Eponentiate: y = = c ec, c > Ans: The hyperbolas y = c, c > 3G. Numerical Integration 3G- Left Riemann sum: ( )(y + y + y + y 3 ) Trapezoidal rule: ( )((/)y + y + y + y 3 + (/)y 4 ) Simpson s rule: ( /3)(y + 4y + y + 4y 3 + y 4 ) a) = /4 and y =, y = /, y = /, y 3 = 3/, y 4 =. Left Riemann sum: (/4)( + / + / + 3/).58 Trapezoidal rule: (/4)((/) + / + / + 3/ + (/)).643 Simpson s rule: (/)( + 4(/) + (/ ) + 4( 3/) + ).657 as compared to the eact answer b) = π/4 y =, y = /, y =, y 3 = /, y 4 =.

12 S. SOLUTIONS TO 8. EXERCISES Left Riemann sum: (π/4)( + / + + / ).896 Trapezoidal rule: (π/4)((/) +/ ++/ +(/) ).896 (same as Riemann sum) Simpson s rule: (π/)( + 4(/ ) + () + 4(/ ) + ).5 as compared to the eact answer c) = /4 y =, y = 6/7, y = 4/5, y 3 = 6/5, y 4 = /. Left Riemann sum: (/4)( + 6/7 + 4/5 + 6/5).845 Trapezoidal rule: (/4)((/) + 6/7 + 4/5 + 6/5 + (/)(/)).88 Simpson s rule: (/)( + 4(6/7) + (4/5) + 4(6/5) + (/)) as compared to the eact answer π/ (Multiplying the Simpson s rule answer by 4 gives a passable approimation to π, of 3.457, accurate to about 5.) d) = /4 y =, y = 4/5, y = /3, y 3 = 4/7, y 4 = /. Left Riemann sum: (/4)( + 4/5 + /3 + 4/7).76 Trapezoidal rule: (/4)((/) + 4/5 + /3 + 4/7(/)(/)).697 Simpson s rule: (/)( + 4(4/5) + (/3) + 4(4/7) + (/)).6935 Compared with the eact answer ln.6935, Simpson s rule is accurate to about 4. b 3G- We have 3 d = b4. Using Simpson s rule with two subintervals, = b/, so 4 that we get the same answer as above: S( 3 ) = b 6 ( + 4(b/)3 + b 3 ) = b 6 ( ) 3 b3 = b4 4. Remark. The fact that Simpson s rule is eact on cubic polynomials is very significant to its effectiveness as a numerical approimation. It implies that the approimation converges at a rate proportional to the the fourth derivative of the function times ( ) 4, which is fast enough for many practical purposes. 3G-3 The sum S = , is related to the trapezoidal estimate of 4 d : y = 3,

13 3. INTEGRATION () But From (), () 4 d = S 4 4 d = 3 3/ S 5 = 3 6 Hence (3) S 666, 77 In (), we have >, as in the picture. Hence in (), we have >, so in (3), we have <, Too high. y = 3G-4 As in Problem 3 above, let S = n Then by trapezoidal rule, 3 n- n Since n n d n = S n d = lnn, we have S lnn + +. (Estimate is too low.) n 3G-5 Referring to the two pictures above, one can see that if f() is concave down on [a, b], the trapezoidal rule gives too low an estimate; if f() is concave up, the trapezoidal rule gives too high an estimate..