Unit #11 - Integration by Parts, Average of a Function

Transcription

1 Unit # - Integration by Parts, Average of a Function Some problems and solutions selected or adapted from Hughes-Hallett Calculus. Integration by Parts. For each of the following integrals, indicate whether integration by substitution or integration by parts is more appropriate. Do not evaluate the integrals. (a sin( d (b (c (d (e (f (g + 3 d e d cos( 3 d 3 + d sin d ln d (a This integral can be evaluated using integration by parts with u =, dv = sin d. (b We evaluate this integral using the substitution w = + 3. (c We evaluate this integral using the substitution w =. (d We evaluate this integral using the substitution w = 3. (e We evaluate this integral using the substitution w = 3 +. (f This integral can be evaluated using integration by parts with u =, dv = sin d. (g This integral can be evaluated using integration by parts with u = ln, dv = d.. To practice computing integrals by parts, do as many of the problems from this section as you feel you need. The problems trend from simple to the more comple. For Questions # to #, evaluate the integral. t sin t We choose u = t and dv = sin t so du = and v = sin t or du = v = cos(t t sin }{{ t } = t ( cos t ( cos t u dv u v v du = t cos t + cos t = t cos t + sin t + C As always, we can check our integral is correct by differentiating: d ( t cos t + sin t + C = cos t t( sin t + cos t = t sin t = original integrand. " 3. te t We choose u = t and dv = e t so du = and v = e t or du = v = e t / t e } t {{ } = t (e t / (e t / u dv u v v du = tet e t = tet ( e t / + C = tet et + C As always, we can check our integral is correct by differentiating: ( d te t et + C = (et + t(e t et = te t + et et = te t = original integrand. "

2 From now on, for brevity, we won t show quite as many steps in the solution, nor will we check the answer by differentiating. However, you should always remember that you can check your antiderivatives/integrals by differentiation. 4. pe.p dp ln d = (ln d = ln d = ln + C 7. y ln y dy We choose u = p so du = dp and dv = e.p dp and v = e.p /(. p e.p dp = p e.p /(. u dv u. (z + e z dz v e.p /(. v = pe.p. + e.p dp = pe.p + ( e.p /(. + C We choose u = (z + = pe.p e.p + C and dv = e z dz so du = dz and v = e z / (z + e z dz = (z + e z / e z /dz (z + ez = e z dz (z + ez = ( e z / + C (z + ez = 4 ez + C There is no need for further simplifications. 6. ln d In this problem, we recall that the only simple calculus formula related to ln is that its derivative is known: d d ln( = /. This means that we have to select u = ln so that it will be differentiated. We choose u = ln so du = d and dv = d and v = dp du In this problem, we recall that the only simple calculus formula related to ln y is that its derivative is d known: dy ln(y = /y. While it might be tempting to keep with our earlier pattern of choosing u = y and dv = ln y dy, that won t work because we won t be able to integrate dv. As a result, we choose u = ln y and dv = y dy so du = y dy and v = y / y ln y dy = (ln y(y / (y / y dy = y ln y y dy = y ln y (y / + C 8. 3 ln d = y ln y We choose u = ln y 4 + C and dv = 3 d so du = d and v = 4 /4 ( 3 ln d = (ln 4 /4 ( 4 /4 d = 4 ln 3 d 4 4 = 4 ln 4 4 (4 /4 + C 9. q ln(q dq = 4 ln C

3 We choose u = ln(q and dv = q dq so du = q ( dq and v = q6 /6 ( q ln(q dq = (ln(q(q 6 /6 (q 6 /6 q dq. t sin t = q6 ln(q 6 = q6 ln(q 6 = q6 ln(q 6 6 q dq 6 (q6 /6 + C 36 q6 + C No further simplification is necessary for a pure evaluate the integral question.. cos(3 d We choose u = so du = d and dv = cos(3 d and v = sin(3/3 cos(3 d = sin(3/3 (sin(3/3 ( d = sin(3 sin(3 d 3 3 We choose u = t so du = t and dv = sin t and v = cos t t sin t = t cos t ( cos t (t = t cos t + t cos t While we have traded our original integral for a slightly simpler one, it is still not simple enough to evaluate by finding an obvious antiderivative. In fact, it is of the form of one of our earlier eamples of integration by parts, so here we must apply integration by parts again to finally evaluate the original integral. We focus on just the new integral part, t cos t : We choose u = t so du = and dv = cos t and v = sin t t cos t = t sin t (sin t = t sin t sin t = t sin t ( cos t + C = t sin t + cos t + C Subbing this result back into the original integral, t sin t = t cos t + t cos t Second by parts step First by parts step = t cos t + (t sin t + cos t + C While we have traded our original integral for a slightly simpler one, it is still not simple enough to evaluate by finding an obvious antiderivative. In fact, it is of the form of one of our earlier eamples of integration by parts, so here we must apply integration by parts again to finally evaluate the original integral. We focus on just the new integral part, sin(3 d: We choose u = so du = d and dv = sin(3 d and v = cos(3/3 sin(3 d = cos(3/3 ( cos(3/3 d = cos(3 + cos(3 d 3 3 = cos( (sin(3/3 + C = cos( sin(3 + C Subbing this result back into the original integral, cos(3 d = sin(3 sin(3 d 3 3 Second by parts step First by parts step = sin(3 ( cos(3 + sin(3 + C = sin(3 3 + sin(3 cos(3 + C 9 7 where C = (/3C is a new integration constant. 3

4 . (ln t Applying integration by parts again to the integral marked will lead to We only know how to differentiate (ln t, so we have to choose it as u. We choose u = (ln t so du = ln t t and dv = and v = t ( ln t (ln t = (ln t t (t t = t(ln t ln t We are left with a simpler integral, but not an easy one (unless you look at the eamples from the course notes! We focus on = ln t, and apply integration by parts once more. We choose u = ln t so du = t and dv = and v = t ( ln t = (ln tt (t t = t(ln t = t ln t t + C Thus, going back to our original integral, (ln t = t(ln t ln t 3. t e t We choose u = t = t(ln t (t ln t t + C and dv = e t so du = t and v = e t / t e t = t e t / (e t /(t = t e t te t = tet et + C so the overall integral will be t e t = t e t te t = t e t = t e t ( te t et + C tet + et + C where C is a multiple of the original C. 4. y y + 3 dy We choose u = y and dv = (y + 3 / d, so du = d and v = 3 (y + 33/ :. y y + 3 dy = 3 y(y + 33/ 3 (y + 33/ dy = 3 y(y + 33/ (y + 3 / + C 3 / = 3 y(y + 33/ 4 (y + 3/ + C. (t + + 3t Let u = t + and dv = + 3t, so du = and v = 9 ( + 3t3/. Then (t + + 3t = 9 (t + ( + 3t3/ ( + 3t 3/ 9 6. = (t + ( + 3t3/ ( + 3t/ + C. (p + sin(p + dp Let u = p + and dv = sin(p +, so du = d and v = cos(p +. (p + sin(p + dp = (p + cos(p + + cos(p + dp = (p + cos(p + + sin(p + + C 4

5 7. z e z dz Rewriting the integral, z e z dz = ze z dz Let u = z, dv = e z d. Thus du = dz and v = e z. Integration by parts gives: ze z dz = ze z ( e z dz = ze z + e z dz 8. ln d Let u = ln, dv = d. Then du = d and v =. = ze z e z + C Integrating by parts, we get: ln d = ln 9. y y dy Let u = y and dv = y, ( d = ln + d = ln + C so du = dy and v = ( y / y y dy = y( y / ( ( y / dy = y( y / + ( y / dy = y( y / + 3 ( y3/ ( + C = y( y / 4 3 ( y3/ + C I can be evaluated using integration by parts. Let u = t and dv = t, so du = d and v = ( t /. t t = t( t / + ( t / = t( t / 4 3 ( t3/ + C. can be integrated directly: 7 = 7(( t / ( + C t = 4( t / + C Adding the two integrals back together, we obtain t + 7 = t( t / 4 t 3 ( t3/ + C I. (ln d + 4( t / + C Select u = (ln( and dv = d, so du = ln d and v = /. ( (ln d = (ln ( ( ln d = (ln ln d This second integral,, can be evaluated with a second application of integration by parts. This was done earlier in Question #7 (using y instead of though: ln d = ln 4 + C. t + 7 t Since we have a fraction in the numerator, we can split the integral into a sum, and then evaluate each term separately. t + 7 t = t t I + 7 t Going back to the original integral (ln d = ( (ln ln +C 4 = (ln ln C

6 . arcsin(w dw We don t know the integral of arcsin, but we do know its derivative. Therefore we pick u = arcsin(w and dv = dw, so du = dw and v = w. w arcsin(w dw = w arcsin(w w w dw The new integral can be evaluate using a substitution. Let z = w, so dz dw = w or dz = dw: w ( w w dw = w z w dz = dz z = (z/ + C = w + C Going back to the original integral, 3. w arcsin(w dw = w arcsin(w dw w arctan(7 d ( = w arcsin(w w + C = w arcsin(w + w + C We don t know the integral of arctan, but we do know its derivative. Therefore we pick u = arctan(7 and dv = d, 7 so du = d and v =. + (7 arctan(7 d = arctan( d The new integral can be evaluate using a substitution. Let w = + 49, so dw d = 98 or d = w = 4 w dw ( 98 dw dw = d: = ln w + C 4 = 4 ln C Going back to the original integral, arctan( d = arctan( 4. arctan( d 7 49 d = arctan( 4 ln C This question starts off best with a substitution, due to the inside the arctan, and the outside: Let w =, so dw = d arctan( d = = ( arctan(w arctan(w dw dw Evaluating, we use by parts, following the same approach as Question #3, but without the 7 factor, arctan(w dw = w arctan(w ln + w + C Going back to the original integral, and using w =, arctan( d = arctan(w dw = ( w arctan(w ln + w + C = ( arctan( ln + ( + C = arctan( 4 ln C 6

7 . 3 e d 9. 3 t ln(t In this problem, we note that we can t integrate e by itself (no closed-form anti-derivative. However, if we package it with one of the s from the 3, we ll get e, and that can be integrated, using substitution (w =. Let u = and dv = e, so du = d and v = e. Using that, 3 e d = e e d = e e + C where is evaluated using the same substitution w =. 6. ln t We integrate by parts, as done in Question #6: 7. 3 ln t = t ln t t from #6 cos d = ( ln ( ln = ln 4. Integrating by parts with u = and dv = cos d gives 3 8. cos d = sin( + cos( ze z dz 3 = sin( + cos( (3 sin(3 + cos( The integral is the same as in Question #7. We use by parts with u = z and dv = e z dz, giving ze z dz = ze z e }{{ z } from #7 = e e ( e = e The integral is the same as in Question #7. We use by parts, with u = ln(t and dv = t t ln(t = t ln t t } {{ 4 } from #7 ( 9 ln 3 = 9 4 = 9 ln 3 arctan(y dy 3 ( ln We follow the work from Question #3, but without the 7 factor, using u = arctan(y and dv = dy. 3. arctan(y dy = y arctan(y ln + y = ( arctan( ( ln( arctan( ln( + t = π 4 ln(.439. We use the solution to Question #6, or applying by parts with u = ln( + t and dv =. 3. ln( + t = ( + t ln( + t ( + t from #6, with =+t arcsin z dz = (6 ln 6 6 ( ln = 6 ln 6.7 Using the solution from Question #, or u = arcsin z and dv = dz, arcsin z dz = z arcsin(z + z = ( arcsin( + ( arcsin( + = π.7 7

8 33. arcsin( d We first simplify the integral with the substitution w =, which leads to the new limits = w = = and = w = =. = = w= arcsin( d = arcsin(w dw w= after substitution At this point, we have returned to the integral in Question #, which can be evaluated using by parts, with u = arcsin(w and dv = dw. = = w= arcsin( d = arcsin(w dw w= = (w arcsin(w + w by parts = [( arcsin( + = π 4.8 ( arcsin( + ] 34. Find the area under the curve y = te t on the interval t. The function te t is always positive on the interval t so the area under the curve is equal to the integral te t Proceeding in the same way as Question #7, using u = t and dv = e t, te t = te t e }{{ t } from #7 = ( e e ( e e = 3e + 3. Find the area under the curve f(z = arctan z on the interval [, ]. On the interval t [, ], the function arctan(z is always positive, so the area equals the integral arctan(z dz. To evaluate the integral, we follow the work from Question #3, but without the 7 factor, using u = arctan(z and dv = dz. arctan(z dz = z arctan(z ln + z = ( arctan( ( ln arctan( = arctan( ln 36. Use integration by parts twice to find e sin( d. There are several ways to evaluate this integral; we ll show just one here. Let u = sin( and dv = e d, so du = cos( d and v = e. e sin( d our goal, I = sin(e cos(e d To evaluate, we select the trig function again as u and the eponential as dv: Let u = cos( and dv = e d, so du = sin( d and v = e. e sin( d our goal, I = sin(e cos(e d ( = sin(e cos(e = sin(e ( sin(e d ( cos(e ( sin(e d Tidying, we obtain e sin( d = sin(e cos(e sin(e d I I Grouping the integrals I, which are what we are looking for, e sin( d = sin(e cos(e or e sin( d = (sin(e cos(e 8

9 37. Use integration by parts twice to find e y cos(y dy. This question is done in the same manner as the previous one. For variety, and to show it works as well, we will select the eponential function as u and the trig functions as dv. (Both choices work, so long as you are consistent in both integration by parts steps. Let u = e y and dv = cos(y dy, so du = e y dy and v = sin(y. e y cos(y dy our goal, I = sin(ye y sin(ye y dy To evaluate, we select the eponential function again as u and the trig as dv: Let u = e y and dv = sin(y dy, so du = e y dy and v = cos(y. e y cos(y dy our goal, I = sin(ye y sin(ye y dy ( = sin(ye y ( cos(ye y = sin(ye y + cos(ye y Tidying, we obtain e y cos(y dy I ( cos(ye y dy cos(ye y dy = sin(ye y + cos(ye y cos(ye y dy I Grouping the integrals I, which are what we are looking for, e y cos(y dy = sin(ye y + cos(ye y or e y cos(y dy = (sin(yey + cos(ye y 38. Use integration by parts to show that n cos(a d = a n sin(a n n sin(a d a We are simply asked to change one integral into another, which can be done here directly with integration by parts. Let u = n and dv = cos(a d, so du = n n d and v = a sin(a Applying the by parts formula, n cos(a d ( = n a sin(a a sin(a n n d ( = n a sin(a n n sin(a d a which is the desired formula. 39. The concentration, C, in ng/ml, of a drug in the blood as a function of the time, t, in hours since the drug was administered is given by C = te.t. The area under the concentration curve is a measure of the overall effect of the drug on the body, called the bioavailability. Find the bioavailability of the drug between t = and t = 3. We have Bioavailability = 3 te.t. We first use integration by parts to evaluate the indefinite integral of this function. Let u = t and dv = e.t, so du = and v = e.t. Then, te.t = (t( e.t ( e.t ( = 7te.t + 7 e.t Thus, 3 = 7te.t 37e.t + C. te.t = ( 7te.t 37e.t = = 4.7. The bioavailability of the drug over this time interval is 4.7 (ng/ml-hours 3 9

10 4. During a surge in the demand for electricity, the rate, r, at which energy is used (in megawatts can be approimated by r = te at where t is the time in hours and a is a positive constant. (a Find the total energy, E, in megawatt hours used in the first T hours. Give your answer as a function of a. (b What happens to E as T? We know that de = r, so the total energy E used in the first T hours is given by E = T te at. We use integration by parts. Let u = t, dv = e at. Then du =, and v = ( /ae at. E = T te at = t T a e at T (/ae at T tidying: = t T a e at + (/a e at [ t ( ] T e at T integrating: = a e at + (/a a [ t tidying: = a e at ] T a e at [ T subbing in T,: = a e at ] a e at [ a ] e tidying: = T a e at a e at + a (b lim E = lim T T Notes: = a T a e at w/l Hop. rule a e at + a Since a >, the e at term is a negative eponential, and so heads towards at T. In the first term, T e at = T e. Although both at the numerator and the denominator go to infinity, the denominator e at goes to infinity more quickly than T does (can verify with l Hopital s rule, so this term also. Thus lim T E = a. In words this means that the total amount of energy in the surge, accounting for over all time (T is a megawatt hours. Average Value 4. (a Using the graph shown below, find 6 f( f( d (b What is the average value of f on [, 6]?

11 The integral represents the area below the graph of f( but above the -ais. (a Since each square has area, by counting squares and half-squares we find 6 f( d = 8.. (b The average value is a vertical quantiy, or (total area divided by (total horizontal length: ( 6 f( d = 8. 6 =.7 4. (a Using the graph below, estimate 3 3 f( d. (b Which of the following average values of f( is larger? (i Between = 3 and = 3, or (ii Between = and = 3? f( (a The integral is the area above the -ais minus the area below the -ais. Thus, we can see that 3 3 f( d is about 6 + = 4 (the negative of the area from t = 3 to t = plus the area from t = to t = 3. (b Since the integral in part (a is negative, the average value of f( between = 3 and = 3 is negative. From the graph, however, it appears that the integral of f( from = to = 3 is positive overall, meaning that the average value will also be positive. Hence (ii is the larger quantity. 43. (a Using the graphs of f( and g( shown below, find the average value on of (i f( (ii g( (iii f( g( f( (b Is the statement that g( Average(f Average(g = Average(f g true or not? Eplain your answer. (i Since the triangular region under the graph of f( has area, we have Average(f = (ii Similarly, Average(g = f( d = = 4 g( d = = 4 (iii Since f( is nonzero only for < and g( is nonzero only for <, the product f(g( = for all. Thus Average(f g = f(g( d = d =. (b Since the average values of f( and g( are nonzero, their product is nonzero. Thus the left side of the statement, Avg(f Avg(g, is nonzero. However, the average of the product f(g( is zero. Thus, the right side of the statement is zero, so the statement is not true. 44. For the function f below, write an epression involving one or more definite integrals of the original f( which would denote: (a The average value of f( for. (b The average value of f( for. f( 3 4

12 (a (Average value of f( on [, ] = f( d. (b (Average value of f( on [, ] requires that we only keep the value of f( as a positive contribution to the overall integral. Since the function switches to negative f( values on [, ], we have to subtract the contributions of the integral on that interval (making them positive overall. = f( d = ( f( d f( d. 4. This question is based on the same function f in the previous question, and the additional fact that f( is an even function. Consider the average value of f over the following intervals: I. II. III. IV. (a For which interval is the average value of f least? (b For which interval is the average value of f greatest? (c For which pair of intervals are the average values equal? We ll show that in terms of the average value of f, I > II = IV > III Using the definition of average value and the fact that f is even, we have Average value of f on II = f( being even: = ( multiplying, = 4 f( d f( d f( d Wait, that s just... = Average value of f on [, ] = Average value of f on interval IV. so the average on II ([, ] = average on IV ([-, ]. Since f is decreasing on [,], the average value of f on some interval [, c], where c, is decreasing as a function of c. The larger the interval the more low values of f are included, lowering the averag e. Hence Average value of f on [, ] > Average value of f on [, ] > Average value of f on [, ] 46. (a Without computing any integrals, eplain why the average value of f( = sin on [, π] must be between. and. (b Compute the eact average of sin on [, π]. (a Since f( = sin over [, π] is between and, the average of f( must itself be between and. Furthermore, since the graph of f( is concave down on this interval, the average value must be greater than the average height of the triangle shown in the figure below, namely y =.. (b Average = π π sin d (a What is the average value of f( = over the interval? (Hint: we don t have the integration tools to evaluate the integral of f( from first principles, but f( describes the graph of a common geometric shape whose area we already know. (b How can you tell whether this average value is more or less than. without doing any calculations? (a Recall that a circle is defined by + y =, so y = is the graph of the top half of the unit circle. Therefore, the average value of f( = = = (area of a quarter circle with r = = π d (b The area between the graph of y = and the -ais is.. Because the graph of y = is concave down, it lies above the straight line y =, so its average value is above.. See the figure below..79. y y = Avg. value or avg on I > avg on II, IV > avg on III

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