Chapter 9 Definite Integrals


 Dustin Boone
 1 years ago
 Views:
Transcription
1 Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished s we try to solve one of the most fmous prolems in mthemtics, finding the re under given curve. 9. Approimting Are nder Curve When it comes to finding the re of sic geometric shpes such s circles, squres, rectngles, tringles, nd trpezoids, we cn rely on geometric formuls to clculte the re. Emple 9. Find the re of the shded regions... Solution. The shded region is hlf of circle with rdius of. Thus, we will use the formul for the re of circle ( A= π r ), multiply it y.5 nd use r =. ( ) A =.5 π() =.5(9 π) = 4.5π region. The shded region is mde up of rectngle ( A lw, l, w ) ( A h,, h ) = = = nd tringle = = =. If we dd the res of the two geometric shpes, we will hve found the re of the shded region.
2 Figure 9. Grph for Emple 9. 5 Ashded = Arectngle + Atringle = ( lw) + h = ( ) + = 6 + = The re of the trpezoid is therefore 7.5 squre units. Finding the re etween the is nd curve f( ) on given intervl is it more chllenging if the region formed is not sic geometric shpe. For emple, the re under the curve f( ) = + on [,4] forms the shpe shown in Figure 9.. Figure 9. Grph of f( ) = +. 5 We cn see from the figure tht the re etween the is nd f( ) is not shpe tht hs fmilir formul for finding the re. When this occurs, we use rectngles to pproimte the re of the region. If we drw four rectngles, s seen in Figure 9., we cn sum up the re of the rectngles (R + R + R + R 4 ) nd otin n pproimtion of the re under the curve. Figure 9. Grph of f( ) = + with 4 right endpoints H. R R R R 4 5 H The rectngles hve een constructed such tht the right endpoint, i, of the intervl touches the curve. Becuse of this, we cll these rectngles right endpoint rectngles
3 We will, however, find n overestimte of the re ecuse the rectngles etend ove the curve. Nonetheless, we will hve some ide of the re under the curve. To find the re of ech rectngle in Figure 9. we need to find the se nd height of ech rectngle. The se of ech rectngle,, is found y tking the length of the given intervl, m min nd dividing it y the numer of rectngles constructed, n. This leds to the following clcultion. m min 4 = = =. n 4 The height of ech rectngle is the vlue of the function from the right end of ech intervl. Figure 9.4 elow shows the dimensions of ech rectngle. Figure 9.4 Dimensions of R, R, R, nd R 4. h = f() h = f() h = f(4) h = f() R R R R 4 = = = = The sum of the rectngles re found in the tle elow. The re, A, is the se times the height. Rectngle # Bse, Right Endpt, i Height, f( i ) A= f ( i ) R f () = + = ()() = R f () = +.4 ()(.4) =.4 R f () = +.7 ()(.7) =.7 R 4 4 f (4) = 4+ = + = ()() = Totl Are (Right).4 Since this is n overestimte, the re under the curve is less then.4 units. We cn lso pproimte the re under the curve using left endpoint rectngles s shown in figure 9.4. This pproimtion will give us n underestimte ecuse the rectngles do not fill the entire re under the curve.
4 Figure 9.5 Grph of f( ) = + with 4 left endpoint rectngles. (renme rectngles!!) R R R R 5 The se of ech rectngle is still unit ut the height of ech rectngle is the vlue of the function from the left end of ech intervl. The sum of the rectngles is found in the tle elow. Rectngle # Bse Left Endpt, i Height, f( i ) A= f ( i ) R f () = + = ()() = R f () = + = ()(.4) =.4 R f () = +.4 ()(.7) =.7 R 4 f () = +.7 ()() = Totl Are (Left) 9.4 Thus, the re must e greter thn 9.4 units. In generl, we cn find n pproimtion for the re under continuous curve f( ) on [, ] y drwing n eqully spced right (or left) endpoint rectngles under the curve nd then finding the sum of the re of the rectngles. If is the width of ech rectngle nd i n endpoint where = nd n =, then the sum of the re of n rectngles is re of st re of nd re of rd re of n th A = rectngle rectngle rectngle rectngle For right endpoint rectngles the sum of the re rectngles cn e denoted s H n Totl Are (Right) = f ( i) = f( ) + f( ) + f( ) f ( n) i= For left endpoint rectngles, the sum of the re of the rectngles cn e denoted s Totl Are (Left) = n i= f ( ) = f ( ) + f( ) + f ( ) f ( ) i n H The symol is the nottion for the sum of
5 Emple 9. Approimte the re under the curve f( ) = + on [,] using. 4 right endpoint rectngles. 8 left endpoint rectngles. Stte if the estimte is n overestimte or n underestimte. Solution. The grph of f ( ) = + on [, ] with the 4 right endpoint rectngles is shown in Figure 9.6. Figure 9.6 Grph of 4 right endpoint rectngles 6 R R R R 4 The se of ech rectngle is = = = =.5 n 4 nd the height is f( i ) where i is the right endpoint of ech intervl. The clcultion elow shows the sum of the res of the four rectngles. 4 Totl Are (Right)= f( i ) =.5 f +.5 f +.5 f +.5 f i= =.5 f f +.5 f f Thus, the re under the curve is less thn 7.75 squre units. ( ) ( ) ( ) ( ) 4 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = = The grph of f ( ) = + on [, ] with the 8 left endpoint rectngles is shown in Figure 9.7. Figure 9.7 Grph of 8 left endpoint rectngles. (renme rectngles!!) 6 R R R R R 4 R 5 R 6 R 7
6 The se of ech rectngle is = = = =.5 n 8 4 nd the height is f( i ) where i is the left endpoint of ech intervl. The tle elow shows the sum of the res of the eight rectngles. 8 Totl Are (Left)= f( i ) i= =.5 f +.5 f +.5 f +.5 f +.5 f +.5 f +.5 f +.5 f ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ( ) ( ) ( ) ( ) ( ) f ( ) f ( ) =.5 f +.5 f f f f +.5 f ( ) ( ) ( ) = = Since these rectngles ll lie elow the curve, the estimte for the re under the curve is n underestimte. There re numerous methods of using rectngles to pproimte the re under curve. A few of the other methods re shown in Figure 9.8 elow. Figure 9.8 Other methods to pproimte f ( ) on [,] = +. () Lower Sum Method ll rectngles lie elow the curve. () Midpoint Method the midpoint of ll rectngles re touching the curve. (c) pper Sum Method ll rectngles lie ove the curve. 9. Definite Integrls nd the Fundmentl Theorem of Clculus The methods used in the previous section llow us to otin good pproimtion of the re under curve, ut cn we mke this pproimtion etter? If we tke thinner nd thinner rectngles, we cn mke the pproimtion of the re under the curve more ccurte. In Emple 9. we found n pproimtion of squre units for the re under the curve using 8 left endpoint rectngles. If we
7 would hve used 4 left endpoint rectngles our pproimtion would hve een 5.75 squre units. The pproimtion with 8 rectngles ws more ccurte simply ecuse more rectngles were used. Compre the rectngles in Figure 9.9. Figure 9.8 Grphs of f( ) = + with 8 nd 6 right endpoint rectngles. 5 5 It ppers tht the mount of ecess re mde y the 6 rectngles is considerly less thn the ecess re mde y the 8 rectngles. One cn imgine tht the pproimtion would e even etter if we could fit rectngles or even rectngles under the curve. Wht if we hd n infinite numer of rectngles drwn under the curve? As one might hypothesize, the sum of n infinite numer of rectngles does ccurtely find the re under curve, nd we represent the re under curve using the definite integrl. The Definite Integrl For continuous function f on the intervl [, ] H ( ) let = n nd i e the right endpoint of the n intervls. Then the definite integrl of f is f ( ) d = lim f( ) n n i = i Some useful properties of definite integrls re listed in Tle 9.. Tle 9. Properties of Definite Integrls k f ( ) d = k f ( ) d [ ± ] = ± f ( ) g( ) d f( ) d g( ) d c c f ( ) d = f( ) d+ f ( ) d H Note: is referred to s the lower limit nd s the upper limit. Together, nd re known s the limits of integrtion.
8 sing the definition of the definite integrl the re in Figure 9.9 is represented s 4 ( + ) d Emple 9. Represent the re of the shded regions from Emple 9. s definite integrls. Solution. 9 d. ( + ) d From Emple 9. () we found the re to e ectly 5 nd from Emple 9. () we found tht the re cn e represented s definite integrl. We cn put the two of these together nd conclude 5 ( + ) d =. Net we cn connect the notion of n ntiderivtive nd definite integrl. Tke the ntiderivtive of +, Note tht F() now find F() F(), = C nd F () is 6 F( ) = + + C = + + C 5 6 F() = ( ) + () + C = + C 5 5 F() F() = + C C =. We cn see tht finding the ntiderivtive F ( ) of function nd then evluting F ( ) F ( ) gives the ect re under the curve. This process is n importnt theorem in clculus known s the Fundmentl Theorem of Clculus.
9 The Fundmentl Theorem of Clculus If f is continuous function defined on closed intervl [, ] nd F is n ntiderivtive of f, then f ( ) d = F ( ) F ( ) = F( )] Emple 9.4 Drw geometric representtion of ech definite integrl nd then evlute the definite integrl using the Fundmentl Theorem of Clculus.. d. e d Solution. The shded region in the grph elow shows the geometric representtion. 6 se the Fundmentl Theorem of Clculus to find the vlue of the definite integrl. 6 d = ( ) ( ) 9 = = =. The shded region in the grph elow shows the geometric representtion. e se the Fundmentl Theorem of Clculus to find the vlue of the definite integrl. e e d = ln] = lne ln= =
10 Emple 9.5 Grph f( ) = + nd use the grph to find + d. 5 Solution The grph of f( ) = + is shown elow. y f ( ) = f ( ) = + 5 R R To find 5 + d, we need to write n integrl tht represents R nd nother to represent R. This is necessry ecuse R nd R re ounded y different functions d = d + + d = = + = 5 All of the integrls we hve considered thus fr hve een positive. Tht is the grphs of the functions lied strictly ove the is. The net emple demonstrtes wht hppens when shded region lies strictly elow the is. Emple 9.6 Drw geometric representtion of ( 4 + ) d then evlute the definite integrl using the Fundmentl Theorem of Clculus. Solution The shded region in the grph elow shows the geometric representtion.
11 ( 4+ ) d = + () () () () () () = = (9 8+ 9) + = This definite integrl is negtive ecuse the shded re lies elow the is. When definite integrl represents portion of the grph tht lies ove s well s elow the is we cn clculte two types of res, gross re nd net re. The gross re is the totl mount of re tht lies etween the curve nd the is while the net re clcultes how much more re lies ove or elow is. Figure 9.7 shows the different vlues of the net re. Figure 9.?? Net Are.  Net re is positive ecuse more re lies ove the is. Net re is negtive ecuse more re lies elow the is Net re is zero ecuse the re ove nd elow the is is the sme. Emple 9.7 Drw geometric representtion of ( 4 + ) d nd then clculte the net nd gross res. Solution The shded region in the grph elow shows the geometric representtion. Region Region
12 To find the gross re we need to evlute the integrl tht represents ech shded region. Region ( 4 ) d 4 = + = + = + = In Emple 9.6, we found the re of region to e 4. Therefore the gross re is Are of Region + Are of Region = ( 4+ ) d + ( 4+ ) d The net re is just the sum of the two integrls, = + = + = 4 4 ( 4+ ) d= ( 4+ ) d+ ( 4+ ) d = + =. Since the net re is zero, we know there is the sme mount of re ove the is s there is elow the is. Notice tht clculting the function over the entire intervl is nother method of otining net re. 9. Are Between Two Curves Suppose we re to find the re of the shded region shown in Figure 9.??. Figure 9.?? The re etween f() nd g(). f ( ) g ( ) The re under f( ) on [, ] is shown in Figure 9.?? () nd the re under g ( ) on [, ] is shown in Figure 9.?? (). If the re under g ( ) is tken wy from the re under f( ) we otin the re in Figure 9.?? (c) which is the re we were trying to find in Figure 9.??.
13 Figure 9.?? Are Between Two Curves f ( ) f ( ) g ( ) re under f( ) on [, ] re under g ( ) on [, ] re under f( ) with re under g ( ) tken wy Thus, we cn find the re etween two curves if we find the re under the top curve nd sutrct off the re under the ottom curve. Are Between Two Curves On the closed intervl [, ], the re etween two continuous functions f() nd g(), where f( ) g( ), is given y [ f( ) g ( )] d The re etween two curves cn e rememered s (top function  ottom function) d Emple 9.8 Find the re etween f = + nd g ( ) ( ) = on [, ]. Solution First lets grph oth functions over [ ]. 5 f ( ) = + g ( ) = Since f( ) is the top function nd g ( ) is the ottom function, the definite integrl, nd thus the re etween the two curves is 8 8 ( + ) d = + d = + () () = + = + = ( )
14 Sometimes the two given curves will intersect t one or more points, thus forming n re ounded y the curves s shown in Figure 9.??. Figure 9.?? g ( ) f ( ) To find the re ounded y two curves we need to find the limits of integrtion. We do this y locting the points where the curves intersect. The definite integrl for Figure 9. is represented y [ ( ) ( )] f g d Emple 9.9 Find the re of the region ounded y = 8 nd y = 8. y Solution First, we need to grph the two functions on the sme coordinte plne. 6 y = (5, 7)  5 (, 5) 6 y = 8 From the grph we notice tht y = is the top function nd y = 8is the ottom function. In ddition, the points of intersection show tht the lower limit of integrtion is = nd the upper limit of integrtion is = 5. Thus, the definite integrl is
15 5 5 5 ( ) ( 8) d= d = d 5 = + + = (5) (5) 5(5) ( ) () 5( ) 8 8 = + = = 6 Emple 9. Find the re of the region ounded y f( ) = nd g ( ) 8 Solution First, we need to grph the two functions on the sme coordinte plne. =. R g ( ) (, ) f( ) 4  (, ) R 6 4 There re two ounded regions (R nd R ) produced y these curves. Notice tht the top function of R is g() nd the top function of R is f(). Consequently we will need to set up n integrl to find the re of R, nother integrl to find the re of R, nd then dd the results. ( 8 ) R = d ( 9 ) = d 4 9 = 4 9 = (8) (9) 4 8 = =.5 4 ( 8 ) R = d ( 9 ) = d 4 9 = 4 9 = (8) (9) 4 8 = =.5 4 Now dding the two results together we get R+ R =.5+.5= 4.5
16 9.4 Applictions of Definite Integrls Consumers nd Producers Surplus Suppose you worked ll summer nd put wy $8 to uy new stereo system for your dorm room. When you went shopping to uy the stereo system you found ectly wht you wnted for only $65. Thus, we could sy tht you sved $5. If we could find ll the consumers who were willing to py over $65 for this stereo system nd clculte the totl svings of ll consumers, we will hve found the consumers surplus. Figure 9.() shows the grph of supply curve, p = s ( ), nd demnd curve, p = d( ). The dotted lines represents the equilirium price, p, nd the equilirium quntity,. The re ove the dotted line, ut elow d( ), would represent the consumers surplus. Figure 9.?? p Consumers Surplus s () p s () p Producers Surplus p d () d () Now lets sy you re the producer of the stereo systems nd re willing to supply the stereos for $5. If, however, you end up selling the stereos for $65, you hve gined $5. The totl mount gined over ll possile prices is the producers surplus. Figure 9. lso shows the grph of the producers surplus. If p = d( ) is the demnd eqution, p = s ( ) the supply eqution, nd (, p) is the equilirium point then the consumers surplus is given y the producers surplus is given y ( ( ) ) d p d ( ( )) p s d Emple 9. A compny hs determined tht its supply nd demnd equtions cn e modeled y p = d( ) = + 7 nd p = s ( ) = + where represents the numer of units supplied ech week nd p is the selling price (in hundreds of dollrs) for ech unit. Find the consumers nd producers surplus.
17 Solution First we need to grph the supply nd demnd functions nd find the equilirium point. The equilirium point is found y setting d( ) = p ( ). p s ( ) 6 (, 5) d( ) 5 d( ) = s ( ) + = = = 4 = = The consumers surplus is p s ( ) 6 (, 5) d( ) d d (8) () 4 = + = + = + = + = 6 6 So the consumers sved pproimtely $66.67 per week when the selling price ws $5.
18 The producers surplus is p s ( ) 6 (, 5) d( ) ( + ) d = ( + 4) d = 4 (8) 4() 8 + = + = + = ( ) So the producers sved pproimtely $5. per week when the selling price ws $5. Smple Quiz Question 9. Find n pproimtion for the re under f( ) = + on [,] using 4 left endpoint rectngles nd 4 right endpoint rectngles. Which is n overestimte nd which is n underestimte? Question 9. Write definite integrl tht represents the shded re. p 6  Question 9. Evlute + 4 d. Question 9.4 Drw grph of f( ) = nd then find 4 d. Question 9.5 Evlute 4 d.
19 Question 9.6 Clculte the net nd gross res of + d. 8 Question 9.7 Find the re etween f( ) = + nd g ( ) 4 9, 6. = + on [ ] Question 9.8 Find re ounded y f( ) = + 4+ nd g ( ) = 4+ 6 Question 9.9 Find the re ounded y f( ) 9 = nd g ( ) =. Question 9. A compny hs determined its demnd eqution cn e modeled y p = d( ) = nd its supply eqution cn e modeled y p = s ( ) =.5+ where is the numer of units sold per dy nd p is the selling price in hundreds of dollrs. Find the consumers nd producers surplus.
Section 7.1 Area of a Region Between Two Curves
Section 7.1 Are of Region Between Two Curves White Bord Chllenge The circle elow is inscried into squre: Clcultor 0 cm Wht is the shded re? 400 100 85.841cm White Bord Chllenge Find the re of the region
More informationContinuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom
Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive
More information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More informationThe Trapezoidal Rule
_.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion
More informationMath 259 Winter Solutions to Homework #9
Mth 59 Winter 9 Solutions to Homework #9 Prolems from Pges 658659 (Section.8). Given f(, y, z) = + y + z nd the constrint g(, y, z) = + y + z =, the three equtions tht we get y setting up the Lgrnge multiplier
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More informationLINEAR ALGEBRA APPLIED
5.5 Applictions of Inner Product Spces 5.5 Applictions of Inner Product Spces 7 Find the cross product of two vectors in R. Find the liner or qudrtic lest squres pproimtion of function. Find the nthorder
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More informationMath RE  Calculus II Area Page 1 of 12
Mth RE  Clculus II re Pge of re nd the Riemnn Sum Let f) be continuous function nd = f) f) > on closed intervl,b] s shown on the grph. The Riemnn Sum theor shows tht the re of R the region R hs re=
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More information5.1 Estimating with Finite Sums Calculus
5.1 ESTIMATING WITH FINITE SUMS Emple: Suppose from the nd to 4 th hour of our rod trip, ou trvel with the cruise control set to ectl 70 miles per hour for tht two hour stretch. How fr hve ou trveled during
More informationChapter 6 Continuous Random Variables and Distributions
Chpter 6 Continuous Rndom Vriles nd Distriutions Mny economic nd usiness mesures such s sles investment consumption nd cost cn hve the continuous numericl vlues so tht they cn not e represented y discrete
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationThe Trapezoidal Rule
SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion Approimte
More informationWhat Is Calculus? 42 CHAPTER 1 Limits and Their Properties
60_00.qd //0 : PM Pge CHAPTER Limits nd Their Properties The Mistress Fellows, Girton College, Cmridge Section. STUDY TIP As ou progress through this course, rememer tht lerning clculus is just one of
More information10. AREAS BETWEEN CURVES
. AREAS BETWEEN CURVES.. Ares etween curves So res ove the xxis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationx ) dx dx x sec x over the interval (, ).
Curve on 6 For , () Evlute the integrl, n (b) check your nswer by ifferentiting. ( ). ( ). ( ).. 6. sin cos 7. sec csccot 8. sec (sec tn ) 9. sin csc. Evlute the integrl sin by multiplying the numertor
More informationSTRAND J: TRANSFORMATIONS, VECTORS and MATRICES
Mthemtics SKE: STRN J STRN J: TRNSFORMTIONS, VETORS nd MTRIES J3 Vectors Text ontents Section J3.1 Vectors nd Sclrs * J3. Vectors nd Geometry Mthemtics SKE: STRN J J3 Vectors J3.1 Vectors nd Sclrs Vectors
More information8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers.
8. Complex Numers The rel numer system is dequte for solving mny mthemticl prolems. But it is necessry to extend the rel numer system to solve numer of importnt prolems. Complex numers do not chnge the
More information5.2 Volumes: Disks and Washers
4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of crosssection or slice. In this section, we restrict
More informationx = a To determine the volume of the solid, we use a definite integral to sum the volumes of the slices as we let!x " 0 :
Clculus II MAT 146 Integrtion Applictions: Volumes of 3D Solids Our gol is to determine volumes of vrious shpes. Some of the shpes re the result of rotting curve out n xis nd other shpes re simply given
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationFarey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University
U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions
More informationShape and measurement
C H A P T E R 5 Shpe nd mesurement Wht is Pythgors theorem? How do we use Pythgors theorem? How do we find the perimeter of shpe? How do we find the re of shpe? How do we find the volume of shpe? How do
More informationProblem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are:
(x + y ) = y + (x + y ) = x + Problem Set 9 Discussion: Nov., Nov. 8, Nov. (on probbility nd binomil coefficients) The nme fter the problem is the designted writer of the solution of tht problem. (No one
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationCalculus  Activity 1 Rate of change of a function at a point.
Nme: Clss: p 77 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit Rte of chnge of function t point. ) Strt Mths Helper Plus, then lod the file: Clculus
More informationUnit #10 De+inite Integration & The Fundamental Theorem Of Calculus
Unit # De+inite Integrtion & The Fundmentl Theorem Of Clculus. Find the re of the shded region ove nd explin the mening of your nswer. (squres re y units) ) The grph to the right is f(x) = x + 8x )Use
More informationMETHODS OF APPROXIMATING THE RIEMANN INTEGRALS AND APPLICATIONS
Journl of Young Scientist Volume III 5 ISSN 448; ISSN CDROM 449; ISSN Online 445; ISSNL 44 8 METHODS OF APPROXIMATING THE RIEMANN INTEGRALS AND APPLICATIONS An ALEXANDRU Scientific Coordintor: Assist
More informationBig idea in Calculus: approximation
Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:
More informationThe problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests.
ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion
More informationDistance And Velocity
Unit #8  The Integrl Some problems nd solutions selected or dpted from HughesHllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More informationChapter 4 Contravariance, Covariance, and Spacetime Diagrams
Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationTriangles The following examples explore aspects of triangles:
Tringles The following exmples explore spects of tringles: xmple 1: ltitude of right ngled tringle + xmple : tringle ltitude of the symmetricl ltitude of n isosceles x x  4 +x xmple 3: ltitude of the
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationAPPM 1360 Exam 2 Spring 2016
APPM 6 Em Spring 6. 8 pts, 7 pts ech For ech of the following prts, let f + nd g 4. For prts, b, nd c, set up, but do not evlute, the integrl needed to find the requested informtion. The volume of the
More informationCHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 6. VOLUMES USING CROSSSECTIONS. A() ;, ; (digonl) ˆ Èˆ È V A() d d c d 6 (dimeter) c d c d c ˆ 6. A() ;, ; V A() d d. A() (edge) È Š È Š È ;, ; V A() d d 8
More informationImproper Integrals with Infinite Limits of Integration
6_88.qd // : PM Pge 578 578 CHAPTER 8 Integrtion Techniques, L Hôpitl s Rule, nd Improper Integrls Section 8.8 f() = d The unounded region hs n re of. Figure 8.7 Improper Integrls Evlute n improper integrl
More information2. VECTORS AND MATRICES IN 3 DIMENSIONS
2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2dimensionl Vectors x A point in 3dimensionl spce cn e represented y column vector of the form y z zxis yxis z x y xxis Most of the
More information9.4. The Vector Product. Introduction. Prerequisites. Learning Outcomes
The Vector Product 9.4 Introduction In this section we descrie how to find the vector product of two vectors. Like the sclr product its definition my seem strnge when first met ut the definition is chosen
More informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More information4 VECTORS. 4.0 Introduction. Objectives. Activity 1
4 VECTRS Chpter 4 Vectors jectives fter studying this chpter you should understnd the difference etween vectors nd sclrs; e le to find the mgnitude nd direction of vector; e le to dd vectors, nd multiply
More informationHow do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?
XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationONLINE PAGE PROOFS. Antidifferentiation and introduction to integral calculus
Antidifferentition nd introduction to integrl clculus. Kick off with CAS. Antiderivtives. Antiderivtive functions nd grphs. Applictions of ntidifferentition.5 The definite integrl.6 Review . Kick off
More informationThis chapter will show you. What you should already know. 1 Write down the value of each of the following. a 5 2
1 Direct vrition 2 Inverse vrition This chpter will show you how to solve prolems where two vriles re connected y reltionship tht vries in direct or inverse proportion Direct proportion Inverse proportion
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.
More informationFORM FIVE ADDITIONAL MATHEMATIC NOTE. ar 3 = (1) ar 5 = = (2) (2) (1) a = T 8 = 81
FORM FIVE ADDITIONAL MATHEMATIC NOTE CHAPTER : PROGRESSION Arithmetic Progression T n = + (n ) d S n = n [ + (n )d] = n [ + Tn ] S = T = T = S S Emple : The th term of n A.P. is 86 nd the sum of the first
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationCHAPTER 1 CENTRES OF MASS
1.1 Introduction, nd some definitions. 1 CHAPTER 1 CENTRES OF MASS This chpter dels with the clcultion of the positions of the centres of mss of vrious odies. We strt with rief eplntion of the mening of
More informationSTRAND B: NUMBER THEORY
Mthemtics SKE, Strnd B UNIT B Indices nd Fctors: Tet STRAND B: NUMBER THEORY B Indices nd Fctors Tet Contents Section B. Squres, Cubes, Squre Roots nd Cube Roots B. Inde Nottion B. Fctors B. Prime Fctors,
More informationMTH 505: Number Theory Spring 2017
MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of $ nd $ s two denomintions of coins nd $c
More informationThe Fundamental Theorem of Algebra
The Fundmentl Theorem of Alger Jeremy J. Fries In prtil fulfillment of the requirements for the Mster of Arts in Teching with Speciliztion in the Teching of Middle Level Mthemtics in the Deprtment of Mthemtics.
More information12 TRANSFORMING BIVARIATE DENSITY FUNCTIONS
1 TRANSFORMING BIVARIATE DENSITY FUNCTIONS Hving seen how to trnsform the probbility density functions ssocited with single rndom vrible, the next logicl step is to see how to trnsform bivrite probbility
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More information6.5 Improper integrals
Eerpt from "Clulus" 3 AoPS In. www.rtofprolemsolving.om 6.5. IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y =
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationPartial Differential Equations
Prtil Differentil Equtions Notes by Robert Piché, Tmpere University of Technology reen s Functions. reen s Function for OneDimensionl Eqution The reen s function provides complete solution to boundry
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationRiemann Integrals and the Fundamental Theorem of Calculus
Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums
More informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
More informationMath 0230 Calculus 2 Lectures
Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two
More informationTest , 8.2, 8.4 (density only), 8.5 (work only), 9.1, 9.2 and 9.3 related test 1 material and material from prior classes
Test 2 8., 8.2, 8.4 (density only), 8.5 (work only), 9., 9.2 nd 9.3 relted test mteril nd mteril from prior clsses Locl to Globl Perspectives Anlyze smll pieces to understnd the big picture. Exmples: numericl
More informationLecture 2 : Propositions DRAFT
CS/Mth 240: Introduction to Discrete Mthemtics 1/20/2010 Lecture 2 : Propositions Instructor: Dieter vn Melkeeek Scrie: Dlior Zelený DRAFT Lst time we nlyzed vrious mze solving lgorithms in order to illustrte
More information[ ( ) ( )] Section 6.1 Area of Regions between two Curves. Goals: 1. To find the area between two curves
Gols: 1. To find the re etween two curves Section 6.1 Are of Regions etween two Curves I. Are of Region Between Two Curves A. Grphicl Represention = _ B. Integrl Represention [ ( ) ( )] f x g x dx = C.
More informationArea and Perimeter. Area and Perimeter. Curriculum Ready.
Are nd Perimeter Curriculum Redy www.mthletics.com This ooklet shows how to clculte the re nd perimeter of common plne shpes. Footll fields use rectngles, circles, qudrnts nd minor segments with specific
More informationChapter 2. Random Variables and Probability Distributions
Rndom Vriles nd Proilit Distriutions 6 Chpter. Rndom Vriles nd Proilit Distriutions.. Introduction In the previous chpter, we introduced common topics of proilit. In this chpter, we trnslte those concepts
More informationMath 113 Exam 1Review
Mth 113 Exm 1Review September 26, 2016 Exm 1 covers 6.17.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More information1 Error Analysis of Simple Rules for Numerical Integration
cs41: introduction to numericl nlysis 11/16/10 Lecture 19: Numericl Integrtion II Instructor: Professor Amos Ron Scries: Mrk Cowlishw, Nthnel Fillmore 1 Error Anlysis of Simple Rules for Numericl Integrtion
More informationCS12N: The Coming Revolution in Computer Architecture Laboratory 2 Preparation
CS2N: The Coming Revolution in Computer Architecture Lortory 2 Preprtion Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes
More informationThe Bernoulli Numbers John C. Baez, December 23, x k. x e x 1 = n 0. B k n = n 2 (n + 1) 2
The Bernoulli Numbers John C. Bez, December 23, 2003 The numbers re defined by the eqution e 1 n 0 k. They re clled the Bernoulli numbers becuse they were first studied by Johnn Fulhber in book published
More informationMath 017. Materials With Exercises
Mth 07 Mterils With Eercises Jul 0 TABLE OF CONTENTS Lesson Vriles nd lgeric epressions; Evlution of lgeric epressions... Lesson Algeric epressions nd their evlutions; Order of opertions....... Lesson
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationC1M14. Integrals as Area Accumulators
CM Integrls s Are Accumultors Most tetbooks do good job of developing the integrl nd this is not the plce to provide tht development. We will show how Mple presents Riemnn Sums nd the ccompnying digrms
More informationx 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx
. Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute
More informationMATHS NOTES. SUBJECT: Maths LEVEL: Higher TEACHER: Aidan Roantree. The Institute of Education Topics Covered: Powers and Logs
MATHS NOTES The Institute of Eduction 06 SUBJECT: Mths LEVEL: Higher TEACHER: Aidn Rontree Topics Covered: Powers nd Logs About Aidn: Aidn is our senior Mths techer t the Institute, where he hs been teching
More informationu(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.
Lecture 4 Complex Integrtion MATHGA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex
More informationl 2 p2 n 4n 2, the total surface area of the
Week 6 Lectures Sections 7.5, 7.6 Section 7.5: Surfce re of Revolution Surfce re of Cone: Let C be circle of rdius r. Let P n be n nsided regulr polygon of perimeter p n with vertices on C. Form cone
More informationBoolean Algebra. Boolean Algebra
Boolen Alger Boolen Alger A Boolen lger is set B of vlues together with:  two inry opertions, commonly denoted y + nd,  unry opertion, usully denoted y ˉ or ~ or,  two elements usully clled zero nd
More informationCS 311 Homework 3 due 16:30, Thursday, 14 th October 2010
CS 311 Homework 3 due 16:30, Thursdy, 14 th Octoer 2010 Homework must e sumitted on pper, in clss. Question 1. [15 pts.; 5 pts. ech] Drw stte digrms for NFAs recognizing the following lnguges:. L = {w
More informationQuadratic Equations. Brahmagupta gave. Solving of quadratic equations in general form is often credited to ancient Indian mathematicians.
9 Qudrtic Equtions Qudrtic epression nd qudrtic eqution Pure nd dfected qudrtic equtions Solution of qudrtic eqution y * Fctoristion method * Completing the squre method * Formul method * Grphicl method
More information1 Nondeterministic Finite Automata
1 Nondeterministic Finite Automt Suppose in life, whenever you hd choice, you could try oth possiilities nd live your life. At the end, you would go ck nd choose the one tht worked out the est. Then you
More informationCBSEXII2015 EXAMINATION. Section A. 1. Find the sum of the order and the degree of the following differential equation : = 0
CBSEXII EXMINTION MTHEMTICS Pper & Solution Time : Hrs. M. Mrks : Generl Instruction : (i) ll questions re compulsory. There re questions in ll. (ii) This question pper hs three sections : Section, Section
More informationP 1 (x 1, y 1 ) is given by,.
MA00 Clculus nd Bsic Liner Alger I Chpter Coordinte Geometr nd Conic Sections Review In the rectngulr/crtesin coordintes sstem, we descrie the loction of points using coordintes. P (, ) P(, ) O The distnce
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationCHAPTER : INTEGRATION Content pge Concept Mp 4. Integrtion of Algeric Functions 4 Eercise A 5 4. The Eqution of Curve from Functions of Grdients. 6 Ee
ADDITIONAL MATHEMATICS FORM 5 MODULE 4 INTEGRATION CHAPTER : INTEGRATION Content pge Concept Mp 4. Integrtion of Algeric Functions 4 Eercise A 5 4. The Eqution of Curve from Functions of Grdients. 6 Eercise
More informationLine and Surface Integrals: An Intuitive Understanding
Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of
More informationIf u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du
Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find ntiderivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible
More information7.6 The Use of Definite Integrals in Physics and Engineering
Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 7.6 The Use of Definite Integrls in Physics nd Engineering It hs been shown how clculus cn be pplied to find solutions to geometric problems
More informationSTEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.
STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t
More information