Chapter 9 Definite Integrals

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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.

Chapter 6 Notes, Larson/Hostetler 3e

Chapter 6 Notes, Larson/Hostetler 3e Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn