Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find the re etween two grphs. If f(x) g(x) for ll x in [,], then we cn pproximte the re etween f nd g y prtitioning the intervl [,] nd forming Riemnn sum, s shown in the picture. The height of ech rectngle is top ottom, f(c i ) g(c i ) so the re of the i th rectngle is (height). (se) = (f(c i ) g(c i )). x. Adding up this rectngles gives n pproximtion of the totl n re s f ci gci i x, Riemnn sum. The limit of this Riemnn sum, s the numer of rectngles gets lrger nd their width gets smller, is the definite integrl x gx f dx. The re etween two curves f(x) nd g(x), where f(x) g(x), etween x = nd x = is f x g x The integrnd is top ottom. Mke grph to e sure which curve is which. dx Exmple Find the re ounded etween the grphs of f(x) = x nd g(x) = for x 4. This chpter is (c). It ws remixed y Dvid Lippmn from Shn Clwy's remix of Contemporry Clculus y Dle Hoffmn. It is licensed under the Cretive Commons Attriution license.
Chpter The Integrl Applied Clculus 4 Alwys strt with grph so you cn see which grph is the top nd which is the ottom. In this exmple, the two curves cross, nd they chnge positions; we ll need to split the re into two pieces. Geometriclly, we cn see tht the re is + ½ =.5. Writing the re s sum of definite integrls, we get: 4 Are = x dx x dx These integrls re esy to evlute using ntiderivtives: x 9 5 x dx x 9 5. x 4 6 x dx x 9. 4 9 The two integrls lso tell us tht the totl re etween f nd g is.5 squre units, which we lredy knew. Note tht the single integrl dx. 5 4 x is not the re we wnt in the lst exmple. The vlue of the integrl is.5, nd the vlue of the re is.5. Tht s ecuse for the tringle on the right, the grph of y = x is ove the grph of y =, so the integrnd x is negtive; in the definite integrl, the re of tht tringle comes in with negtive sign. In this exmple, it ws esy to see exctly where the two curves crossed so we could rek the region into the two pieces to figure seprtely. In other exmples, you might need to solve n eqution to find where the curves cross. Exmple Two ojects strt from the sme loction nd trvel long the sme pth with velocities t t v B t t 4t meters per second. How fr hed is A fter seconds? v A nd Since vat vb t, the "re" etween the grphs of v A t nd t represents the distnce etween the ojects. v B After seconds, the distnce prt dt 5t t v t v t dt t t 4t 5 t A t B 5 7 9 dt. 5 meters.
Chpter The Integrl Applied Clculus 5 Volume Just s we cn prtition n intervl nd imgine pproximting n re with rectngles to find formul for the re etween curves, we cn prtition n intervl nd imgine pproximting volume with simple shpes to find formul for the volume of solid. While this pproch works for vriety of shpes, our focus will e on shpes formed y revolving curve round the horizontl xis. We strt with n re, the region elow function on the intervl x. We re going to tke tht region, nd rotte it round the x xis, creting the solid shpe shown. Rotte out the xis To find the volume of this solid, we cn strt y prtitioning the intervl [,] nd pproximting the re with rectngles. As efore, the width of ech rectngle would e x nd the height f(c i ). If we took just one of these rectngles nd rotted it out the horizontl xis, it would form cylindricl shpe. The rdius of tht V r h f ( ci ) cylinder would e f(c i ), so the volume would e x The volume of the whole solid could e pproximted y rotting ech of the rectngles out the x xis. Adding up the volume of ech of the little cylindricl discs gives n pproximtion of the totl volume s n i f c i x, Riemnn sum. The limit of this sum s the width of the rectnges ecomes smll is the definite integrl f x dx. The volume of the solid otined y rotting out the x-xis the re ounded y the curve f(x), the x-xis, x =, nd x = is f x dx
Chpter The Integrl Applied Clculus 6 Exmple x Find the volume of the solid formed y rotting the re under out the x-xis. f x e on the intervl [,] This is the region pictured in the erlier exmple. We sustitute in the function nd ounds into the formul we derived to set up the definite integrl. Volume = e x dx Using exponent rules, the integrnd cn e simplified. The constnt π cn e pulled out of the integrl. e x dx Using the sustitution u = -x, we cn integrte this function. e x dx e x () () e e.58 cuic units Averge Vlue We know the verge of n numers,,,..., n, is their sum divided y n. But wht if we need to find the verge temperture over dy s time -- there re too mny possile tempertures to dd them up. This is jo for the definite integrl. The verge vlue of function f(x) on the intervl [, ] is given y f x dx The verge vlue of positive f hs nice geometric interprettion. Imgine tht the re under f (Fig. ) is liquid tht cn "lek" through the grph to form rectngle with the sme re (Fig. ). If the height of the rectngle is H, then the re of the rectngle is H. We know the re of the rectngle is the sme s the re under f so H f xdx. Then H f x dx, the verge vlue of f on [,]. The verge vlue of positive function f is the height H of the rectngle whose re is the sme s the re under f.
Chpter The Integrl Applied Clculus 7 Exmple 4 During 9 hour work dy, the production rte t time t hours fter the strt of the shift ws given y the function r( t) 5 t crs per hour. Find the verge hourly production rte. 9 The verge hourly production is 5 t 7 9 dt crs per hour. A note out the units rememer tht the definite integrl hs units (crs per hour) (hours) = crs. But the /(-) in front hs units /hours the units of the verge vlue re crs per hour, just wht we expect n verge rte to e. In generl, the verge vlue of function will hve the sme units s the integrnd. Function verges, involving mens nd more complicted verges, re used to "smooth" dt so tht underlying ptterns re more ovious nd to remove high frequency "noise" from signls. In these situtions, the originl function f is replced y some "verge of f." If f is rther x x5 5 jgged time dt, then the ten yer verge of f is the integrl x f t of f over 5 units on ech side of x. For exmple, the figure here shows the grphs of Monthly Averge (rther noisy dt) of surfce temperture dt, n Annul Averge (still rther jgged), nd Five Yer Averge ( much smoother function). Typiclly the verge function revels the pttern much more clerly thn the originl dt. This use of moving verge vlue of noisy dt (wether informtion, stock prices) is very common. g dt, n verge Exmple 5 http://commons.wikimedi.org/wiki/file:short_instrumentl_temperture_record.png, CC-BY
Chpter The Integrl Applied Clculus 8 The grph to the right shows the mount of wter in reservoir over hour period. Estimte the verge mount of wter in the reservoir over this period. If V t is the volume of the wter (in millions of liters) fter t hours, then the verge mount is t V dt. In order to find the definite integrl, we ll hve to estimte. I ll use 6 rectngles, nd I ll tke the heights from their right edges. My estimte of the integrl is V t dt 8 9.7 8. 9.9 79. 6 The units of this integrl re millions of liters feet. So my estimte of the verge volume is 79.6 5 millions of liters. Your estimte might e little different. In the figure elow, you cn see the sme grph with the line y 5 drwn in. The re under the curve nd the re under the rectngle re (pproximtely) the sme. In fct, tht would e different wy to estimte the verge vlue. We could hve estimted the plcement of the horizontl line so tht the re under the curve nd under the line were equl..
Chpter The Integrl Applied Clculus 9.6 Exercises In prolems 4, use the vlues in the tle to estimte the res. x f x g x h x 5 5 6 6 6 8 4 6 4 5 5 4 4 6. Estimte the re etween f nd g, etween x = nd x = 4.. Estimte the re etween g nd h, etween x = nd x = 6.. Estimte the re etween f nd h, etween x = nd x = 4. 4. Estimte the re etween f nd g, etween x = nd x = 6. 5. Estimte the re of the islnd shown In prolems 6 5, find the re etween the grphs of f nd g for x in the given intervl. Rememer to drw the grph! 6. f(x) = x +, g(x) = nd x. 7. f(x) = x +, g(x) = + x nd x. 8. f(x) = x, g(x) = x nd x. 9. f(x) = (x ), g(x) = x + nd x.. f(x) = x, g(x) = x nd x e.. f(x) = x, g(x) = x nd x 4.. f(x) = 4 x, g(x) = x + nd x.. f(x) = e x, g(x) = x nd x. 4. f(x) =, g(x) = x nd x. 5. f(x) =, g(x) = 4 x nd x.
Chpter The Integrl Applied Clculus For prolems 6-8, find the volume of the solid otined y rotting the specified region out the x xis. 6. Region under f(x) = x + for x. 7. Region under f(x) = 4 x for x. 8. Region under f(x) = x for x. In prolems 9 nd use the vlues in the tle to estimte the verge vlues. x f x g x 5 6 6 4 4 5 4 6 9. Estimte the verge vlue of f on the intervl [, 6].. Estimte the verge vlue of g on the intervl [, 6]. In prolems 6, find the verge vlue of f on the given intervl.. f(x) from the grph for x.. f(x) from the grph for x 4.. f(x) from the grph for x 6. 4. f(x) from the grph for 4 x 6. 5. f(x) = x + for x 4. 6. f(x) = x for x. 7. The grph shows the velocity of cr during 5 hour trip. () Estimte how fr the cr trveled during the 5 hours. () At wht constnt velocity should you drive in order to trvel the sme distnce in 5 hours? 8. The grph shows the numer of telephone clls per minute t lrge compny. Estimte the verge numer of clls per minute () From 8 m to 5 pm. () From 9 m to pm.