Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F (x) (so tht F (x) is n ntiderivtive of f(x)) on intervl [, b], then the totl chnge F (b) F () in F (x) s x chnges from to b is given by f(x) dx. This sttement is known s the Fundmentl Theorem of Clculus. f(x) dx = F (b) F () This sttement reltes the definite integrl b f(x)dx with n ntiderivtive F (x) so it reltes definite nd indefinite integrl. Moreover, it provides us with the wy of clculting definite integrl without computing the sums of ny rectngles nd finding their limits. When evluting the ntiderivtive F (x) t b nd nd subtrcting the nswers, the nottion F (x) b is used for F (b) F (). Thus, to evlute f(x) dx. find n ntiderivtive F (x) of f(x),. evlute F (x) b,. obtin the numericl nswer F (b) F (). Exmple. Evlute the integrl x dx computing the re under x from to. Solution. The function x hs n ntiderivtive x. Thus x dx = x = = 8 =.666... Keep in mind the difference between the definite nd indefinite integrl. If f(x) is continuous function with ntiderivtive F (x), then
the indefinite integrl f(x) dx is the fmily of functions F (x) + c, the definite integrl f(x) dx is the number F (b) F (). The fct tht the Fundmentl Theorem of Clculus enbles you to compute the totl chnge in ntiderivtive of f(x) when x chnges from to b is referred lso s the Totl Chnge Theorem. Thus, the definite integrl cn be used to find the totl chnge in quntity on n intervl given its rte. Exmple. Suppose tht the velocity of n object is given by the function v(t) =.t where t is the time in seconds nd v is the velocity in feet per second. Determine the totl movement of the object between nd seconds. Solution. The totl movement is the totl chnge in position s(t) from to seconds is s() s(). This cn be found s the definite integrl from to of the rte v(t) = s (t). Thus, s() s() = v(t) dt =.t dt =. t =.5( ) =.5() = 45 feet. Units of the totl chnge. If [x] denotes the units of quntity x nd [f(x)] denotes the units of f(x), the definite integrl f(x)dx is in units [f(x)] [x]. For instnce, in Exmple bove, the units of the definite integrl re the units of velocity multiplied by the units of time. Thus, the nswer is in meters since the product meters second second produces meters. In some cses you my be instructed to use the clcultor to pproximte the definite integrl. Exmple. The size of certin bcteri culture grows t rte of f(t) = te t/ milligrms per hour. Use the Left-Right Sums clcultor progrm to pproximte the bcteri size fter the first hours to the first two nonzero digits. Solutions. The bcteri size fter the first hours cn be found s tet/ dt. Enter the function in your clcultor s xe x/, strt the progrm nd enter tht =, b =. With n = for exmple, both the left nd the right sums round to. Thus, in hours the size incresed by mg. Recll lso the following properties of the definite integrl. Assume tht f(x) is continuous functions on [, b] with ntiderivtives F (x).. f(x) dx = F () F () =. f(x) dx = F (b) F () = (F () F (b)) = f(x) dx. b. If c is ny number from [, b], then f(x) dx = F (b) F () = F (b) F (c) + F (c) F () = (F (c) F ()) + (F (b) F (c)) = c f(x) dx + f(x) dx. This property cn be shown to hold c for ny number c, not necessrily between nd b. 4. The vrible nme used in the definite integrl does not mtter. In other words, f(x) dx = b f(t) dt = f(u) du =...
Finding the re between f(x) nd x-xis on given intervl. So fr we relted the definite integrl nd re under the grph of continuous function just in the cse the function is nonnegtive. Recll tht in this cse the definite integrl is exctly equl to the re. Let us consider now the other computing the re if the function is not nonnegtive on entire intervl. First, let us now consider the cse when given function is less or equl to zero on n intervl [, b]. If f(x) then f(x) = f(x) nd the size of the re A between x-xis nd f(x) on [, b] is the sme s the size of the re under the curve of f(x) = f(x) on [, b]. Thus A = f(x)dx = f(x)dx We cn unify the cses f(x) or f(x) by writing tht in either cse f(x) dx = f(x) dx but note tht this does not hve to hold if the function f(x) chnges sign on intervl [, b]. Let us ssume tht f(x) is chnging the sign just once t c in [, b]. Sy tht f is chnging from negtive to positive t c s in the figure below. On intervl [, c], f(x) so the re A between f(x) nd x-xis cn be found s A = c f(x) dx. On intervl [c, b], f(x) so the re A between f(x) nd x-xis cn be found s A = f(x) dx. The totl re A cn be obtined s the sum A + A. c Thus A = A + A = c f(x)dx + c f(x)dx. Note tht the totl re cnnot be evluted using single definite integrl. The property. does not pply to the sums of two integrls in the lst formul since the integrnd in the first integrl is f(x) nd the integrnd in the second integrl is f(x). Similrly, if f is chnging from positive to negtive t c in [, b] we would hve tht A = A +A = c f(x)dx + f(x)dx. c If f(x) hs more thn one x-intercept in [, b], one would need to divide the totl re in more thn two regions. This brings us to the following procedure for finding the totl re under the curve.
Finding Are. To find the totl re A between the grph of continuous function f(x) nd x-xis on [, b],. Grph the function on [,b] nd check if f is positive, negtive or it chnges the sign.. Consider the following cses. Cse. If f is positive on [, b], then A = f(x) dx. Cse. If f is negtive on [, b], then A = f(x) dx. Cse. If f is chnging sign on [, b] proceed with the following steps.. Find ll x-intercepts c, c... c k of f(x) which re in [, b] nd divide the intervl into subintervls such tht f does not chnge sign on ech subintervl. 4. Finding the re between f nd x-xis on ech subintervl flls under either cse or. 5. The totl re A is the sum of the res on ech subintervl. Exmple 4. Find the re between the grph of f(x) = 4 x nd x-xis on [, ]. Solution. Consider the grph of the function on the intervl [,]. Note tht function chnges the sign from positive to negtive t point between nd. Set the function to zero to find the x-intercepts. 4 x = x = 4 x = ±. Since - is not in [,], just is relevnt. Let A denote the re on [,] nd A the re on [,]. The function is positive on [,) nd negtive on (,] so A = (4 x )dx nd A = (4 x )dx = (4 x )dx. Find the ntiderivtive 4x x of 4 x nd evlute the two res s follows. A = A = (4 x )dx = (4x x ) (4 x )dx = (4x x ) = = (4() ) (4() ) = 8 8 = 6. ( (4() ) (4() ) ) Thus the totl re A is A = A + A = 6 + 7 = 7.67. Prctice Problems.. Evlute the following. () x dx (b) x + dx (c) 4 x dx 4 = ( 9 8+ 8 ) = 7.
. Find the re between the grph of f(x) nd x-xis on [, b]. () f(x) = x ; =, b =. (b) f(x) = x ; =, b = (c) f(x) = x 9; =, b = 4 (d) f(x) = x x =, b = (e) f(x) = x 4; =, b = 9. From pst records, botnist knows tht certin species of tree hs rte of growth tht cn be modeled by f(t) = t, t 4, where t is the ge of the tree in yers nd f(t) is the growth rte in feet per yer. Determine how much did the tree grow from the time when it ws yer old to the time it ws four yers old. 4. Suppose tht the velocity of n object is given by the function v(t) = t t where t is the time +9 in seconds nd v is the velocity in feet per second. Determine the totl movement of the object between nd 5 seconds. 5. The rte of chnge in the U.S. popultion cn be modeled by g(x) =.e.t, t where t represents the number of yers since 9 nd g represents the rte of chnge in popultion mesured in millions per yer. Determine the totl increse in the U.S. popultion from 9 to 95. Solutions.. () x dx = x = x = 4 = 4. (b) x + dx = ( x + x) (c) 4 x dx = x 4 = + () = 7. = 4 = =. x 4 4. () Since f(x) is positive on the given intervl, the re is A = x dx = x = 8. (b) Grph the function f(x) = x nd note tht it is negtive on [,]. Thus, the re is A = x (x )dx = ( x) = + 6 + = + =. (c) Grph the function nd check the sign of f(x) on [-,4]. Notice tht f(x) chnges the sign. Find x-intercepts: f(x) = x 9 = when x = 9 x = ±. The relevnt zero is since - is not in the intervl [-,4]. From the grph, you cn see tht f(x) is negtive on [-, ) nd positive on (, 4]. Thus, the totl re cn be obtined s the sum of A = f(x)dx nd A = 4 f(x)dx. A = (x 9)dx = ( x + 9x) ( x 9x) 4 = 9 + 7 + 9 = 8 = 6.67. A = 4 (x 9)dx = = 64 6 9 + 7 = =. The totl re is A = A + A = 8 + =. Creful: the totl re is not 4 f(x)dx =.. (d) Check the sign of f(x) on [,] nd note tht f(x) chnges the sign. Find x-intercepts: f(x) = x x = x(x ) = x =, x =. The relevnt zero is since is not in the 5
intervl [,]. From the grph, you cn see tht f(x) is negtive on [, ) nd positive on (, ]. Thus, the totl re cn be obtined s the sum of A = f(x)dx nd A = f(x)dx. A = (x x)dx = ( x + x ) = 8 + 4 + = =.67. A = (x x)dx = ( x x ) = 9 9 8 + 4 = 4 =. The totl re is A = A + A = + 4 =. Creful: the totl re is not f(x)dx =.67. (e) Check the sign of f(x) on [,9] nd note tht f(x) chnges the sign. Find x-intercepts f(x) = x 4 = when x = 4 x = x = 4. From the grph, you cn see tht f(x) is negtive on [, 4) nd positive on (4, 9]. Thus, the totl re cn be obtined s the sum of the re A = 4 f(x)dx nd A = 9 f(x)dx. Similrly s in previous problems, you cn 4 clculte tht A = 6 = 5. nd A = 6 = 5.. The totl re is A = A +A = =.67. Creful: the totl re is not 9 f(x)dx =.. The totl growth cn be found by integrting the growth rte from to 4. 4 4 t 4 = 8 4 = 4 feet. t dt = t/ 4 = 4. If s(t) denotes the distnce trveled, the problem is sking for s(5) s() = 5 v(t)dt. Using the substitution with u = t + 9 find the ntiderivtive of the velocity function to be t + 9. Thus s(5) s() = t + 9 5 = 4 8.588 ft. 5. The totl increse in the number of people from t = to t = 5 cn be found s 5.e.t dt. Use the substitution u =.t to find the ntiderivtive nd get. e.t 5 =... (e.(5) ) 79.(.96) = 7.54 millions. / 6