# Section 4.3. By the Mean Value Theorem, for every i = 1, 2, 3,..., n, there exists a point c i in the interval [x i 1, x i ] such that

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1 Difference Equtions to Differentil Equtions Section 4.3 The Fundmentl Theorem of Clculus We re now redy to mke the long-promised connection between differentition nd integrtion, between res nd tngent lines. We will look t two closely relted theorems, both of which re known s the Fundmentl Theorem of Clculus. We will cll the first of these the Fundmentl Theorem of Integrl Clculus. Suppose f is integrble on [, b] nd F is n ntiderivtive of f on (, b) which is continuous on [, b]. In prticulr, F () = f() for ll in (, b). Let P = {,,,..., n } be prtition of [, b] nd, s usul, let i = i i, i =,, 3,..., n. Now F (b) F () = F ( n ) F ( ) = F ( n ) + (F ( n ) F ( n )) + (F ( n ) F ( n )) + + (F ( ) F ( )) F ( ) = (F ( n ) F ( n )) + (F ( n ) F ( n )) + + (F ( ) F ( )) n = (F ( i ) F ( i )). i= (4.3.) By the Men Vlue Theorem, for every i =,, 3,..., n, there eists point c i in the intervl [ i, i ] such tht Since F (c i ) = f(c i ) nd i i = i, it follows tht Hence, putting (4.3.3) into (4.3.), F (c i ) = F ( i) F ( i ) i i. (4.3.) F ( i ) F ( i ) = f(c i ) i. (4.3.3) F (b) F () = n f(c i ) i. (4.3.4) i= Thus F (b) F () is equl to the vlue of Riemnn sum using the prtition P, nd so must lie between the upper nd lower sums for P. Tht is, we hve shown tht for ny prtition P, L(f, P ) F (b) F () U(f, P ). (4.3.5) Copyright c by Dn Sloughter

2 The Fundmentl Theorem of Clculus Section Figure 4.3. Region beneth the grph of f() = over the intervl [, ] But, since f is integrble, there is only one number tht hs this property, nmely, f()d. In other words, we hve shown tht f()d = F (b) F (). (4.3.6) Fundmentl Theorem of Integrl Clculus If f is integrble on [, b] nd F is n ntiderivtive of f on (, b) which is continuous on [, b], then f()d = F (b) F (). (4.3.7) This result revels sense in which integrtion is the inverse of differentition: The definite integrl of function f my be evluted esily, using (4.3.7), provided we cn find function F whose derivtive is f. It is common to write F () b for F (b) F (). With this nottion, (4.3.7) becomes f()d = F () b (4.3.8). Emple Since is n ntiderivtive of f() =, we hve F () = 3 3 d = 3 3 = 3 = 3. Thus the re under the prbol y = nd bove the intervl [, ] on the -is is ectly 3. See Figure 4.3..

3 Section 4.3 The Fundmentl Theorem of Clculus Figure 4.3. Region beneth the grph of y = sin() over the intervl [, π] Note tht F in the previous emple is but one of n infinite number of ntiderivtives of f. We cn in fct use ny ntiderivtive of f we wnt in pplying (4.3.7), lthough we typiclly choose the simplest one we cn find. Emple Since G() = is n ntiderivtive of g() = + (you my check by differentiting G), we hve ( + )d = s we climed in Section 4.. ( ) = ( ) ( 3 ) = 6, Emple If A is the re under one rch of the curve y = sin(), then A = sin()d. Since F () = cos() is n ntiderivtive of f() = sin(), we hve A = sin()d = cos() π = cos(π) ( cos()) = + =. See Figure Emple Since F () =

4 4 The Fundmentl Theorem of Clculus Section 4.3 is n ntiderivtive of f() = 4 + (gin, you my check this by differentiting F ), we hve 3 ( 4 (4 + )d = 3 3 ) 3 + ( 8 = 3 9 ) + 6 ( 33 ) 4 = Emple Since is n ntiderivtive of f(t) = t, we hve 4 F (t) = 3 t 3 t dt = 3 t 3 4 = 6 3 = 6 3. As cn be seen from these emples, the Fundmentl Theorem of Integrl Clculus provides us with powerful tool for evluting definite integrls ectly. However, to utilize the theorem we must first find n ntiderivtive for the function we re integrting. This turns out to be difficult problem in generl, nd we will devote the net two sections, s well s prts of Chpter 6, to developing techniques to id in finding ntiderivtives. For emple, F () = 3 cos() sin() cos() 3 8 sin() is n ntiderivtive of f() = 3 sin(), s my be checked by differentition, but t this point it is not cler how to find such n ntiderivtive in the first plce. Moreover, there re integrble functions, even reltively simple ones such s f() = sin(), which do not hve ntiderivtives epressible in terms of the elementry functions studied in clculus. The Fundmentl Theorem of Integrl Clculus tells us tht if function f hs n ntiderivtive, then we my use tht ntiderivtive to evlute definite integrl of f, but it does not tell us which functions hve ntiderivtives. The Fundmentl Theorem of Differentil Clculus will tell us, in prt, tht every continuous function hs n ntiderivtive. Before beginning tht discussion, we need to etend the definition of the definite integrl slightly. The definition of f()d in Section 4. implicitly ssumes tht < b. For the work we re bout to do, we need to etend the definition to include b, s we did in Problem 9 of Section 4.. First of ll, if = b, it would seem resonble for the vlue of the definite

5 Section 4.3 The Fundmentl Theorem of Clculus 5 integrl to be since the region between the grph of the function nd the -is hs been reduced to line segment. Hence we mke the following definition. Definition For ny function f defined t point, we define f()d =. (4.3.9) Note tht with this definition, the sttement f()d = f()d + c f()d, (4.3.) which we discussed in Section 4. in the cse < c < b, holds true even if = c, b = c, or = b = c. Now suppose we hve < b < c. Then from which it follows tht If we define f()d = f()d = c f()d + f()d b b f()d, (4.3.) f()d. (4.3.) f()d = f()d, (4.3.3) b then we my rewrite (4.3.) in the form of (4.3.). For this reson, we mke the following definition. Definition If b < nd f is integrble on [b, ], we define f()d = f()d. (4.3.4) b You my check tht with these two etensions to the definition of the definite integrl, we my now stte the following proposition. Proposition If f is integrble on closed intervl contining the points, b, nd c, then f()d = f()d + c f()d. (4.3.5)

6 6 The Fundmentl Theorem of Clculus Section 4.3 y = f( t) b Figure F () = f(t)dt is the re from to We my now return to our discussion of ntiderivtives nd the Fundmentl Theorem of Differentil Clculus. Suppose f is continuous on the intervl [, b]. We wnt to construct n ntiderivtive for f on (, b). From the Fundmentl Theorem of Integrl Clculus, we know tht if F is n ntiderivtive of f on (, b) which is continuous on [, b], then for ny in (, b) we would hve tht is, f(t)dt = F () F (), (4.3.6) F () = F () + f(t)dt. (4.3.7) Hence, if we re seeking n ntiderivtive for f, it mkes sense to define F () = f(t)dt (4.3.8) nd verify tht F () = f() for ll in (, b). Note tht F (), geometriclly, is the cumultive re between the grph of f nd the -is from to, s shown in Figure We need to compute ( F F ( + h) F () +h ) () = lim = lim f(t)dt f(t)dt (4.3.9) h h h h for in (, b). Now so +h +h f(t)dt = f(t)dt f(t)dt + f(t)dt = +h +h f(t)dt, (4.3.) f(t)dt. (4.3.)

7 Section 4.3 The Fundmentl Theorem of Clculus 7 y = f( t) + h b Figure h f(t)dt f(t)dt = +h f(t)dt See Figure Thus F +h () = lim f(t)dt. (4.3.) h h Suppose h >. Since f is continuous, f hs minimum vlue m(h) nd mimum vlue M(h) on the intervl [, + h]. Hence m(h) f() M(h) for ll in [, + h], from which it follows tht +h m(h)dt +h f(t)dt Since m(h) nd M(h) re constnts, (4.3.3) implies Thus m(h)h m(h) h +h +h +h M(h)dt. (4.3.3) f(t)dt M(h)h. (4.3.4) f(t)dt M(h). (4.3.5) Now m(h) = f(c) for some c in [, + h]. Moreover, s h pproches, + h pproches, nd so c must lso pproch. Hence, since f is continuous, lim h h m(h) = lim f(c) = f(). (4.3.6) + + Similrly, lim M(h) = f(). (4.3.7) h +

8 8 The Fundmentl Theorem of Clculus Section 4.3 It now follows from (4.3.5) tht A similr rgument shows tht nd so we my conclude tht +h lim f(t)dt = f(). (4.3.8) h + h +h lim f(t)dt = f(), (4.3.9) h h F +h () = lim f(t)dt = f(). (4.3.3) h h Tht is, F () = is n ntiderivtive of f on (, b). f(t)dt Fundmentl Theorem of Differentil Clculus intervl [, b] nd F is the function on (, b) defined by If f is continuous on the closed F () = f(t)dt, (4.3.3) then F is differentible on (, b) with F () = f() for ll in (, b). In other words, for ll in (, b). d f(t)dt = f() (4.3.3) d It is worth noting tht (4.3.3) holds for < s well, s long s f is continuous on closed intervl which contins both nd. Emple Let sin(), if, f() =, if =. Then f is continuous on (, ), so F () = f(t)dt

9 Section 4.3 The Fundmentl Theorem of Clculus Figure Grphs of f() = sin() nd F () = sin(t) t dt is n ntiderivtive of f on (, ). In prticulr, F () = sin() for ll. The grphs of F nd f re shown in Figure Geometriclly, F () is the cumultive re between the grph of f nd the -is from to nd F () is the rte t which re is ccumulting s increses. Since the rte t which re is ccumulting depends on the height of the curve, it is nturl to epect, nd the Fundmentl Theorem of Differentil Clculus confirms, tht F () = f(). The function F is known s the sine integrl function. It my be shown tht it is not representble in closed form in terms of the elementry functions of clculus. Emple Emple Using Leibniz nottion, Suppose d d G() = sin(t )dt = sin( ). 3 Then G() = F (h()), where h() = 3 nd sin(t )dt. Hence, using the chin rule, F () = sin(t )dt. G () = F (h())h () = sin((3) )(3) = 3 sin(9 ). Emple Suppose H() = + t 4 dt.

10 The Fundmentl Theorem of Clculus Section 4.3 Then, using (4.3.4), so H() = H () = d d + t 4 dt, + t 4 dt = + 4. Emple Suppose Then, using (4.3.5) nd (4.3.4), F () = + t4 dt + F () = + t4 dt. + t4 dt = + t4 dt + + t4 dt. Note tht there is nothing specil bout using in this decomposition, other thn the requirement tht the function f(t) = + t 4 be integrble on ll of the relevnt intervls. Now we hve F () = d + t4 dt + d + t4 dt d d = + () 4 () + + ( ) 4 () = To summrize this section, the Fundmentl Theorem of Integrl Clculus provides us with n elegnt method for evluting definite integrls, but is useful only when we cn find n ntiderivtive for the function being integrted. The Fundmentl Theorem of Differentil Clculus tells us tht every continuous function hs n ntiderivtive nd shows how to construct one using the definite integrl. Unfortuntely, this brings us in circle nd does not provide us with n effective mens for finding ntiderivtives to use in pplying the Fundmentl Theorem of Integrl Clculus. For emple, we know tht F () = sin(t) t is n ntiderivtive of f() = sin(), but this is of no help in evluting, sy, 4 sin() Hence in order to fully utilize the Fundmentl Theorem of Integrl Clculus in the evlution of definite integrls, we must develop some procedures to id in finding ntiderivtives. We will turn to this problem in the net section. d. dt

11 Section 4.3 The Fundmentl Theorem of Clculus Problems. Evlute the following definite integrls using the Fundmentl Theorem of Integrl Clculus. () (c) (e) (g) (i) 4 d 3 d (b) (d) ( )d (f) t dt sin()d (h) (j) π ( + )d 3 d d t + dt cos(z)dz. Evlute the following definite integrls using the Fundmentl Theorem of Integrl Clculus. () (c) (e) (g) (i) (k) 4 3 π π ( + ) d (b) + t dt sin()d (d) 4 (f) 4 sin(3)d (h) sin( )d sin( )d (j) π 3. For ech of the following functions, grph both f nd together over the given intervl. F () = (l) f(t)dt ( + ) d sec ()d 5 cos(3)d 8 cos(5θ)dθ ( + ) 5 d ( + ) 5 d () f() = sin() on [ π, π] (b) f() = sin( ) on [, ] (c) f() = on [ 3, 3] (d) f() = on [, ] + 4 +

12 The Fundmentl Theorem of Clculus Section Find the derivtives of ech of the following functions. () F () = (c) F () = (e) f() = sin (4t)dt (b) g() = 5. Evlute the following derivtives. () (c) d d d dt 5 cos 3 (t)dt (d) G(t) = ds (f) h(z) = + s + t dt (b) d d t sin (3)d (d) d d 3 3 t 3z z 3 t + dt 4 z dz sin(3t) t + t dt dt + t dt 6. Find the re of the region beneth one rch of the curve y = 3 sin(). 7. Let R be the region bounded by the curves y = nd y = ( ) nd the -is. Find the re of R. 8. Eplin why the integrl ( )d is the re of the region bounded by the curves y = nd y =. Find this re. 9. Eplin why the integrl ( )d is the re of the region bounded by the curves y = nd y =. Find this re.

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