The Fundamental Theorem of Calculus

Transcription

1 MATH 6 The Fundmentl Theorem of Clculus The Fundmentl Theorem of Clculus (FTC) gives method of finding the signed re etween the grph of f nd the x-xis on the intervl [, ]. The theorem is: FTC: If f is continuous function on [, ] nd F is ny ntiderivtive of f, then Are( f, [, ]) = F() F(). Recll tht the ntiderivtive of f is lso clled the indefinite integrl denoted y f (x). An ntiderivtive is nother function F such tht F (x) = f (x). Becuse the derivtive of constnt is, ny constnt C my e dded to F(x) to otin nother ntiderivtive. Thus we write f (x) = F(x) + C. Then since G(x) = F(x) + C is lso n ntiderivtive, the FTC sttes tht we cn lso use G(x) to find Are( f, [, ]) y computing G() G(). However, G() G() = ( F() + C) ( F() + C) = F() F(). Thus the dditionl constnt C is irrelevnt when computing Are( f, [, ]). Definite Integrl Nottion Becuse Are( f, [, ]) is computed y using n ntiderivtive f (x) = F(x), we use the sme integrl sign nottion to denote the signed re, ut we dd limits of integrtion nd to denote the intervl. Thus we shll strt writing f (x) to denote the signed re Are( f, [, ]). This expression is clled the definite integrl of f from to. We use the following nottion to write the evlution nd computtion of the definite integrl: f (x) = F(x) = F() F(). Exmple. Evlute the signed re etween the grph of f (x) = 4 x nd the x-xis on the intervl [, 4]. Explin the result pictorilly. 4 Solution. We evlute the definite integrl (4 x ) y finding n ntiderivtive F(x), nd then computing F(4) F() :

2 4 (4 x ) = 4x x 4 64 = 6 7 = 9 64 = 7 units. Thus, the net re is 7/ sq. units. There is more re elow the x-xis resulting in negtive net re. 4 Exmple. Evlute the signed re etween the grph of f (x) = 4 sin(x) nd the x- xis on the intervl [, ]. Explin the result pictorilly. Solution. We find n ntiderivtive F(x) of 4 sin(x) nd then evlute from to y computing F() F(): 4sin(x ) = 4 cos( x) = cos(x ) = cos() ( cos() ) = ( ) = Grphing f (x) = 4 sin(x) on [, ], we see tht there is n equl mount of re ove the x-xis s elow the x-xis. Thus, the net signed re is. An definite integrl cn e computed numericlly in two different wys on your clcultor. With either wy, first type the function into Y. We shll demonstrte with 4 (4 x ). The fnint Commnd On TI-84: After typing the function 4 x into Y, press Quit to return to the Home screen. Press MATH nd scroll down to fnint( nd press ENTER. In generl, enter the commnd fnint(y, X,, ). Here use the commnd fnint(y, X,, 4) to otin.. On TI-89 : After typing the function 4 x into y, press Quit to return to the Home screen. Press F then press for (. In generl, enter the commnd ( y(x), x,, ). Here use the commnd ( y(x), x,, 4) to otin.. You cn lso use the nint commnd in F: nint(y(x), x,-, 4).

3 Evluting the Integrl After Grphing On TI-84: After grphing the function 4 x on the intervl [, 4], press CALC, then press 7. Type for Lower Limit nd press ENTER, then type 4 for Upper Limit nd press ENTER. On TI-89 or TI-9: After grphing the function 4 x on the intervl [, 4], press F5 then press 7. Type for Lower Limit nd press ENTER, then type 4 for Upper Limit nd press ENTER. Are Between Curves Let f ( x) nd g(x ) e continuous functions over [, ] with f ( x) g(x ). Then the re etween the grphs of f nd g is given y Are = ( g( x) f ( x) ). g(x ) f ( x) Are If it is not lwys the cse tht f ( x) g(x ), then the re etween the curves is Are = g(x ) f (x ). Exmple. () Find the re etween the grphs of y = 4 cos x nd y = 4sin x on the intervl [, /]. () Find the re etween these grphs on the intervl [, ]. Solution. () On [, /], we hve 4 sin x 4 cos x. Thus, the re etween the curves is / / Are = ( 4cos x ( 4 sin x) ) = ( 4 cos x + 4sin x) = ( 4sin x 4 cos x) / = ( 4sin( / ) 4 cos( / ) ) ( 4sin 4 cos ) 4 sin x 4 cos x = (4 ) ( 4) = 8 squre units. () On [, ], it is not lwys the cse tht one grph is greter thn the other. So we must find the point of intersection of the two grphs y solving 4 cos x = 4 sin x, which gives = tn x, nd x = / 4. From to /4, we hve 4 sin x 4 cos x. But from /4 to, we hve 4 cos x 4 sin x. So the totl re etween the curves is

4 /4 Are = 4cos x ( 4 sin x) + 4sin x 4cos x /4 /4 = 4cos x + 4sin x + 4 sin x 4 cos x /4 /4 = ( 4sin x 4cos x) + 4 cos x 4sin x /4 = ( 4sin( /4) 4cos( /4)) ( 4sin 4 cos) + ( 4cos 4sin) ( 4cos( /4) 4 sin( /4)) = 4 4( ) + 4( ) 4 4( ) 4 = = 8 squre units. Exmple 4. Find the re etween the grphs of k (x ) = x + nd h(x ) = x on the intervl [, ]. Solution. From the grphs of the line k (x ) nd the prol h(x ), we see tht no one function is lwys greter thn the other over the entire intervl [, ]. So we must find the points of intersection of the two grphs y solving x + = x. Multiplying y gives x + = x, then = x + x = (x + )(x ). Thus, the grphs intersect t x = nd x =. From to nd from to, we hve x + x. But from to, we hve x x +. Thus the re etween the curves is given y

5 Are = k( x) h( x) = x x + + x + x + x x + = x + x + x + x + x + x = 6 x + 4 x x + 4 x + x 6 x + 6 x + 4 x x 8 = = = 7.5 squre units.