SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is not immeditely cler how. To more fully explore this connection, let us consider Riemnn sums in more depth. Our first step will be to give the Riemnn sum new nme nd nottion. Rther thn write out the sum in full detil ech time we compute it, we will use the nottion f(x) = lim n n f(x i ) x. (1) This is clled the definite integrl nd it is one of the foundtionl concepts of clculus. We will see lter in this lecture tht the definite integrl cn be defined in multiple wys, of which the Riemnn sum formultion (1) is only one possibility. (The other definition will provide very importnt connection with ntidifferentition!) Consider the question of how the re under function f(x) chnges s we move one of the bounds to the right or the left. For the ske of illustrtion, let s keep the bound fixed nd llow the bound b to be given by the vrible x. Given this set up, we will let nd g(x + h) = +h i=1 f(x) f(x). Tht is to sy, we let g(x) correspond to the re under the curve from to x, nd g(x + h) to be the re under the curve from to x + h. Now we sk the question: wht is the difference between these two res? It is esy to see tht the mount of re dded (or subtrct, if below the xis) by extending the right bound from x to x+h is g(x+h) g(x). We cn see tht this re is very well pproximted by the rectngle between x nd x + h where we tke the height to be f(x). This rectngle hs dimensions h nd f(x) so tht we hve g(x + h) g(x) hf(x) 1
or, written nother wy, g(x + h) g(x) h f(x). This is lredy strting to tke fmilir form! But let s mke one more considertion. We cn see tht for well-behved functions, this pproximtion becomes better nd better s h becomes smller nd smller, so tht in the limit we hve g(x + h) g(x) lim = f(x). h h We immeditely recognize the left-hnd side s the derivtive of g(x). In other words, we hve g (x) = f(x)! Wht does this men? The function g(x) corresponded to the re under the curve f(x) from the left-point to the right-point x. This tells us tht if we tke the function f(x), compute its re g(x), nd tke the derivtive of tht quntity, we recover the originl function f(x). In other words, derivtion undoes integrtion. This is clled the First Fundmentl Rule of Clculus. Theorem.1 (First Fundmentl Theorem of Clculus). Suppose f is continuous on [, b]. Then the function g defined by f(t) dt x b is continuous on [, b] nd differentible on (, b), nd g (x) = f(x). The sttement g (x) = f(x) cn be restted in the equivlent, nd often more insightful, form f(x) = d f(t) dt. This is the most explicit wy of stting the intuition of the min result, tht differentition reverses integrtion. Exmple 1: Find the derivtive g (x) of the function Solution: We hve tht g (x) = d sin(t) t sin(t) t 2 dt. dt = sin(x) x
by the First Fundmentl Theorem of Clculus. Exmple 2: Find the derivtive g (x) of the function 5 t 2 ln(πt + 7) dt. Solution: When pplying the First Fundmentl Theorem of Clculus, it does not mtter wht the lower bound is. Consequently, we hve tht g (x) = d 5 t 2 ln(πt + 7) dt = x 2 ln(πx + 7). Now we sk the reverse question. If we differentite function, does integrtion recover the originl function? And if so, in wht mnner does this mnifest itself? Consider the function F (x) given by F (x) = f(x). We know from the First Fundmentl Theorem of Clculus tht F (x) = f(x). In other words, F (x) is n ntiderivtive of f(x)! This mens tht ntidifferentition nd integrtion (in the form of Riemnn sums) were essentilly the sme process. Now consider wht hppens when we tke the fixed bounds of integrtion nd b. We hve F (b) F () = f(x) f(x) = f(x) since the second integrl hs no width (nd therefore zero re). This gives rise to the Second Fundmentl Theorem of Clculus. Theorem.2 (Second Fundmentl Theorem of Clculus). Suppose f is continuous on [, b]. Then f(x) = F (b) F () where F is the ntiderivtive of f (i.e. F (x) = f(x)). In other words, s we expected, integrtion reverses differentition. More importntly, however, is tht we now hve wy to integrte which does not rely on Riemnn sums. All we need to do is find the ntiderivtive of
the given function, nd evlute tht t the upper nd lower bounds nd b! Exmple : Reconsider the question of finding the re under the curve f(x) = x 2 between x = nd x = 1. Solve using the Second Fundmentl Theorem of Clculus. Solution: We cn now stte this s the integrl question x 2 nd solve it using ntidifferentition. We know tht the ntiderivtive of f(x) = x 2 is F (x) = x. We use the Second Fundmentl Theorem of Clculus. The stndrd nottion for the solution is [ ] x x 2 1 = [ (1) = () ] = 1. In other words, we hve very esily recovered the result from lst week. Exmple 4: Evlute the definite integrl sin(πt) dt. Solution: We hve d cos(πt) = π sin(πt) so tht dt [ sin(πt) dt = cos(πt) ] 1 π [ = cos(π) + cos() ] π π = 2 π. Exmple 5: Evlute 1 1 t dt. 4
Solution: We re tempted to use the First Fundmentl Theorem of Clculus; however, the upper bound is x 2, not x, so this integrl does not fit the required form for the theorem. We cn, however, solve this directly using the Second Fundmentl Theorem. Since F (t) = ln(t) stisfies F (t) = 1/t, by the Second Fundmentl Theorem of Clculus, we hve 1 1 t dt = d [ln(t)]x2 1 = d [ ln(x 2 ) ln(1) ] = d ln(x2 ) = 2 x. This nswer is different thn wht we would hve gotten if we hd pplied the First Fundmentl Theorem of Clculus. Clerly, for integrls with bounds other thn constnt to x, differentition undoes integrtion in different wy thn presented in the First Fundmentl Theorem. It is not cler yet, however, wht this difference is. To nswer this, we consider the generl form d h(x) f(t) dt. (2) g(x) Tht is to sy, consider tking n integrl where both the lower nd upper bounds vry in x, nd then tking the derivtive with respect to x. Exmple 5 flls into this ctegory, tking x 2 nd h(x) = 1. Suppose tht F (x) is n ntiderivtive of f(x), i.e. function such tht F (x) = f(x). We cn use the Second Fundmentl Theorem of Clculus to give d h(x) g(x) f(t) dt = d [F (g(x)) F (h(x))] (FTCII) = F (g(x))g (x) F (h(x))h (x) (Chin rule) = f(g(x))g (x) f(h(x))h (x). In other words, to evlute derivtive of n integrl of the form (2) we need to evlute the function t the upper bound (g(x)) then tke the derivtive of tht bound, nd evlute the function t the lower bound (h(x)) then tke the derivtive of tht bound. This is clled Liebniz s Rule. 5
It is importnt to note tht we do not need to know the ntiderivtive F (x) in order to evlute (2). Tht is to sy, we cn evlute these integrls even when the function itself cnnot be integrted! Exmple 6: Evlute t dt. x Solution: We hve tht h(x) = x, x 2, nd f(t) = t. It follows tht h (x) = 1, g (x) = 2x, f(h(x)) = x, nd f(g(x)) = x 2, so tht t dt = x 2 (2x) x(1) = 2x x. x Exmple 7: Find p (x) given tht p(x) = 2 cos(x) sin(x) e t2 dt. Solution: We hve tht h(x) = sin(x), 2 cos(x), nd f(x) = e t2. It follows tht h (x) = cos(x), g (x) = sin(x), f(h(x)) = e sin2 (x), nd f(g(x)) = e (2 cos(x))2 so tht p(x) = d 2 cos(x) sin(x) e t2 dt = e (2 cos(x))2 sin(x) e sin2 (x) cos(x). It is interesting to note tht we did not hve to compute the ntiderivtive of e t2 in order to evlute this. This is prticulr fortunte for this exmple... becuse e t2 does not hve one! 6