Overview of Calculus I


 Egbert Mason
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1 Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things, nd understnd how they relte. You lso lerned lot of nottion, nd pplictions for the derivtive. You lso need to be ble to turn the crnk nd find the derivtive nd integrl of functions like f(x) = x 2 + 3x + 2 or g(x) = e x sin(e x ). The discovery of clculus is one of the crowning chievements of Western civiliztion, nd I hope you hve gotten feeling of how powerful these three concepts (the limit, derivtive nd integrl) re. Plese spend the time to go over this overview nd hopefully get the big picture of clculus. It will py off on the finl exm. The limit: lim f(x) = L, or f(x) L s x, mens tht f(x) is x rbitrrily close to L for ll x sufficiently close to, but not equl to,. Thus, the definition of lim f(x) hs nothing to do with f(). A function f is continuous t x iff lim f(x) = f(). If we know function is continuous t we cn use this fct to x compute the limit of f s x pproches. Almost ll the functions tht we re fmilir with re continuous on their domin. Exmple: f(x) = x2 is undefined t x =, so f is not continuous t x. We cn still consider lim f(x) since this limit hs nothing to do with x f(). Since (x + )(x ) f(x) = = x + x for ll x except, lim f(x) (x + ) = + = 2. This lst limit ws x x evluted by replcing x with, since x + is continuous function, s re ll polynomils. Thus, lim f(x) = 2, even though f() is undefined. x The grph of f is the line y = x + with the point (, 2) removed. On the other hnd, if we do not know tht function is continuous t, we need to use the definition of continuity to determine if it is continuous. Exmple: Let f be defined by f(x) = sin(π/x) if x, nd f() =. Let g be defined by g(x) = x sin(π/x) if x nd g() =. Determine if f nd g re continuous t. Answer: Since f(x) tkes on vlues, nd everything in between for rbitrrily smll x, lim f(x) Does Not x Exist (DNE). Hence, f is not continuous t. On the other hnd, since x x sin(π/x) x for ll x, nd lim( x ) x =, the x x squeeze theorem sys tht lim g(x) =. Since g() =, g is continuous x t.
2 Even though lim does not exist, we cn write lim =. This mens tht x2 x x2 is rbitrrily lrge nd positive for ll x sufficiently close to, but not equl to,. x2 x We lso studied onesided limits. For exmple, lim = nd lim x x =. An importnt reltionship between the regulr limit nd the onesided limits is ( ) lim f(x) = L lim f(x) = L nd lim f(x) = L x x x + x + x x x The derivtive: Intuitively, the derivtive of f t x =, denoted f () is the slope of the tngent line to y = f(x) t x =. The limit definition of the derivtive is f f(x) f() f( + h) f() (). x x h h The derivtive function f is defined by f f(x + h) f(x) (x). There re two h h interprettions of the derivtive: () f () is the rte of chnge of f with respect to x, t x =. (2) As stted previously, f () is the slope of the tngent line to y = f(x) t (, f()). Exmple: Let f(x) = x 2. Then f f(x) f() () x x (The lst limit ws computed in the first exmple.) x x 2 x = 2. You re expected to be ble to differentite just bout ny function using the rules nd fcts given on the hndout entitled Differentition Shortcuts. Implicit differentition is just n ppliction of the chin rule. Exmple: Find y for the curve x 2 +xy+2y 2 =. Tke the d/dx of both sides to get 2x+y+xy +4yy =. Then solve for y (2x + y) = x + 4y. An eqution of the tngent line to y = f(x) t x = is y = L(x) := f() + f ()(x ). The tngent line pproximtion is f(x) L(x) ner x =. Exmple: Let f(x) = x 2. Find n eqution of the tngent line to y = x 2 t x =, nd use this to pproximte. 2. f() = nd f (x) = 2x so f () = 2. So L(x) = + 2(x ) nd n eqution of the tngent line is y = + 2(x ). The tngent line pproximtion gives. 2 = f(.) + 2(.) =.2. This is rther close to the exct squre,. 2 =.2. Relted rtes problems: The key here is to write n eqution connecting the two quntities tht holds for ll times. Then tke the derivtive of both sides with respect to time, using the chin rule. Shpe of Grphs: The signs of f nd f determine the shpe of the grph of f. We won t define incresing nd decresing here. We sy tht f is concve up on n intervl I if f is incresing on I. There is similr definition for concve down. An inflection point is vlue of x where the concvity of f(x) chnges. Some uthors define the inflection point s the (x, y) point on the grph where the concvity of
3 y = f(x) chnges. The rules for the shpe of y = f(x) cn be summrized in this tble: f > f is incresing f < f is decresing f > f is incresing f is concve up f < f is decresing f is concve down Exmple: Consider f(x) = 2x 3 3x 2. f (x) = 6x 2 6x = 6x( x) nd f (x) = 2x 6 = 6(2x ). We cn grph f nd f to determine their signs. We cn conclude tht f is decresing on (, ), incresing on (, ), nd decresing on (, ). Furthermore, f is concve down on (,.5) nd concve up on (.5, ). Thus, x =.5 is n inflection point of f. A function f with domin D s globl minimum t x = c if f(c) f(x) for ll x in D, in which cse f(c) is the globl minimum vlue of f. A function f hs locl minimum t x = c if f(c) f(x) for ll x in the domin sufficiently close to c. (Note tht n endpoint of D cn be locl minimum.) A criticl point of f is number c in the domin of f such tht f (c) = or f (c) is undefined. An importnt theorem is: If f hs locl minimum or locl mximum t c, then c is criticl point of f. The converse is not true: If c is criticl point of f, then it is not necessrily true tht f hs locl extremum t c. Exmple: It is cler from the grphs of f(x) = x 2 nd g(x) = x tht both hve locl minimum t. According to the theorem, must be criticl point of both f nd g. Indeed it is, becuse f () = nd g () is undefined. However, consider h(x) = x 3. Since h (x) = 3x 2, is criticl point of h, but is neither locl minimum nor locl mximum of h. Therefore, criticl point of f is just cndidte for locl extremum of f. The first derivtive test for locl extrem tells us for sure if the criticl point is locl extremum: If c is criticl point of f nd f (x) chnges from positive to negtive t c, then f hs locl mximum t c. There s similr sttement for locl minim. Furthermore, if f does not chnge sign t the criticl point c, then c is not locl extremum of f. Exmples: f(x) = x 2 hs f (x) = 2x. Since f (x) chnges from negtive to positive t x =, f hs locl minimum t x =. Furthermore, f hs globl minimum t nd f() = is the globl minimum vlue of f. On the other hnd, is criticl point of h(x) = x 3, but since h (x) = 3x 2 is positive on both sides of, the first derivtive test for locl extrem sys tht the the function h does not hve locl extremum t. Recll tht x = c is n inflection point of function f if f chnges concvity t x = c. In this cse c is criticl point of f, nd furthermore f hs locl extremum t c. Also, f chnges sign t c.
4 Exmples: f(x) = x 3 3x 2 hs f (x) = 3x 2 6x nd f (x) = 6x 6 = 6(x ). Then, x = is n inflection point of f since f (x) chnges sign t x =. On the other hnd, g(x) = (x ) 4 hs f (x) = 4(x ) 3 nd f (x) = 6(x ) 2. Even though g () =, g does not chnge sign t x =, nd g hs no inflection point. You should know L Hôpitl s Rule nd how to pply it. Exmple: sin(x) lim x x [sin(x)] x [x] x cos(x) = cos() = For optimiztion problems, you need to get function of one vrible whose output you wnt to mximize or minimize. Then pply clculus techniques to find globl mximum or minimum vlue of the function f. Find the criticl points of the function nd evlute the function t the criticl points nd the endpoints to find the globl mximum or minimum vlue (= output) of f on closed intervl. The Integrl: Intuitively, the definite integrl of f from to b, denoted b f or b f(x) dx is the signed re A A 2 with x b. The re under the curve y = f(x) nd bove the y = is A, nd the re bove the curve below the xxis is A 2. For continuous functions f, the definite integrl is defined s b n f(x) dx f(x i ) x. n i= Here, x = (b )/n, x =, nd x i = +i x so x n = b. We cll f(x) the integrnd. n b The Riemnn sum using right endpoints is R n = f(x i ) x, so f(x) dx R n n by definition. We cn lso compute Riemnn sums using left endpoints (L n ). These Riemnn sums cn be used to pproximte definite integrl. The Connection between the Derivtive nd the Integrl: If we cn find n ntiderivtive of the integrnd of definite integrl we cn use the evlution theorem (lso clled the Fundmentl Theorem of Clculus, or FTC) to compute the integrl: b i= f(x) dx = F (b) F (), where F = f. (In other words, F is n ntiderivtive of f.) The most generl ntiderivtive of f is written s the indefinite integrl: f(x) dx = F (x) + C. Exmple: 2x dx = x 2 + C is the most generl ntiderivtive of f(x) = 2x. We cn use this to evlute 3 2x dx = x 2 3 = 32 2 = 9.
5 Does every continuous function hve n ntiderivtive? Yes. The FTC2 sys tht the ntiderivtive of f(x) tht stisfies F () = is F (x) = the ntiderivtive of f(x) = x 2 with F () = is F (x) = x f(t) dt = x t 2 dt = t3 3 x x = x f(t) dt. For exmple It is esy to verify tht F (x) = x 2 nd F () =. We cn write the two fundmentl theorems of clculus with single function f: b FTC: f d x (x) dx = f(b) f() FTC2: f(t) dt = f(x) dx An elementry function is one tht involves sums, products, inverses nd compositions of power functions, trigonometric functions, nd exponentils. Unfortuntely, mny elementry functions do not hve ntiderivtives which re elementry functions. Exmples of such functions re e x2, sin(x) x, cos(x2 ), nd + x 3. We sy tht e x2 dx is not n elementry integrl. The simplest expression for the ntiderivtive of f(x) = e x2 tht stisfies F () = is F (x) = x e t2 dt, nd F (x) is not n elementry function. We hve rules tht llow us to differentite just bout ny function. On the contrry, we do not hve enough rules to llow us to integrte ny function. So how do we do n integrl? There re three techniques we lerned in this clss: () recognize, (2) simplify, nd (3) substitute. (You will lern more techniques in Clculus II.) First of ll, try to recognize the integrnd s the result of differentiting some function. For exmple, sec 2 (x) dx = tn(x) + C, since we know tht d dx tn(x) = sec 2 (x). The second technique is to use lgebric simplifiction to write the integrnd s sum or difference of terms tht we cn integrte. Here re some exmples: x(2 3x) dx = (2x 3x 2 ) dx = x 2 x 3 + C x 2 + dx = (x + x2 ) dx = + ln x + C x x 2 Wrning: The integrl of product is not the product of integrls: x(2 3x) dx x2 3x2 (2x 2 2 ) + C.
6 We lerned one more technique of integrtion: usubstitution. Exmple of usubstitution: To integrte 2x cos(x 2 ) dx, let u = x 2, so du = 2xdx. Then 2x cos(x 2 ) dx = cos(u) du = sin(u) + C = sin(x 2 ) + C. Wrning: While it is true tht u 2 du = u3 3 + C, sin 2 x dx sin3 x 3 + C If u = sin(x), then du = cos(x)dx so proper use of this u substitution is: sin 2 x cos x dx = u 2 du = u3 3 + C = sin3 x + C 3 To do usubstitution in definite integrl there re two methods: Method is to do the substitution first for the indefinite integrl. For exmple, using the exmple of u substitution bove we found tht 2x cos(x 2 ) dx = sin(x 2 ) + C. Therefore π 2x cos(x 2 ) dx = sin(x 2 ) π = sin( π 2 ) sin() = sin(π) sin() =. Method 2 is to rewrite the limits in terms of u. For exmple with u = x 2, the lower limit x = corresponds to the lower limit u = 2 =, nd the upper limit x = π corresponds to the upper limit u = π 2 = π. Therefore π Wrning: Note tht 2x cos(x 2 ) dx = π cos(u) du = sin(u) π = sin(π) sin() =. since π π 2x cos(x 2 ) dx = while π 2x cos(x 2 ) dx π cos(u) du, cos(u) du = sin(u) π = sin( π) sin() = sin( π).98.
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