Antiderivatives/Indefinite Integrals of Basic Functions


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1 Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second formul until clc ). dx = ln x + C Integrl of Constnt: k dx = kx + C Trigonometric Functions: sin(x) dx = cos(x) + C sec (x) dx = tn(x) + C sec(x) tn(x) dx = sec(x) + C cos(x) dx = sin(x) + C csc (x) dx = cot(x) + C csc(x) cot(x) dx = csc(x) + C Constnt Multiple Rule: k f(x) dx = k f(x) dx (This pplies to definite integrls s well.) Sum/Difference Rule: f(x) ± g(x) dx = f(x) dx ± g(x) dx (This pplies to definite integrls s well.) Properties Prticulr to Definite Integrls: b f(x) dx = f(x) dx f(x) dx = 0 b f(x) dx + c b b f(x) dx = c f(x) dx If f is even [i.e., f( x) = f(x)], then If f is odd [i.e., f( x) = f(x)], then f(x) dx = f(x) dx f(x) dx = 0 0
2 Integrtion Theorems nd Techniques Riemnn Sums nd Approximting Definite Integrls Approximte the net re under the grph of f(x) over the intervl [, b] using n rectngles nd the indicted end/midpoint, where x = b n ; x 0 =, x i = + i x, x n = b; x i = x i + x i Left Endpoint Approximtion: b Right Endpoint Approximtion: b Midpoint Approximtion: b f(x) dx L n = xf(x 0 ) + xf(x ) xf(x n ) = x [f(x 0 ) + f(x ) f(x n )] n i=0 f(x) dx R n = xf(x ) + xf(x ) xf(x n ) = x [f(x ) + f(x ) f(x n )] i= f(x) dx M n = xf(x ) + xf(x ) xf(x n ) = x [f(x ) + f(x ) f(x n )] i= Forml Definition of Definite Integrl (the midpoint of the intervl [x i, x i ]) Let f be defined on the intervl [, b], nd let [, b] be prtitioned into n subintervls, x i be the length of the i th subintervl nd x i ny point in tht subintervl. If the limit lim n, x 0 f(x i ) x i, exists, then we sy f in integrble on [, b] nd denote tht limit by b i= f(x) dx = lim n, x 0 f(x i ) x i, To mke this definition little esier to work with, we force the size of x to be uniform nd dependent on n, nd then define x i in terms of x nd i: x = b n, x i = + i x nd then the definition becomes b f(x) dx = lim f(x i ) x i= n i=
3 The Fundmentl Theorem of Clculus: Suppose f is continuous on the closed intervl [, b]. x. If g(x) = f(t) dt, then g (x) = f(x). [Sometimes clled the Second Fundmentl Theorem of Clculus.]. b f(x) dx = F (b) F (), where F is ny ntiderivtive of f, i.e., F = f. [Sometimes clled the Evlution Theorem.] usubstitution: If u = g(x) is differentible function whose rnge is n intervl I nd f is continuous on I, then f (g(x)) g (x) dx = f(u) du Generl Steps:. Let u = (some function of x). u is function whose derivtive (up to constnt multiple) ppers in the integrnd nd/or usully the inside of composition. It my not lwys be immeditely obvious wht u should be or tht it s derivtive is in the integrnd, but with prctice you ll strt to build librry of common tricks nd it will become esier/fster.. Rewrite the integrl in terms of u (chnging expressions with x s nd dx to expressions with u s nddu). If you hve definite integrl, then you cn convert the limits of integrtion to u s nd evlute everything in terms of u without chnging bck to x s: b f (g(x)) g (x) dx = g(b) g() f(u) du If you choose not to do this, then you should write x = in your limits of integrtion until you chnge bck to x s t the end. Either wy, you should end up with n integrnd in terms of only u, nd this integrl should be esier to evlute thn the one with which you strted. 3. Evlute this simpler u integrl. 4. If you hve n indefinite integrl, reverse the substitution to chnge your finl nswer bck to x s by replcing ll u s with whtever they re in terms of x (bsed on your originl substitution). If you hve definite integrl nd chnged your limits to u s bck in step, then simply use the Fundmentl Theorem s usul (without chnging bck to x s!). If you left your limits s x s in step, then you need to reverse the substitution (get nswer bck to x s) nd then use the Fundmentl Theorem. 3
4 Antiderivtives/Indefinite Integrls of Bsic Functions, Clc II Exponentil Functions: With bse : With bse e, this becomes: x dx = x ln() + C e x dx = e x + C If we hve bse e nd liner function in the exponent, then e x+b dx = ex+b + C AntiDerivtives Involving Inverse Trigonometric Functions: More generlly, dx = rctn(x) + C + x dx = rcsin(x) + C x x dx = rcsec(x) + C x + x dx = ( x ) rctn + C Integrtion Theorems nd Techniques, Clc II Integrtion by Prts: u dv = u v v du When choosing u nd dv, we wnt u tht will become simpler (or t lest no more complicted) when we differentite it to find du, nd dv wht will lso become simpler (or t lest no more complicted) when we integrte it to find v. If you re hving trouble deciding wht u nd dv should be to ccomplish this, you cn use LIATE to choose u (choose s high on the list s possible):. Logrithmic. Inverse Trigonometric 3. Algebric (such s polynomils [including powers of x] nd rtionl functions) 4. Trigonometric 5. Exponentil nd then whtever is left is dv. This doesn t lwys work, but it s good plce to strt. Trigonometric Integrls For integrls involving only powers of sine nd cosine (both with the sme rgument): If t lest one of them is rised to n odd power, pull off one to sve for usub, use the Pythgoren identities to convert the remining (now even) power to the other trig function, then mke usub with u =(whtever trig function you didn t sve), nd then the trig function you set side erlier will be prt of du. If they re both rised to n even power, use the hlfngle formuls (cos (x) = ( + cos(x)), sin (x) = ( cos(x))) to convert to cosines, expnd the result nd pply hlfngle formul gin if needed (keep doing this until you no longer hve ny powers of cosine), then integrte (my need simple usub). 4
5 Trigonometric Substitutions If the integrl contins n...then mke the...nd... expression of the form... substitution... x x = sin θ dx = cos(θ) dθ + x x = tn θ dx = sec (θ) dθ x x = sec θ dx = sec(θ) tn(θ) dθ Prtil Frction Decomposition Given rtionl function to integrte, follow these steps:. If the degree of the numertor is greter thn or equl to tht of the denomintor perform long division.. Fctor the denomintor into unique liner fctors or irreducible qudrtics. 3. Split the rtionl function into sum of prtil frctions with unknown constnts on top s follows: For exmple: A B + x + b cx + d + C (cx + d) }{{}}{{ } for liner fctor for repeted liner fctor x + 7 (x + )(x 3) (x + 3x + ) = A x Multiply both sides by the entire denomintor nd simplify. + B x 3 + Dx + E ex + fx + g }{{} for n irreducible qudrtic C (x 3) + Dx + E x + 3x + 5. Solve for the unknown constnts by using system of equtions or picking pproprite numbers to substitute in for x. 6. Integrte ech prtil frction. (You my need to use usubstitution nd/or x + dx = ( x ) tn + C). 5
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