Anti-derivatives/Indefinite Integrals of Basic Functions

Transcription

1 Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: x n+ x n n + + C, dx = ln x + C, if n if n = In prticulr, this mens tht dx = ln x + C x nd x 0 dx = dx = dx = x + C Integrl of Constnt: k dx = kx + C 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 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 Anti-Derivtives 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 Constnt Multiple Rule: k f(x) dx = k f(x) dx (This pplies to definite integrls s well.)

2 Sum/Difference Rule: f(x) ± g(x) dx = f(x) dx ± g(x) dx (This pplies to definite integrls s well.) Other Properties of 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 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/mid-point, where x = b n ; x 0 =, x i = + i x, x n = b; x i = x i + x i Left End-point Approximtion: b Right End-point Approximtion: b Mid-point 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= (the midpoint of the intervl [x i, x i ])

3 Forml Definition of Definite Integrl 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 The Fundmentl Theorem of Clculus: Suppose f is continuous on the closed intervl [, b].. If g(x) = i= n i= x f(t) dt, then g (x) = f(x). [Also 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. [Also clled the Evlution Theorem.] u-substitution: 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. 3

4 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. 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 u-sub, use the Pythgoren identities to convert the remining (now even) power to the other trig function, then mke u-sub 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 hlf-ngle formuls (cos (x) = ( + cos(x)), sin (x) = ( cos(x))) to convert to cosines, expnd the result nd pply hlf-ngle formul gin if needed (keep doing this until you no longer hve ny powers of cosine), then integrte (my need simple u-sub). 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. 4

5 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 u-substitution nd/or x + dx = ( x ) tn + C). 5