Further integrtion Stndrd derivtives nd integrls The following cn be thought of s list of derivtives or eqully (red bckwrds) s list of integrls. Mke sure you know them! There ren t very mny. f(x) f (x) f(x) f (x) x n nx n sinh x cosh x log x /x cosh x sinh x e x e x tn x sec 2 x sin x cos x tn x /( + x 2 ) cos x sin x Notes: (i) log x mens the sme s ln x (log bse hs no importnce in clculus); (ii) for sin, cos, etc, x must be in rdins (gin, degrees re n rbitrry mesure with no significnce in clculus); (iii) sec 2 x = + tn 2 x; (iv) tn mens the sme s rctn. Recll the chin rule: dy dx = dy du d log(x + ) = dx x +, nd corrsponding integrls in reverse, e.g. e 2x dx = 2 e 2x + c, du. This gives exmples like dx d dx ex = e x, d cos x = sin x, dx (x + ) 2 dx = Definite integrls We define b f(x) dx (where b) to be the re between y = f(x), y =, x = nd x = b (with res below y = counted negtively). We lso define f(x) dx to be b f(x) dx. b Bsic fct. If F (x) = f(x) for ll x nd f(x) is continuous, then b f(x) dx = F (b) F (). Note. In clcultions, one often writes the RHS first s [F (x)] b.
Sketch of reson. For ech x, let I(x) = x f(t) dt. Then I() =, I(b) = b f(t) dt nd I(x + δx) I(x) = re of thin strip f(x)δx, hence I (x) = f(x). But F (x) = f(x), so I (x) F (x) = for ll x. Hence I(x) F (x) is constnt, sy I(x) = F (x) + c. So s stted. Exmple. b x 2 + dx = f(t) dt = I(b) I() = F (b) F (), Some elementry fcts. () b c dx = c(b ), simply becuse this is the re of the rectngle shown: there is no rel need for the [cx] b step (nd this step will often be omitted in these notes). (2) If f(x) throughout [, b], then b f(x) dx. In other words, if the integrnd is positive throughout the rnge, then the nswer must be positive. This very simple fct is often useful check to one s clcultions! Similrly, if m f(x) M then m(b ) b f(x) dx M(b ). throughout [, b], For instnce, in the lst exmple, this principle shows t once tht the integrl must be less thn, so tht 45 would be nonsensicl nswer! (3) The integrnd must exist throughout the intervl (but see below). The following is bsolute nonsense! x 2 dx = Integrtion by prts Recll the product rule: (uv) = u v + uv. Stted in terms of integrls, this sys (u v) = uv (uv ), b u (x)v(x) dx = [u(x)v(x)] b 2 b u(x)v (x) dx.
Exmple. To find 2 log x dx. This equls 2. log x dx = Rewriting the integrnd Sometimes this must be done before we cn recognize the integrl. Some exmples: 2 x Exmple. Find dx. This is x + 2 ( ) dx = x + Exmple. Find dx. By prtil frctions, the integrnd equls x(x ) so the integrl is x x, Exmple. (An importnt one!) cos 2 x dx. Recll tht cos 2 x = ( + cos 2x). Hence 2 the integrl is x + sin 2x + c. 2 4 Even nd odd functions We sy tht f is n even function if f( x) = f(x) for ll x, nd n odd function if f( x) = f(x) for ll x. For exmple, x 2 nd cos x re even, while x 3 nd sin x re odd, nd e x is neither. If f is even, then by symmetry so f(x) dx = f(x) dx = 2 f(x) dx, f(x) dx. If f is odd, then by cncelltion f(x) dx = f(x) dx, so f(x) dx =. 3
For exmple, with no further effort, we hve π π sin x x 2 + dx =. Two further types derived from the chin rule so () By the chin rule, d dx log f(x) = f (x) f(x), f (x) f(x) dx = log f(x) + c. One could do this type by substituting f(x) = u, but it rises so often tht it is worth regrding it s stndrd type to remember. Exmples: 6x + dx = 2 π/4 (2) Since x x 2 + dx = tn x dx = x log x dx = for n. Exmples: (x 2 + ) 3 x dx = π/4 sin x cos x dx = d dx f(x)n+ = (n + )f(x) n f (x), it follows tht f(x) n f (x) dx = f(x)n+ n + + c cos x sin 2 x dx = log x x dx = Reduction formule (log x) x dx = This technique is best illustrted by n exmple. Let I n = 4 x n e x dx
for n =,, 2,.... We express I n in terms of I n (for n ); in this cse (but not lwys), this is done by integrting by prts: I n = We now evlute I directly, nd use (*) to deduce the vlues of I, I 2,.... Since x =, I = So by (*), Improper integrls f(x) dx mens the limit s R of R Exmple. To find R (x + 2) 2 dx = (x + 2) 2 dx. f(x) dx, if the limit exists. so the required integrl exists nd equls (Note tht we relly re considering limit, not just tking the vlue t.) Similrly, if f(x) is undefined t (for exmple, if it tends to infinity there), we define b f(x) dx to be the limit (if it exists) of b f(x) dx s δ. +δ Exmple. Let < r <. Then /x r s x +. However, δ x r dx = so x dx = r r. Integrtion by substitution This is the chin rule in reverse. We replce x by function x(t) (which we hve to 5
think of for ourselves!). The rule (for definite integrls) is x2 x f(x) dx = t2 t f[x(t)] dx dt dt, where x(t ) = x nd x(t 2 ) = x 2. Sketch of reson: if F (x) = f(x), then the required integrl is But by the chin rule, so tht t2 t Exmple. Find I = F (x 2 ) F (x ) = F [x(t 2 )] F [x(t )]. d F [x(t)] = f[x(t)]dx dt dt, f[x(t)] dx dt dt = [F [x(t)]]t 2 t = F (x 2 ) F (x ). 3 x /2 + x dx. Substitute x = t 2. Then dx dt = 2t, nd: x 3 t 3. So I = 3 t 2t dt = 2 + t2 3 ( ) + t 2 dt To integrte qudrtic expressions rised to some power, substitutions involving sin, tn or sinh re often effective, becuse of the identities sin 2 t = cos 2 t, + tn 2 t = sec 2 t, + sinh 2 t = cosh 2 t. Exmples: Exmple. Let I = ( 2 x 2 ) /2 dx (the re of semicircle of rdius ). Substitute x = sin t, so tht 2 x 2 = 2 cos 2 t nd dx dt = cos t. Then x t π/2 π/2. For t in this intervl, cos t, so ( 2 x 2 ) /2 = cos t (not cos t). So I = π/2 π/2 cos t. cos t dt Exmple. Find I = dx. (x 2 + 2 ) 3/2 6
So Substitute x = tn t, so tht x 2 + 2 = 2 sec 2 t nd dx dt = sec2 t. x t π/2. Exmple. Find I = 2 dx. (x 2 + 4) /2 Substitute x = 2 sinh t, so tht x 2 + 4 = 4 cosh 2 t. sinh = log( + 2). So x 2 t c, where c = Exmple. Find dx (note this is the sme s e x + e x Substitute e x = u, or x = log u. The integrl is 2 cosh x dx). Also, to find I = dx. e x + e x Exmple. Find I = π/4 + 2 cos 2 θ dθ. First, rewrite this s π/4 sec 2 θ sec 2 θ + 2 dθ. We substitute tn θ = t. Then sec 2 θ + 2 = t 2 + 3 nd dt dθ = sec2 θ. For integrtion purposes, it is legitimte to rewrite this s sec 2 θ dθ = dt, becuse ccording to the originl instruction dθ is replced by dθ dt dt = sec 2 θ dt. To evlute this, you cn either remember tht t 2 + dt = t 2 tn, or substitute gin, s follows. Let t = 3 tn φ, so tht 3 + t 2 = 3 + 3 tn 2 φ = 3 sec 2 φ. 7
π/2 Exmple. Find I = 2 + cos θ dθ. We express this in terms of 2 θ: 2 + cos θ = It is possible (but tricky!) to combine both steps by substituting t = tn 2 θ. Volume of revolution A three-dimensionl figure is formed by rotting the curve y = f(x) round the x-xis between x = x nd x = x 2 (we ssume tht f(x) > ). The re of the circulr cross-section is πy 2, so the volume of thin slice is pproximtely πy 2 δx, nd hence the totl volume of our figure is V = x2 x πy 2 dx = x2 x πf(x) 2 dx. Exmple. A cone. This is formed by rotting the stright line y = (/h)x round the x-xis, for x h. The volume is Exmple. An ellipsoid. The ellipse x2 + y2 = is rotted round the x-xis, between 2 b2 x = nd x =. The volume is The cse b = gives the volume of sphere: 4 3 π3. 8
Length of curve Consider smll enough piece of the curve for it to be nerly stright. By Pythgors, we hve (δl) 2 (δx) 2 + (δy) 2 = (δx) 2 ( + ( ) ) 2 δy. δx Proceeding to the limit s δx tends to, we conclude tht the length of curve between x = x nd x = x 2 is L = x2 x ( + ( ) ) 2 /2 dy dx. dx It hs to be sid tht this integrl is often hrd to evlute! Of course, the integrnd is not the sme s + dy dx. length. Exmple. The curve y = cosh x, for x. This is the esiest exmple on curve Exmple. The curve y = x 3/2, for x. If the curve is given prmetriclly, by x = x(t), y = y(t), then [ (δx ) 2 ( ) ] 2 δy (δl) 2 (δx) 2 + (δy) 2 = + (δt) 2, δt δt nd the length of the curve between t nd t 2 is [ t2 (dx ) 2 L = + dt t ( ) ] 2 /2 dx dt. dt Exmple. The circle of rdius, given by x = cos t, y = sin t. The length of the rc from t to t 2 is Hence the length of the whole circle is 2π (s you well knew!). An ellipse (x = cos t, y = b sin t) lredy leds to n integrl for which there is no elementry expression. 9
Exmple. The curve y = x 2 for x. We now need to evlute both c nd sinh 2c, given tht sinh c = 2. Are of surfce of revolution Agin rotte the curve y = f(x) round the x-xis for x between x nd x 2. As before, let δl denote the length of smll section of the curve. The contribution to the surfce re of thin strip is roughly 2πyδL, nd hence the totl surfce re is S = x2 x 2πy dl = x2 x 2πy ( + ( ) ) 2 /2 dy dx. dx Exmple. The curve y = cosh x for x. Exmple. A sphere of rdius. This is formed by rotting the upper hlf of the circle x 2 + y 2 = 2 round the x-xis, with x vrying from to.
Also, consider the slice between the plnes x = x nd x = x 2 (where x < x 2 ). The surfce re of the slice is 2π(x 2 x ). It is rther remrkble tht this depends only on x 2 x, the thickness of the slice, not on its position. The volume of pyrmid Consider the figure formed by joining plne re A to point P. Here A is drwn s tringle, but it could be nything. Let h be the length of the perpendiculr from P to A (i.e. the ltitude of P ). At distnce x from P, the cross-section is copy of A, with its liner mesurements scled down by fctor x/h. Hence its re is (x 2 /h 2 )A, nd the volume of our figure is V = h x 2 h 2 A dx = A h 2 h 3 3 = 3 Ah. A prticulr cse is the circulr cone, considered in previous exmple. A digression: the pint prdox Rotte the curve y = /x round the x-xis for x [, ), to form n infinitely long trumpet-shped figure, which we regrd s continer of pint. volume is V = π dx = π. x2 The Let s try to use the pint to pint the surfce. The surfce re is ( 2π + ) /2 dx > 2π x x 4 x dx, which is, since the integrl is divergent. So we cn t possibly pint it! Cn you resolve this?
Applictions of integrtion to inequlities We will describe some pplictions of the following elementry principle: if f(x) g(x) for ll x in [, b], then b f(x) dx b g(x) dx. In prticulr, if m f(x) M on [, b], then m(b ) Exmple. Recll tht n+ n b When n x n +, we hve f(x) dx M(b ). x dx = log(n + ) log n = log( + n ). n + x. So, by the principle just stted, n n + log( + n ) n. By joining together successive intervls [r, r + ], we rrive t the following further development of the sme principle. THEOREM. Suppose tht f(x) is decresing on [p, q], where p,q re integers. Then f(p + ) + f(p + 2) + + f(q) q p f(x) dx f(p) + f(p + ) + + f(q ). The opposite inequlity pplies if f(x) is incresing. Proof. Suppose tht f(x) is decresing. The integrl, by definition, is the re under the curve. Clerly, it is not greter thn the totl re of the outer rectngles shown. Consider the first outer rectngle: its height is f(p) nd its width is, so its re is f(p). Similrly, the re of the second outer rectngle is f(p+), nd the totl re of the outer rectngles is f(p) + f(p + ) + + f(q ). In the sme wy, the totl re of the inner rectngles is f(p + ) + + f(q). If f(x) is incresing, the expressions for inner nd outer rectngles re interchnged. 2
In mny cses, we hve no precise expression for discrete sum of the form S n = f() + f(2) + + f(n), but we do hve one for the corresponding integrl, nd we cn use it to give useful upper nd lower estimtions for S n. Exmple. Let S n = + + +. There is no esy formul for S 2 n n which you cn prove by induction! But we cn give upper nd lower estimtions, s follows. Exmple. Let S n = n r= r/2. We give bounds for S n by considering f(x) = x /2. Exmple. Tke f(x) = log x, so tht S n = n r= log r = log(n!). Now n log x dx = Cll this I n. Since f(x) is incresing, we hve So we deduce the following pir of inequlities for n!: 3
The gmm function The gmm function is defined s follows s function of p for ll p > : Γ(p) = x p e x dx. The nme gmm function is entrenched for historicl resons. So is the fct tht the stted integrl is defined to be Γ(p) rther thn Γ(p ) (though this eventully does hve dvntges when it comes to stting vrious formule). The definition presupposes tht the integrl converges. cse. Results in this subsection will be numbered Γ, Γ2, etc. We strt with the simplest Γ. We hve Γ() =. Proof. By definition, Γ() = e x dx. This equls, becuse Recll tht e x = n= xn /n!, so tht for x > nd ny fixed n, we hve e x > x n+ /(n + )! (just one term of the series), hence e x x n > Hence lso for non-integer p >, x p e x s x, since it is less thn x n e x, where n is n integer greter thn p. Γ2. If the integrl defining Γ(p) converges, then so does the integrl defining Γ(p + ), nd Γ(p + ) = pγ(p). Proof. Integrting by prts, we hve R x p e x dx = When R, we hve R p e R, so the RHS tends to pγ(p). So the LHS lso tends to this limit, showing tht x p e x dx exists nd equls pγ(p). Γ3 COROLLARY. For integers n, we hve Γ(n) = (n )!. Proof. Induction. The cse n = is result Γ. Assuming the sttement true for certin n, we hve t once from Γ2 tht For non-integers p >, it now follows tht the integrl defining Γ(p) converges, by 4
comprison with n integer n > p s bove. So the gmm function interpoltes fctorils: it is function defined for ll positive rel numbers tht grees with fctorils t integers. For < p <, the integrl is lso improper on [, ], becuse x p s x from bove. However, we sw erlier tht xp dx converges. Since x p e x x p, it follows tht the gmm integrl converges on [, ], nd indeed tht the identity Γ(p + ) = pγ(p) still holds. We will not lbour this point! Finlly, substitution equtes Γ( ) to nother interesting integrl (though t this stge 2 we don t know how to evlute either of them): Γ4. We hve Γ( 2 ) = 2 e x2 dx. Proof. Substitute u = x 2 to get Bet integrls For p > nd q >, we define B(p, q) = x p ( x) q dx. The B is cpitl bet. If p < or q <, this is n improper integrl, becuse the integrnd tends to infinity t or ; however, it converges in the sme wy s xp dx. We will show how to evlute these integrls in certin cses. We will lso show how they re equivlent, through substitutions, to some other interesting integrls. evlution will be chieved by putting together number of fcts, ech (on its own) very simple to estblish. B. We hve B(q, p) = B(p, q). Proof. Substitute x = y to obtin The B2. For ll p >, we hve B(p, ) = B(, p) = p. Proof. B3. We hve B(p, q) = B(p +, q) + B(p, q + ). 5
Proof. B4. If q >, then Proof. Integrte by prts: B(p, q) = q B(p +, q ). p B5. For q >, B(p, q) = q B(p, q ). p + q Proof. By B4 nd B3 (with q substituted for q), We cn now evlute B(p, q) when p or q is n integer: B6 PROPOSITION. If q is n integer ( ), then B(p, q) = (q )! p(p + )... (p + q ). Proof. By repeted ppliction of B5, strting with B(p, 2) = By B2, we hve B(p, ) = /p, hence the sttement. Note. One cn lso prove B6 directly from either B3 or B4, without the other (exercise!). Exmple. x 3 ( x) 4 dx 6
Since Γ(q) = (q )! nd Γ(p + q) = p(p + )... (p + q )Γ(p) for integers q, one cn write B6 in the form B(p, q) = Γ(p)Γ(q) Γ(p + q). This formul is meningful when neither of p,q is n integer, nd in fct one cn show tht it equls B(p, q) in this cse too. We now consider the cse where p nd q re hlf-integers. B7. We hve B( 2, 2 ) = π. Proof. Substitute x = sin 2 θ: dx = x /2 ( x) /2 Note. As mentioned before, this is n improper integrl, but the substitution process cn be justified by considering limits. Exmple. x /2 dx ( x) /2 Using B5, we cn derive generl formul for the hlf-integer cse. B8 PROPOSITION. For integers m, n : B(, n +.3.... (2n ) ) = π, 2 2 2.4.... (2n) B(m +, n + [.3.... (2m )][.3.... (2n )] ) = π. 2 2 2.4.... (2m + 2n) Proof. Repeted ppliction of B5, strting with B(, 3 ), gives 2 2 Remember tht B(q, p) = B(p, q)! Further ppliction of B5 gives 7
Now we describe some other integrls tht equte to bet integrls by substitutions. B9. We hve ( x 2 ) p dx = 2 B(p, 2 ). Proof. Substitute x = u /2 to get Exmple. ( x 2 ) 3/2 dx For generl p nd q, the substitution used in B7 gives: B PROPOSITION. For ny p, q >, B(p, q) = 2 Proof. Put x = sin 2 θ to obtin x p ( x) q dx = π/2 In prticulr, from the formul in B8, we hve π/2 cos 2n θ dθ = B( 2, n + 2 sin 2p θ cos 2q θ dθ..3.... (2n ) ) = π, 2.4.... (2n) nd similr formul for the integrl of cos 2n+ θ is found from B6. Of course, one cn derive these integrls directly nd use them to deduce results like B. Exmple. π/2 sin 5 θ cos 9 θ dθ B PROPOSITION. For ny p > nd q >, we hve B(p, q) = y p dy. ( + y) p+q Proof. Substitute x = + y. Then x = y + y nd x y Exmple. y 4 ( + y) 8 dy 8