15 DEFINITE INTEGRALS
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1 5 DEFINITE INTEGRAL DEFINITION OF A DEFINITE INTEGRAL Let f(x) e defned n n ntervl 5 x 5. Dvde the ntervl nto n equl prt of length Ax = ( )/n. Then the defnte ntegrl of f(x) etween z = nd x = defned 5. f(x)dx = lm {f(u) Ax + f( + Ax) Ax f f( + 2Ax) Ax f( + (n ) Ax) Ax} nm The lmt wll certnly ext f f(x) pecewe contnuou. If f(x) = &g(), then y the fundmentl theorem of the ntegrl clculu the ove defnte ntegrl cn e evluted y ung the reult 5.2 f(x)dx = d g(x) dx = g(x) (I dx = c/() () If the ntervl nfnte or f f(x) h ngulrty t ome pont n the ntervl, the defnte ntegrl clled n mproper ntegrl nd cn e defned y ung pproprte lmtng procedure. For exmple, 5.3 m f(x) dx = lm f(x) dx tm Cc f(x) dx = 5.4 m f(x) dx m nrm m c 5.5 f(x) dx = lm f(x) dx f ngulr pont t 5.6 f(x) dx = lm c f(x) dx f ngulr pont +E GENERAL F6RMULA INVOLVING DEFINITE INTEGRAL 5.7 {f(x) g()*h()*...}dx = f(x) dx * g(x) dx * h(x) dx 2 * * * cl 5.8 cf(x)dx = c f (4 dx where c ny contnt 5.9 f(x) dz = 5. f(x)dx = f(x)dx 5. C f(x) dx + j c f(x) dx 5.2 = ( 4 f(c) where c etween nd Th clled the merl vulzce theorem for defnte ntegrl nd vld f f(x) contnuou n x. 94
2 DEFINITE INTEGRAL f(x) ) dx = f(c) f g(x) dx where c etween nd * Th generlzton of 5.2 nd vld f j(x) nd g(x) re contnuou n 5 x Z nd g(x) $ dlz() F(x,) dx = 6,() LEIBNITZ RULE FOR DIFFERENTIATION OF lntegral m,() F xdx f F($2,~) 2 F(+,,Y) 2 m,() APPROIMATE FORMULA FOR DEFINITE INTEGRAL In the followng the ntervl from x = to x = udvded nto n equl prt y the pont = ~, l, 22,...,,l, x, = nd we let y. = f(xo), y = f(z,), yz = j(@,..., yn = j(x,), h = ( )/%. Rectngulr formul 5.5 f (xl dx (I = h(y, + Yl + Yz +.. * + Yn) Trpezodl formul 5.6 j(x) dx = $(Y, + 2y + ZY, %,lt Y?J mpon formul (or prolc formul) for n even I 5.7 f(z) dz = ; (y. + 4y, + 2Y, + 4Y, Y,2 + 4Ynl f Yn) DEFINITE INTEGRAL INVOLVING RATONAl OR IRRATIONAL EPRI!ON 5.8 m zg o x2 dx + 2 y; xpldx 5.9 ~ =? +x n p7r O<p<l 5.2 = xmdx,n+ln o<m+<n o ~ xn + n = n n [(m + ),/n] 5.2 xm dx 77 n m/3 o + 2x co p + x2 = n m7 n / ,,md =?$ 5.24 xm(n xn)p dx m+*+n~l?[(m+l)ln]~(p+l) = nl [(m + )/n + p + l] (l)r7rm+nrr[(m + )/n] 5.25 n n \(m + l)nln](r l)! l [(m + )/n T + l] o<m+<nr
3 96 DEFINITE INTEGRAL DEFINITE IM fegrdl~jnvdlvno TRONOMETRIC FUNCTION All letter re condered 5.26 n mx n nx dx potve unle otherwe ndcted. = m, n nteger nd m f n r/2 m, n nteger nd m = n 5.27 D co mx co nx dx = m, n nteger nd m # n 7~2 m, n nteger nd m = n 5.28 TT n mx co nx dx 5.29 T/2 /2 m, n nteger nd m + n odd II 2mf (m2 4) m, n nteger nd m + n even n2 x dx = cot325 dx = ; 5.3??I2 n2mx dx = /2 co2 x dx = rnl 246..* 2m 2 m=,2 )... n/2 n/2 2*4*6..*2m 5.3 $m+l x dx = co+ +2 dx = m+l m=l,2, jr2 n2p x co29z dx = UP) r(4) 2 r(p + 9) xl2 p > 5.33 "dx = p=o %I2 p < p>q>o 5.34 m n px co qx dx = d2 < p < q TI4 p = q > 5.35 m n p:;n qx dx = pl2 <p 5 q uql2 p 2 q > 5.36 m n2px dx = m x n mx dx = :em 5.37 "lopxdx = 2 x m nmx o (x2+ 2) dx = (lem)l 5.38 m co px co qx dx = ln 9 2 P dx + n x 5.39 m~o~p~/q~ dx = 49 P) * comx o x2 + u2 dx = kem dx + co x /2 dx = co (l) + cox $23
4 2r; 27r d d o ( + n x)2 = o ( Jr co x)2 = (z ) dx 27r O<<l l2cox+z = 5.48 T x n x dx o 2 co x + 2 = 5.49 Tr co mx dx rm o l2cox+2 = l2 5.5 w DEFINITE INTEGRAL 97 (57/) In ( + ) lj < 77 In ( + l/) Il > r n x2 dx = co x2 dx = 2 II nxn dx = r(lln) n &, n>l nyn 5.52 m co xn dx = & rfl/n) co 2, n>l 5.53 jc n dx= 6 m co x dx = 6 = 5.55!?$?dx = 2Iyp) k (pn/2) 2l3p) c,, (p/2) 5.56 m n x2 co 2x dx = k 5.57 m co x2 co 2x dx = 5.58 * n3 x x3 &y = $f 2 <, m =,,2,.. O<p<l O<p<l 5.6 * tnxdx = T x z VT/2?r/z dx =T + trp tn' x dx = $ _ 32 '+$A n'x dx = ;ln llcoxdx _ m dx co x = y 5.66 : (h cox)'$ = y 5, tnl px tnlqx dx = p 5.67
5 excox dx = m ez n x dx = ~ m ez n x dx = tnl k 5.7 mcz ez dx = In!! m ecz2 co x dx = 2/ e(z2tz+c) dz = erfc 2f where m co 5.75,&tztc) d = cc xnezdx = Iyn + ) 5.76 n cc 5.78 m mezz dx = r[(m + )/2] 2(mfl)/Z ek&+/z2) dx = ; d ;e 2' "g+ = A+$+$+$+ *** = f xnl L+&+$ dx = l'(n) ln ( > For even n th cn e ummed n term of Bernoull numer [ee pge 89 nd 45. = m xdx ez + m xnl 5.82 o ez+l dx = r(n) $+$$+..* = ( $ &+ & *** > For ome potve nteger vlue of n the ere cn e ummed [ee pge 89 nd cdl: = +coth; & co ez2e*dx = & 5.86
6 DEFINITE INTEGRAL m ez x et px m e~x _ ez x cc px dx = tn tnl% 5.89 m e (lx; O ),jx = cotl ; In (2 + ) 5.9 xm(ln x) dx = (l)%! m >, n =,,2,... (m + l)n+l If n#,,2,... replce n! y r(n. + ). 5.9 l lnx dx = $ o +x & = $ 5.93 In ( + x) dx = $ ln(lx) dx =? x In x In ( + x) dx 5.96 = 22ln22 In x In (lx) dx = 2 c WC pn cot p O<p<l F dx = In m exlnxdx = y = 5(y + 2 ln2) n/2 5.2 In n x dx = dx = $ RI2 /2 5.3 (ln n x)2 dx = n/z lncox dx = (In co x)2 dx = l In2 5.4 rxlnn x dx = $ln n x In n x dx = In 2 2 2n 5.6 In ( + n x) dx = In(+coz)dx = 2rrIn(+dn)
7 DEFINITE INTEGRAL 5.7 7r ln( + cox)dx = T In U+@=G ( 2 ) 7 2~ In, 2 > In (2 2 co x + 2) dx = 5.8.( 2~ In, 2 > T/4 5.9 In ( + tn x) dx = In2 dx = +{(co~u)~ (co )2} ee lo 5.2. y n 2 + T+ n (. : DEFNlT t!tthral. NVOLVlNG NYPERBQLIC FUNCTtC?N 5.2 m nz nh x 5.3 p co x o ch x dx = $ tnh $ 7 dx = & ech% = $ 5.5 m xndx = o nh z r(n+ ) If n n odd potve nteger, the ere cn e ummed [ee pge m nh x dx = 2 cc $ 2 ez * nh ux ez dx = & 5 cot % 5.8 m ftux) ftx) & = {f(o) f(m)} ln Th clled Frullun ntegrl. It hold f f (x) contnuou nd dx 5.9 = 22 f(x) f(m) dx x converge. 5.2 I (u+x)ml(x)l& = (2)m+n;;'f;;
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