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    Hari B

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DEFINITE INTEGRALS

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    CHAPTER 9 DEFINITE INTEGRATION TOPICS: 1.The definite integral 2.1nterpretation of definite integral 3.Fundamental theorem of integral calculus 4.Properties of definite integral 5. Reduction formulae
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    DEFINITE INTEGRAL Let f(x) be a function defined on [a, b]. If = F(x) , then F(b) — F(a) is called the definite b integral of f(x) over [a, b]. It is denoted by . The real number 'a' is called the lower limit and the real number 'b' is called the upper limit. This is known as fundamental theorem of integral calculus. THEOREM b b = i.e., definite integral is independent of its variable. Geometrical interpretation of definite integral b If f(x)>0 for all x in [a, b] then ff (x)dx is numerically equal to the area bounded by the b curve y =f(x), the x-axis and the lines x=a and x=b i.e., ff (x)dx . b In general, ff (x)dx represents to algebraic sum of the areas of the figures bounded by the curve y =f(x) , the x axis and the lines x=a and x=b.. the areas above x-axis are taken with plus sign and the areas below x-axis are taken with minus sign i.e., x=a c b D y=tlx) area ABC — area CPD — x=b x area EQE + area EEG.
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    PROPERTIES OF DEFINITE INTEGRALS b 1. b a Proof : let = F(x) then b b b b 2 . If a < c < b then = . Proof : Let = F(x) . b Then = b R.H.S= = b = F(c) + F(b) -F(c) = F(b) -F(a) = = L.H.S = If (a —x)dx 3. 0 0 Proof : Put a — x = t. dx = —dt U.L. t=0. 0 0 R.HS.= ff(a -x)dx = = - =L.H.S. 0 0 THEOREM b b = ff(a Proof : Put a + b — x = t, then —dx = dt dx = —dt L.L,x=a t = b U.L. t=a. b b b R.HS. = ff(a +b-x)dx = = - -LHS b b
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    THEOREM = 2 , if f(x) is an even function= 0, if f(x) is an odd function. 0 Proof: Since —a < 0 < a, = 0 In the 1st integral of RHS, Put x = —t, then dx = —dt, L.L, x = -a t = a U.L. t=O. 0 0 0 a 0 0 0 0 Case I : If f(x) is an even function then f(—x) = f(x) Then from (1), a 0 0 Case Il :lf f(x) is an odd function then f(—x) = —f(x) From (1), 0 0 THEOREM Since 0 < a < 2a, = ---(1) = 2 if f(2a — x) = f(x)s= 0 if f(2a — x) = —f(x). 0 Proof : 0 In the 2nd integral of rhs, 0 Put 2a — t = x, then —dx = dt dx = —dt
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    U.L. t=0. o o = Jf(2a —Jf(2a —t)dt — t)dt = Jf(2a — x)dx —t) (—dt) = o o From (1), = + ff(2a — x)dx ----(2) 0 0 0 Case I : if f(2a — x) = f(x) From (2), 0 0 0 Case Il : if f(2a — x) = —f(x) From (2), 0 0 THEOREM : If f(x) is a periodic function with period 'T' then = n . 0 Proof : Let S(n) be the statement that = n for n€N. 0 Let n=l, then 0 0 . S(l) is true. Assume that S(k) is true. 0 (k+1)T (k+1)T f(x)dx= f(x)dx Now 0 In the 2nd integral of rhs 0 (k+1)T f(x)dx 0
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    Put x = kT + t, then dx = dt. x = kT, (k + I)T t = 0, t = a. (k+1)T f(x)dx = = = 0 [ f(x) is a periodic function with period 'T'] (k+1)T (k+1)T f(x)dx = k f(x)dx 0 0 0 0 . S(k + 1) is true. By principle of Mathematical Induction S(n) is true, V n e N. 0 THEOREM If f(x) is an integrable function on [a, b] and g(x) is derivable on [a, b] then b f(x)dx. EXERCISE - 9(A) Evaluate the following definite integrals. I. 1. a f(a2x — x3)dx 0 a Sol. f(a2x —x3)dx = a 4 4 a x 4 2 2 Il =1n10-1n5- 4 a 4 4 a 4 3 2xdx 2. 2 2 3 2xdx 2 2 — [In | 1 + x -In(10/ 5) = In 2
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    2+2 cos ede Sol. Sol. — 4sin— = 4 Sin sin x • cos xdx sin x • cos xdx — Jsin (Tt—x)cos (Tt— x)dx — sin x cos x dx — 21 = I = O 111— x I dx Sol. 111— x I dx = —l)dx+ — + l)dx + —l)dx cos x dx 7t/2 2 — l)dx 2 sol. Let cos x dx ...(i) —7t/2
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    It/2 COS(TC/ 2 —TC/ 2— x)dx = If (a + b— x)dx Tt/2 Tt/2 ex cos xdx —It/2 Adding (1) and (2) , It/ 2 It/ 2 cos + e —It/ 2 TC/2 21=2 cosxdx(•.• Tt/2 dx = cos xdx —It/ 2 cos x is even function I — [Sin x] I = I dx 3—2x Sol. - ( 3-2.1 3-2k Sol. ax + 2 —8a 3
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    sec 4 Ode It/4 sec4 Ode = sec e. sec 7t/4 7t/4 sec2 + tan 2 0)dO Ode = Tt/4 Sol. Let 10. sec e + sec Otan e 7t/4 tan e = tan O)oTt/4 + dx o x +16 ) de = 7t/4 sec2 Ode + tan e sec2 Ode Asn: 11. sol. 12. II. 1 dx dx = J2xe-x dx, put— —2xdx = dt 2xdx= dt J —etdt = dx 2x — 1 Ans:2 Evaluate the following integrals. dx 1+x sol. dx = x —1+1 dx x — l)dx + dx
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    —x + [log(l + x)] dx sol. dx dx — jdx—2 : tan : 3-Jî tan-l(O-tan dx — tan : 3 + JE tan dx Sol. dx x +1—1 — tan — jdx — dx — dx ffl2 x 71/2 sol. x 1 — tan tan- x sin xdx TC/ 2 TC/2 sin xdxz x (—cos x)] (2x)(—cos x)dx 0) +2 xcosxdx
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    TC/2 Tt/2 (2)(sin x)dx = sin x] It/ 2 = 2 —>
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    It/ 2 sin sin Sol. let 1= It/2 x —cos x dx x + cos x It/ 2 sin x —cos x sin sin TC/2 cos cos sin x+cos x (TC/2—x) (71/2 —x) + x —sin x dx x + sin x dx If (a 2dt —x)dx = If (x)dx ...(2) Adding (1) and (2), It/ 2 Odx dx I = O cos x + sin x Ill. Evaluate the following integrals. Tt/2 dx 4 + 5 cos x Tt/2 dx sol. 4 + 5 cos x put tan Tt/2 dx 1 — tan 4+5 1 + tan Tt/2 0 4 tan dx +1 +5 1—tan2 x tan +1 —sec —dx and t
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    4t2 +4+5—5t2 9—t2 Í (x—a)(b— Sol. (x —a)(b— b—a b—a 2 dt = — In 2-3 -In 2 x)dx x)dx = 2dt 1-kt —x + (a +b)x —abdx dx —ab = TC x (b —a)2 sin 4-2 sin TC 2
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    x sin x dx put sin- x = t — dx = dt Sol. and x=sint x=0 t=0 and x=— t x sin x dx = t. sin tdt = t Isin tdt)6 0 —0+ 1. cos t)dt 0 =t(—cost)6 +(sin 12 It/4 sin x + cos x dx 9+16 sin 2x It/4 sin x + cos x dx = 9+16 sin 2x It/4 sol. sin x + cos x dx 9 + 16[1 — (sin x — cos x)2] put sin x —cosx = t (cos x + sin x)dx = dt x t = —1 and t = 0 dt I dt 25-16t2 - 16 J 25 2 16 16 1/4 — — —In —.21n.3= —In 3 40 9/4 40 20
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    a sin x + b cos x 5. dx sin x + cos x Sol. a sin x + b cos x let I = sin x + cos x 7t/2 a sin —— x +bcos —x 2 2 sin o —x + cos —x 2 2 a cos x + b sin x (2) dx sin x + cos x 1= a a = Jf(a — x)dx 21 = (a+b)dx a Jx(a — dx 6. a a(sin x + cos x) +b(sin x + cos x) dx cos x + sin x 4 Sol. letl= fx(a — x y dx a — dx a ax ndx — xn+ldx 0 n -1-1 a a x a ...(1) ...(2) a
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    sol. x 2—xdx Ans : 15 x sin xdx 3sinx —sin 3x x sin xdx — J(lt— x) sin3(1t— x)dx = Tf (a — x)dx — J(lt— x) sin xdx = Jltsin xdx— x sin xdx Itsin xdx = I 21 = Itsin xdx = 7c cos 3x —3 cosx+ —.16/3 lt 16 21C 2.4 3 dx I + sin x sol. dx I + sin x (T — x)dx 1 + sin(7t — x) ltdx 1 + sin x ltdx 1 + sin x xdx 1+ sin x
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    Itdx 1 + sin x 2 (1 — sin x) dx = 2 2 1 —sin x cos x sec xdx — sin x cos x dx 1 + sin x 1 — sin x dx cos x 1 dx cos x sec x • tan xdx ([tan x]; — [ sec x] 10. sol. x sin 1 + cos x dx x x sin x dx 1 + cos x (7t—x) sin (It—x) dx 1+ cos2(1t— x) (It—x) sin x dx 1 + cos x sin x dx — 1 + cos x sin x dx-l 1 + cos x sin xdx 1 + cos x x sin x dx 1 + cos x Put t = cos x dt = —sin x dx dt= 21 =
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    dt — [—t+ 2tan t TC TC 11. Sol. = [—1+2 tan 1 flog(l + x) dx flog(l + x) dx Put x = tan e dx = sec2 e de x = 1 —e=TC/4 It/4 log(l + x) dx = Tt/4 log(l + tan O) sec2 Ode (I + tan2 0) log(l + tan 0)d9 7t/4 log(l + tan 0)d9 let log 1+ tan tan = JIog 1+ — tan e de 1 + tan — tan e 1 — tan e log 1+ de 1 + tan e 7t/4 1+ tan e +1 — tan e log de 1 + tan e
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    7t/4 log de 1 + tan e [log 2 — log(l + tan 7t/4 7t/4 = log2 J de- log(l + tan O)dO 7t/4 = log2 J de-1 21 = log = (log 2) 1 = —log 2 x sin x 12. dx 1+cos x sol. 13. sol. x sin x dx = 1 + cos x — x) sin x dx = 1 + cos x = It{tan-l (—cos x)) (It— x) sin(Tt— x)dx 1+ cos2 x) sin xdx x sin xdx 1+ cos x 1 + cos x 21 = —+ x sin x dx = 1 + cos x sin x dx cos x + sin x sin x dx cos x + sin x sin o cos dx + sin
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    cos xdx 2. sin x + cos x 0 Adding 1. and 2. sin x +cos x dx sin x + cos x 1 dx 2 sin x + cos x dx Consider sin x + cos x 0 Put tan(x/2) = t I—t2 2dt dx = sin x = cos x = dx 2 1 2tdt sin x + cosx 0 2t + (I—t2) 1 dt 1 0 uä-l 1 logl — log 1 2 1 14. Suppose that f : R R is a continuous periodic function and T is the period of it. Let a a+11T e R. Then prove that for any positive integer n, f(x)dx=n f(x)dx. 0 0 a+2T f(x)dx f(x)dx+ Sol. dx . f (x)dx sx+ a 4-rT ...(1) 0 0 Consider ( r+l)th integral of RHS f (x)dx Let x = y + rT dx = dy
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    x = a + rT y = a x = a + (r + I)T y = a + T a -1-04-1) T f(x)dx = f(y)dy (f is periodic) f(x)dx Similarly we can show that each integral of (1) is equal to f(x)dx+ f(x)dx ...n terms = n f(x)dx 0 0 0 0 REDUCTION FORMULAE THEOREM 1 : sinn xdx then In If In 0 Proof : n 0 sinn xdx = sin x. sin x dx 0 7t/2 7t/2 = —sinn- xcosx + (n —1) sinn x. coe xdx 0 — sin x)dx xdx— sin xdx 7t/2 n—2 sin 0 TC/2 sin 0 = (n - (n -l)ln n -l ln_2 n Inn = (n —l)ln 2
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    Note In (1), replace n by n-2,n-3, I In 2 = 2—— In 4 In 4 n—2 —4 —2 then n—l n —3 n —5 n n—2 n It/ 2 10 or 11 according as n is even or odd. 71/2 TC/2 TC But 10 - sin xdx= sin x dx — [ — —cosx = —cos—+ cos 0 — sinn xdx = n It/ 2 sinn x dx = n THEOREM 2 : if n is even. n n —•1 if n is odd. —2 It/ 2 If In Tt/2 cosn xdx then 11/2 cosn xdx = cos n—2 • Tt/2 dx = sin xdx then In x dx THEOREM 3 : If 1 Proof : 71/4 tann x dx = 71/4 tann 2 x(sec tan tan x —l)dx — n—2 x tan xdx 7t/4 tan n—2 Tt/4 tann 2 xdx x sec xdx— tan In x
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    THEOREM 4 : If ln Proof : Tt/ 4 secn xdx then n—2 • secn x dx = sec sec n 2 x tan x n—2 x sec xdx -(11-2) J sec sec — (n — 2)In + (n —2) secn- x sec x tan xdx x(sec2 x —1) dx 71/4 secn 2 xdx xdx— -2)1n 2 -2)1n 2 -2)1n 2 ln (n — l) n—l THEOREM 5: If 1 Proof : Tt/ 2 sin Tt/ 2 x cosn xdx then I m,n sin x cosn x dx . m,n sin m x cos n x dx m,n sin m 1 (sin x cos n x)dx TC/ 2 —sin m x cosn+l x Tt/2 cosn+l x (m — 1) sinm 2 x cos x dx m—1 sin m—l sin n -1-1 xcosn x cos xdx x cosn —sin x)dx
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    m—l n -1-1 m—l m—2 x cosn x —sin sin m—l m—2,n m,n n -1-1 m—l m—l x cos x)dx n+m sin m—2,n m—l m—2,n x cos m+n m—2,n m+n Note: replacing m by m-2,m-4, m—l m—l m —3 (1) m—l m —3 m —5 O,n m + n m+n—2m+n—4 m—2,n m,n m + n m+n—2 or 11,11 according as n is even or odd. But - l,n sin xcosn xdx= cosn x dx TC/ 2 cosn+l x sin x cosn xdx = n -1-1 m—l m —3 —if m is Odd m,n m+nm+n—2 n+l m—l m —3 m+n m+n—2 COROLLARY 2: If 1 m,n cosn xdx if m is even sin xcosn xdx then n—l 1 m,n—2 • m,n

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