Definite Integrals PDF

60 322 Definite Integral 6.1 Definition. integral of f(x) over [a, b] is denoted by  b a d [ (x )]  f (x ). Then the definite dx E3 Let  (x ) be the primitive or anti-derivative of a function f(x) defined on [a, b] i.e., f (x )dx and is defined as [ (b)   (a)] i.e., b  f (x)dx  (b)  (a)....

60 322 Definite Integral 6.1 Definition. integral of f(x) over [a, b] is denoted by  b a d [ (x )]  f (x ). Then the definite dx E3 Let  (x ) be the primitive or anti-derivative of a function f(x) defined on [a, b] i.e., f (x )dx and is defined as [ (b)   (a)] i.e., b  f (x)dx  (b)  (a). This is also a ID called Newton Leibnitz formula. The numbers a and b are called the limits of integration, ‘a’ is called the lower limit and ‘b’ the upper limit. The interval [a, b] is called the interval of integration. The interval [a, b] is also known as range of integration. Important Tips b  f x dx   x   c, then U  In the above definition it does not matter which anti-derivative is used to evaluate the definite integral, because if  f x d x   x   c a   (b)  c    (a)  c    (b)  (a). b a  D YG In other words, to evaluate the definite integral there is no need to keep the constant of integration. Every definite integral has a unique value.  1 x x2  1  tan 1  dx =  tan 2 x  1  x 1   Example: 1 (a)  (b) 2  x   1 x 1  tan x 2  1  cot x 2  1  dx  I  [Karnataka CET 2000] (c) 3 (d) None of these 1   I 3  1 2   sin 2  dx  I  2 [ x ]31   2 [3  1]  2. x dx is equal to ST Example: 2 3 U Solution: (b) 3 [MP PET 1999] 0 (a)  Solution: (b) I (b) /2  (c) 0 (d) None of these    1 1 2 sin 2 x dx  [1  cos 2 x ]dx 2 0 2 0  I  1  1 sin 2 x  x  I  [ ] . 2 2 2  2  0 6.2 Definite Integral as the Limit of a Sum. Let f(x) be a single valued continuous function defined in the interval a  x  b, where a and b are both finite. Let this interval be divided into n equal sub-intervals, each of width h by inserting (n – 1) points a  h, a  2h, a  3 h......a  (n  1)h between a and b. Then nh  b  a. Definite Integral 323 Now, we form the sum hf (a)  hf (a  h)  hf (a  2h) ........  hf (a  rh) ......  hf [a  (n  1)h] = h[ f (a)  f (a  h)  f (a  2h) .....  f (a  rh) ....  f {a  (n  1)h}] n 1 =h  f (a  rh) r 0 60 where, a  nh  b  nh  b  a n 1 The lim h h 0  f (a  rh), if it exists is called the definite integral of f(x) with respect to x between the limits a and b r 0 and we denote it by the symbol b  f (x )dx. Thus,  E3 a b a f (x )dx  lim h[ f (a)  f (a  h)  f (a  2h) ......  f {a  (n  1)h}]  h 0 where, nh  b  a, a and b being the limits of integration.  b a f (x )dx  lim h h 0 n 1  f (a  rh) r 0 ID The process of evaluating a definite integral by using the above definition is called integration from the first principle or integration as the limit of a sum. Important Tips b a f (x )dx  where, nh = b – a.      b  b c x     x   a a c       x a  dx  b  a bx 2 f x  c  dx   b f x  dx or a  b c ac f x  c  dx   b      b  a x     x  f x  dx  1 n  nb dx   8    2 f x  dx na f x  dx a U  dx n b lim h[ f (a  h)  f (a  2h) .....  f (a  nh)] ,   f (x )dx  h  f (a  rh) a h0 r 1 D YG Then we have, U  In finding the above sum, we have taken the left end points of the subintervals. We can take the right end points of the sub-intervals throughout. ST Some useful results for evaluation of definite integrals as limit for sums n(n  1)(2n  1) n(n  1) (i) 1 + 2 + 3 + …….+ n = (ii) 1 2  2 2  3 2 ........  n 2  6 2  n(n  1)  (iii) 1 3  2 3  3 3 .......  n 3     2  (v) a  ar  ar 2 ...............  ar n 1  2 (iv) a  ar  ar 2 ........  ar n 1  a (1  r n ) , r  1, r  1 1r   n  1    nh  sin a    h sin    2   2  (vi) sin a  sin( a  h) ........  sin[ a  (n  1)h]  [sin( a  nh)]  h r 0 sin   2 n 1  a(r n  1) ,r  1 , r  1 r 1 324 Definite Integral   n  1    nh  cos a    h sin    2   2  (vii) cos a  cos(a  h)  cos(a  2h) .....  cos[a  (n  1)h]  [cos(a  nh)]  h r 0 sin   2 n 1  1 1 2  ....................   8 32 52 (xii) cos   1 1 1 2   ....................   24 22 4 2 62 (xi) e i  e i e i  e i and sin   2 2 (xiii) cosh   e   e  e   e  and sinh   2 2 E3 (x) 1  1 1 1 1 1 1 1 1 1 1 2 2 (ix) 1          ....................................     12 6 22 32 4 2 5 2 62 22 32 4 2 5 2 6 2 60 (viii) 1  6.3 Evaluation of Definite Integral by Substitution. For example, if we put  (x )  t in the integral ID When the variable in a definite integral is changed, the substitutions in terms of new variable should be effected at three places. (i) In the integrand (ii) In the differential say, dx (iii) In the limits  f {(x )} ' (x )dx , then  b b a U a f { (x )} ' (x )dx   (b )  (a ) f (t) dt. Important Tips   0  dx 2 1  sin x  /2   0 log tan x  dx  0 0     a  a 0 a 0   /2 D YG  dx a2  x 2   2  0  [ f(g[h(x)])] b If h(a)  h(b ), then Example: 3 1 dx  2 log sin x  cos x  2  1 dx a  , where f a  x    f x  1  e f x  2  dx  x  a 2 2a  2  a a 2  x 2 dx  0 f ' (g[h(x )]) g' [h(x )] h' (x ) dx is equal to a 2 4 [MP PET 2001] U a (a) 0   )]) t 1dt  log( t)ff ((gg[[hh((ba)]) 0 The value of the integral (a) 3/2 Solution: (b) f [g(a)]  f [g(b)] (c) f ( g[h(b )]) f ( g[h(a)]) Example: 4 f (a)  f (b ) (d) None of these Put f (g[h(x )])  t  f ' (g[h(x )]) g' [h(x )] h' (x )dx  dt ST Solution: (a) (b)  e2 e 1 [ h(a)  h(b )] log e x dx is x [IIT 2000] (b) 5/2 (c) 3 (d) 5 Put log e x  t  e  x t  dx  e t dt and limits are adjusted as –1 to 2  I  2 1  2 t t e dt  | t | dt  I  t 1 e 0 2 2  t2    t2   tdt  tdt  I        I  5/2 1 0  2  1  2  0  0  Definite Integral 325  /2  dx equals 1  sin x [MNR 1983; Rajasthan PET 1990; Kurukshetra CEE 1997] (a) 0  I Solution: (b) dx  /2  dx (sin x /2  cos x /2) 0   I2 2 1 dt t2 2   /2  0 sec 2 x /2 dx (1  tan x /2) 2 1 sec 2 x /2 dx  dt 2 Put (1  tan x /2)  t  2 1  1 1  2    2     1  t 1 2 1 6.4 Properties of Definite Integral. b b a a  f (x )dx   f (t) dt i.e., The value of a definite integral remains unchanged if its variable is replaced by any other symbol. 6 3 1 dx is equal to x 1 log( x  1) 63 (a)  I Solution: (c) 6 1 dx  [log( x  1)]63 , x 1  b (c) Both (a) and (b) I  6 3  (d) None of these 1 dt  [log( t  1)]63 t 1 D YG 3 (2) (b) [log( t  1]63 U  Example: 6 ID (1) (d) 2 sin 2 x /2  cos 2 x /2  2 sin x /2 cos x /2 0 I (c) –1 (b) 1  / 2 60 0 E3 Example: 5 a f(x)dx   f(x)dx i.e., by the interchange in the limits of definite integral, the sign of the integral is changed. a b Suppose f is such that f(–x) = –f(x) for every real x and Example: 7 (a) 10 (b) 5  f(x) dx  5 , then  f(t) dt  1 0 0 1 (c) 0 [MP PET 2000] (d) –5  f (x )dx  5 1 Solution: (d) Given, 0 U Put x  t  dx  dt  I  1 ST 0 (3)  b f (x )dx  a or  b a f (x )dx   c a  c1 a  0 f (t)dt   f (t)dt  I  5  1 b f (x )dx  f (x )dx , (where a < c < b) f (x )dx  c  c2 c1 f (x )dx .....   b cn f (x )dx ; (where a  c 1  c 2 ........ c n  b ) Generally this property is used when the integrand has two or more rules in the integration interval. Important Tips   ( x a  b a x  b ) dx  (b  a)2 326 Definite Integral Note :  Property (3) is useful when f (x ) is not continuous in [a, b] because we can break up the integral into several integrals at the points of discontinuity so that the function is continuous in the sub-intervals.  The expression for f (x ) changes at the end points of each of the sub-interval. You might at times be puzzled about the end points as limits of integration. You may not be sure for x = 0 as the limit of the first integral or the next one. In fact, it is immaterial, as the area of the line is always zero. Therefore, even if you 0 (1  2 x ) dx , whereas 0 is not included in its domain it is deemed to be understood that you are 1 60  write approaching x = 0 from the left in the first integral and from right in the second integral. Similarly for x = 1.  2 | 1  x 2 | dx is equal to Example: 8 [IIT 1989; BIT Ranchi 1996; Kurukshetra CEE 1998] (a) 2  1    1  I   (1  x 2 )dx  2  [x 1.5 Example: 9 2  1 2 1 1 1 2 (1  x 2 )dx  (1  x 2 )dx  I  1  1.5  1 [x 2 ]dx  [x 2 ]dx  0  3 1.5 [x 2 ]dx  I  0  f (x )dx  2  3 2  f (x )dx  2 f (x )dx  2 a  f (a  x )dx  3 cos x  2 1dx  1 2  1.5 2 1 (d) 2dx  2  1  3  2 2  I  2  2 2 f (x )dx  D YG  I 0   [DCE 2000, 2001; IIT 1988; AMU 1998] 1 2 (c) [IIT 2000] (b) 1 | x |  2  2  x  2 and f (x )  e Solution: (c)  [x 2 ]dx  1 (a) 0 a 2 e cos x. sin x , | x |  2 If f (x )   , then  2, otherwise Example: 10 (4) 2 2 (b) 0 4 4 4   4. 3 3 3 U 2 2 I Solution: (b) (d) 8 ]dx , where [.] denotes the greatest integer function, equals 0 (a) 2 | dx  | 1  x 2 | dx  | 1  x 2 | dx  | 1  x 2 | dx 2 2  (c) 6 1 ID  | 1x I Solution: (b) (b) 4 2 E3 2 (c) 2 (d) 3 sin x is an odd function. f (x )dx  I  0  2  2dx  [2 x] 3 2 3 2  f (x)  0 if f(x) is odd and in (2, 3) f(x) is 2] a 2 [ a : This property can be used only when lower limit is zero. It is generally used for those 0  /2  sin n x dx  sin n x  cos n x ST (i) U complicated integrals whose denominators are unchanged when x is replaced by (a – x). Following integrals can be obtained with the help of above property. (ii) 0  /2  0 (iv)  (v)   2 0 (vii)  2 0 tan n x dx  1  tan n x 0 0 0 f (sin 2 x ) sin xdx  f (tan x )dx   /2  0  /2  0 cos n x  dx  n n 4 cos x  sin x cot n x  dx  n 4 1  cot x sec n x dx  sec n x  cosec n x  /2    /2   /2   /2 0  /2 (iii)  (vi)  0 1 dx  1  tan n x  2  0 1  dx  n 4 1  cot x cosec n x  dx  n n 4 cosec x  sec x f (sin 2 x ) cos xdx f (cot x )dx (viii)  /2 0 1  0 f (sin x )dx  f (log x )dx   /2  0 1  0 f (cos x )dx f [log(1  x )]dx Definite Integral 327  /2  0  /2  (xi) 0  /2  (xii) 0 Example: 11 log tan xdx   /2  log sin xdx  0   /2   1 log 2  log 2 2 2 a sec x  b cosec x dx  sec x  cosec x  /4  0 log(1  tan x )dx   /2   /2  0 0 2 x cos 3 xdx = 0 OR For any integer n,  e sin 2 x cos 3 (2n  1)xdx = [MP PET 2002; MNR 1992, 98] 0 (a) –1 I = 0.  f (x ) dx is equal to f (x )  f (2a  x ) 2a (a) a I  f (x ) dx  f (x )  f (2a  x ) 0   0 f (x )  f (2a  x ) dx  f (x )  f (2a  x ) 2a (c) 2a 0 (d) 0 f (2a  x ) dx f (2a  x )  f (x ) 2a  dx  [x ] 2a 2a 0  2a D YG 2I  [IIT 1988; Kurukshetra CEE 1999; Karnataka CET 2000] (b) a/2 2a (d) None of these ID  0 Solution: (a) (c) 1 f2 (x )  cos 3 (2n  1)x   f (  x ) and Example: 12 (b) 0 Let, f1 (x )  cos 3 x   f (  x ) U Solution: (b) log 2 E3  8 a tan x  b cot x  dx  (a  b) tan x  cot x 4  e sin  log cos xdx  0 a sin x  b cos x dx  sin x  cos x  (x) log cot xdx 60 (ix) 0  I  a.  /2 Example: 13  0 tan x  1  tan x  /2  0 sin x sin x  cos x (a) /4 [MNR 1989, 90; Rajasthan PET 1991, 95] (b)   /2 Solution: (a) dx is equal to We know,  0 (c) –1 (d) 1 tan xdx   for any value of n 1  tan n x 4 n (5)  a U  I  /4. a f (x ) dx   a 0 f (x )  f ( x ) dx. a   f (x ) dx  2 0 f (x ) dx , if f (x ) is even function or f ( x )  f (x ) a  0 , if f (x ) is odd is odd function or f ( x )   f (x )  This property is generally used when integrand is either even or odd function of x. a ST  In special case : Example: 14 The integral   1  x   [ x ]  ln   dx equal to 1 / 2   1  x   1/2 1 2 (a) Solution: (a)  (b) 0 (c) 1 (d) 1  x  log   is an odd function of x as f ( x )   f (x ) 1  x  I  1/2 [x ]dx  0  I  1 / 2  0 1 / 2 [x ]dx   1/2 0 [x ]dx  I   0  1dx  0   [ x ]01 / 2  1 / 2 1. 2 1 2 ln  2 328 Definite Integral  1 The value of the integral 1 (a) 0   (1  x The value of 2 (c) log 1/2 [Roorkee 1992; MNR 1998] (b) 1 (c) 2 Let, f1 (x )  (1  x ) , f2 (x )  sin x and f3 (x )  cos x 2 2 Now, f1(x )  f1( x ) , f2 (x )   f2 (x ) and f3 (x )  f3 ( x )       f (x )dx  [ f1 (x ). f2 (x ). f3 (x )]dx =  [ f1 ( x ). f2 ( x ). f3 ( x )] dx     I0  2a 0  f (x )dx  a f (x )dx  0  a f (2a  x ) dx 0 , if f (2a  x )   f (x )  0 f (x ) dx   a 0 2 f (x ) dx , if f (2a  x )  f (x )   0 It is generally used to make half the upper limit. 2a  Example: 17 If n is any integer, then  e cos 2 x ID  cos 3 (2n  1)x dx is equal to U  In particular, 0 (a) x  e cos 2 (  x ) (c) 0 [IIT 1985; Rajasthan PET 1995] (d) None of these. cos 3 (2n  1)(  x ) dx D YG I Solution: (c) (b) 1  (d) 3 E3  I (6) (d) None of these ) sin x cos 2 xdx is  (a) 0 Solution: (a) [MP PET 2001] f (x )  log[ x  x 2  1] is a odd function i.e. f ( x )   f (x )  f (x )  f ( x )  0  I = 0. Solution: (a) Example: 16 log  x  x 2  1 dx is   (b) log 2 60 Example: 15 0  I  e cos 2 x. cos 3 (2n  1)x dx  I  I 0  2 I  0  I  0. Example: 18 If I1  (a)  3 0 f (cos 2 x ) dx and I 2  I1  I 2   f(cos 2 x ) dx then 0 (b) I1  2I 2 (c) I1  3I 2 (d) I1  4 I 2 f (cos x )  f (cos (3  x )) 2 Solution: (c) 2   U  I1  3 f (cos 2 x ) dx  I1  3I 2 0 (7) b b  f (x ) dx   f (a  b  x )dx a ST a f (x ) dx 1  (b  a) f (x )  f (a  b  x ) 2 b Note :  a Example: 19  3 / 4  /4 dx is equal to 1  cos x (b) –2 (a) 2 Solution: (a) I  3 / 4 1 dx 1  cos x  /4  2I  [IIT 1999]  3 / 4  /4  2I  2  (d) –1/2    3    x    cos x  [  cos   4 4     2 dx 1  cos 2 x 3 / 4  /4 (c) 1/2 co sec 2 xdx  2 I  2[cot x ]3/4/4  4  I = 2. Definite Integral 329 Example: 20  The value of 3 x dx is 5x  2 (a) 1 I Solution: (d)  [IIT 1994; Kurukshetra CEE 1998] x (c) –1 (b) 0 3 x dx 5x  2 (d) 1/2 x Put x  2  3  t  dx  dt  I 5t 2 5t  t 3  2I  (dt )  5x 2 5x  x dx and 2 I   x f (x)dx is equal to ab (b) f (x )dx 2   b x f (x )dx and I  b (a  b  x ) f (a  b  x )dx a  b  I  (a  b  x ) f (x )dx  I  0 x f (x )dx   Example: 22  If 1 a 2  a 0 0     (b) /2 D YG (a)  Solution: (b)  Given, x f (sin x )dx  k 0  f(x)dx (c) /4 (d) 1   f (sin x )dx 0     0 0 0 0    0 0 0 x f (sin x )dx  k   f (sin x )dx 0    2k  0  k   /2. U (9) If f (x ) is a periodic function with period T, then  nT 0 f (x )dx  n  T 0 f (x )dx , Deduction : If f (x ) is a periodic function with period T and a  R  , then  a  nT a f (x ) dx  n  T 0 f (x ) dx where n  I (a) In particular, if a = 0  (b) If n = 1,  nT 0 aT 0 nT  (ii)  (i) mT  f (x ) dx   f (x ) dx  n a  nT f (x ) dx  0 T 0 f (x ) dx  (n  m) b  nT T  b a  T 0 f (x ) dx , where n  I f (x ) dx , f (x ) dx , f (x ) dx ,  a nT nT is a periodic function with period T, then f (x ) ST (10) (i) If where n, m  I where n I b a [IIT 1982]  (  x)f(sin(  x))dx  k  f(sin(  x))dx    f(sin x)dx     f (sin x )dx  2k f (sin x )dx  0  (  2k ) f (sin x )dx  0     (d) None of these f (x )dx a f (sin x )dx , then the value of k will be 0   b b  b  ab (a  b) f (x )dx  x f (x )dx  2 I   f (x )dx  (a  b)  I  2 a a   f (x )dx if f (a  x )  f (x ) x f (sin x )dx  k b a 2 U a  b a a  (c) a a 2 [Kurukshetra CEE 1993; AIEEE 2003] b f (b  x )dx a (8) 5x  x 3 ID  I Solution: (b)  b 2  dx  1 dx E3 ab 2 x  5x b a (a) 3  1  I 1/2 [ x ]22 If f (a  b  x )  f (x ) then Example: 21 3  60  f (x )dx   a 0 f (x )dx 330 Definite Integral (11) If f (x ) is a periodic function with period T, then  aT a (12)  b a f (x ) is independent of a.  f (b  a) x  a dx 1 f (x ) dx  (b  a) 0 x  (x )   f (t) dt is an even function (13) If f (t) is an odd function, then (14) If f (x ) is an even function, then  (x )  Note  x f (t) dt is on odd function. 0 :  If f(t) is an even function, then for a non zero ‘a’,  x f (t) dt is not necessarily an odd function. It will be 0 a E3  f (t) dt  0 odd function if 0 For n  0,  2 x sin 2 n x sin 0 2n x  cos 2 n x (a)  2 I  2 x sin 2 n xdx and I  sin 2 n x  cos 2 n x 0  2 I  2  2 0  nT using  T f (x )dx 0  I  4 0  2 0  a 0 f (x )    f(a  x) a 0 sin 2n x dx sin x  cos 2n x 2n If f (x ) is a continuous periodic function with period T, then the integral I  (b) Equal to 3a Consider the function g(a)  U aT  f(x)dx is a (c) Independent of a a T 0 T aT a a 0 T (d) None of these  f(x )dx =  f(x)dx   f(x)dx   f(x)dx Putting x  T  y in last integral, we get  g(a)     sin 2n x dx  I  4  ( /4 )   2. sin 2n x  cos 2n x (a) Equal to 2a Solution: (c) 0 4 2 (d) (2  x ) sin 2 n (2  x )dx sin 2 n (2  x )  cos 2 n (2  x ) D YG  2 sin 2 n  dx  I   sin x  cos 2 n x f (x )  n  /2  3 2 (c) 2n 0 Example: 24 2 2 (b) [IIT 1996] U Solution: (a) dx is equal to ID Example: 23 60 a  aT f (x )dx  T  a f (y  T ) dy  0  f(y)dy a 0  f(x)dx   f(x)dx   f(x)dx =  f (x )dx 0 T a T a 0 0 0 ST Hence g(a) is independent of a. Important Tips   Every continuous function defined on [a, b] is integrable over [a, b]. Every monotonic function defined on [a, b] is integrable over [a, b].  If f(x) is a continuous function defined on [a, b], then there exists c  (a, b) such that  b f (x )dx  f (c).(b  a). a The number f (c)   1 (b  a)  f (x )dx b is called the mean value of the function f (x ) on the interval [a, b]. a  f(t)dt x If f is continuous on [a, b], then the integral function g defined by g(x) = a all x  [a, b]. for x  [a, b] is derivable on [a, b] and g ( x )  f ( x ) for Definite Integral 331   b If m and M are the smallest and greatest values of a function f(x) on an interval [a, b], then m(b  a)  f (x )dx  M (b  a) a  If the function  (x ) and  (x ), are defined on [a, b] and differentiable at a point x  (a, b) and f(t) is continuous for  (a)  t   (b) , then   (x )   f (t)dt   f ( (x )) ' (x )  f ((x ))' (x )  ( x )    b f (x )dx  | f (x ) | dx a a  f (x ) g(x ) dx   a   b  f 2 (x ) dx  a   b 1/2 60  b  b 2   g (x ) dx  a    1/2 If f 2 (x ) and g 2 (x ) are integrable on [a, b], then  Change of variables : If the function f(x) is continuous on [a, b] and the function x  (t) is continuously differentiable on the interval [t1 , t 2 ] and a  (t1 ), b  (t 2 ), then f (x )dx  a  f ((t))' (t)dt. t2 t1 f (x , ) be continuous for a  x  b and c    d. Then for any   [c, d] , if I( )  Let a function I' ( )  b  f(x,)dx , b then a ID   E3   f '(x,)dx , b a Where I' ( ) is the derivative of I( ) w.r.t.  and f ' (x , ) is the derivative of f (x , ) w.r.t. , keeping x constant. For a given function f (x ) continuous on [a, b] if you are able to find two continuous function f1 (x ) and f2 (x ) on [a, b] such that  b a f1 (x )dx   b f (x )dx  a  b a D YG f1 (x )  f (x )  f2 (x )  x  [a, b], then U  f2 (x )dx 6.5 Summation of Series by Integration. We know that  b a f (x )dx  lim h n  n  f (a  rh) , Now, put a = 0, b = 1,  nh  1 or h  1. Hence n :  Express the given series in the form  U Note where nh  b  a r 1  1 Example: 25 If S n  1 1 n  1 2  2n ........  (a) log 2 Solution: (b) 1 1 0 n  1 r r . Replace by x, by dx and the limit of the sum is n h n n 1 f f (x ) dx. ST 0 r  f (x )dx  lim n  f  n   lim r   lim S n  n  n  n2 (b) 2 log 2 1 n  1  rn  1 0  lim n  1 x (1  x ) 1 r r n   n n   dx = 2[log(1  x )]10  2 log 2 then lim S n is equal to [Roorkee 2000] n  (c) 3 log 2 (d) 4 log 2 332 Definite Integral 1/n Example: 26 (n! )1 / n  n!  or lim  n  n  n  n  n is equal to lim [WB JEE 1989; Kurukshetra CEE 1998] (b) e–1 (a) e (c) 1 (d) None of these (n! )1 / n n  n Let A  lim 1/n 1/n n 1 2 3  log A  lim log ......  n  n n n n  1.2.3...... n   log A  lim log   n  nn    log A   log xdx  [x log x  x] 1  log A  1  A  e 1 1 0 0   /2 sin m r   log n  r 1 m 1 n 1      2   2  n x cos xdx  m  n  2 2   2   ID 0  n E3 6.6 Gamma Function. If m and n are non-negative integers, then 1 n  n  log A  lim 60 Solution: (b) where (n) is called gamma function which satisfied the following properties (n  1)  n(n)  n! i.e.  (1)  1 and (1 / 2)   / 2  0 U In place of gamma function, we can also use the following formula : sin m x cos n xdx = (m  1)(m  3).....(2 or 1)(n  1)(n  3).....(2 or 1) (m  n)(m  n  2)....(2 or 1) D YG It is important to note that we multiply by (/2); when both m and n are even.  /2 Example: 27 The value of  sin 4 x cos 6 xdx = [Rajasthan PET 1999] 0 (a) I Solution: (c) 3 /312 (b) 5 / 512 (c) 3 / 512 (d) 5 / 312 (4  1).(4  3).(6  1).(6  3).(6  5)  3.1.5.3.1  3. .  (4  6)(4  6  2)(4  6  4 )(4  6  6)(4  6  8) 2 10.8.6.4.2. 2 512 6.7 Reduction formulae Definite Integration.  0 U  e ax sin bxdx   b a b 2 Example: 28 If In  (a) Solution: (c) (2) 2  ST (1) e  x x n 1dx , then 0  e x   0 a e ax cos bxdx  a b 2 2 (3)  0 (b) 1 In  (c) In (d)  n I n n 1  n   0 e t t n 1dt = 1  n  e 0 x x n 1 dx  In n 6.8 Walli's Formula.   /2 0 sin n xdx    /2 0 0 e ax x n dx  x n1 dx is equal to Put, x  t, dx  dt , we get, e x x n 1dx   0 In   2 n 1 n  3 n  5........ ,  3 cos n xdx   n n  2 n  4 n 1 n  3 n  5 3 1  ........... , 4 2 2  n n2 n4 when n is odd when n is even n! a 1 n Definite Integral 333 sin m x cos n dx  (m  1)(m  3)..........(n  1)(n  3)........ [If m, n are both odd +ve integers or one odd +ve integer] (m  n)(m  n  2)   /2 Example: 29 (m  1) (m  3)............(n  1) (n  3) . (m  n) (m  n  2) 2  sin7 xdx has value (a) 37 184 [If m, n are both +ve integers] [BIT Ranchi 1999] 0 (b) (c) 16 35 7  1 7  3 7  5 6.4.2 16..   7 7  2 7  4 7. 5.3 35 (d) 16 45 E3 Using Walli’s formula,  I  Solution: (c) 17 45 60  /2 0 6.9 Leibnitz’s Rule. (1) If f(x) is continuous and u(x), v(x) are differentiable functions in the interval [a, b], then, d v( x ) d d f (t)dt  f {v(x )} {v(x )}  f {u(x )} {u(x )} dx u( x ) dx dx (2) If the function  (x ) and  (x ) are defined on [a,b] and differentiable at a point x (a, b), and d  dx   (x )  Example: 30 (x ) f (x , t) dt    Let f (x )   x  (x )  (x ) 2  t 2 dt. Then the real roots of the equation x 2  f ' (x )  0 are 1 1 (b)  D YG (a) 1 (c)  2 Solution: (a) f (x )   x f (x , t) is continuous, d  d  (x )   d  (x )  f (x , t) dt    f (x , (x ))    f (x ,  (x )) dx  dx   dx  U then, ID  1 2 [IIT 2002] (d) 0 and 1 2  t 2 dt  f ' (x )  2  x 2.1  2  1.0  2  x 2 1  x 2  f ' (x )  2  x 2  x 4  x 2  2  0  (x 2  2)(x 2  1)  0  x  1 (only real). Example: 31 Let f : (0, ) R and f (x )   f(t)dt. If f(x )  x (1  x), then f (4) equals x 2 2 [IIT 2001] 0 (b) 7 U (a) 5/4 Solution: (c) By definition of f(x) we have f (x 2 )   f (t)dt  x x (c) 4 (d) 2 2 2  x3 (given) 0 ST Differentiate both sides, f (x 2 ).2 x  0  2 x  3 x 2 Put, x  2  4 f (4 )  16  f (4 )  4 6.10 Integrals with Infinite Limits (Improper Integral). b  f (x )dx A definite integral is called an improper integral, if a The range of integration is finite and the integrand is unbounded and/or the range of integration is infinite and the integrand is bounded. 1 1 1 dx is an improper integral, because the integrand is unbounded on [0, 1]. Infact, 2  as e.g., The integral 0 x2 x  1 dx is an improper integral, because the range of integration is not finite. x 0. The integral 0 1 x2 There are following two kinds of improper definite integrals:   334 Definite Integral (1) Improper integral of first kind : A definite integral b  f (x )dx is called an improper integral of first kind if the a range of integration is not finite (i.e., either a  or b  or a  and b  ) and the integrand f(x) is bounded on [a, b]. 1 x 1 dx , 2   0 1 dx , 1 x2  1 dx ,  1  x 2    1 3x dx are improper integrals of first kind. (1  2 x )3 60   Important Tips  In an improper integral of first kind, the interval of integration is one of the following types [a, ), (–, b], (–, ).  The improper integral  a integral. If lim  k k  a   f(x )dx k lim exists finitely and this limit is called the value of the improper k  a E3  f (x )dx is said to be convergent, if f (x ) dx is either + or –, then the integral is said to be divergent.   f (x )dx is said to be convergent, if both the limits on the right-hand side exist finitely and are independent of each other. The improper integral f (x )dx is said to be divergent if the right hand side is + or –  The improper integral  ID   (2) Improper integral of second kind : A definite integral b  f (x )dx is called an improper integral of second kind if U a the range of integration [a, b] is finite and the integrand is unbounded at one or more points of [a, b]. b  f (x )dx is an improper integral of second kind, then a, b are finite real numbers and there exists at least one point a D YG If c  [a, b ] such that f (x )  or f (x )  as x c i.e., f(x) has at least one point of finite discontinuity in [a, b]. For example : (i) The integral  (ii) The integral 1  1  dx , is an improper integral of second kind, because lim  . x 2 x  2  x 2 3 1 1  log xdx ; is an improper integral of second kind, because log x  as x 0. 0  2 0 1 1 becomes infinite dx , is an improper integral of second kind since integrand 1  cos x 1  cos x U (iii) The integral at x    [0, 2 ]. 1  sin x sin x  1. dx , is a proper integral since lim x 0 x x ST (iv) 0 Important Tips  Let f(x) be bounded function defined on (a, b] such that a is the only point of infinite discontinuity of f(x) i.e., f(x)  as x a. Then the improper integral of f(x) on (a, b] is denoted by   b f (x )dx and is defined as b a a side exists. If l denotes the limit on right hand side, then the improper integral f (x )dx  lim  0  b  b a  f (x )dx. Provided that the limit on right hand f (x )dx is said to converge to l, when l is finite. If l = +  or l a = –, then the integral is said to be a divergent integral.  Let f(x) be bounded function defined on [a, b) such that b is the only point of infinite discontinuity of f(x) i.e. f(x)  as x b. Then the improper integral of f(x) on [a, b) is denoted by  f (x )dx and is defined as  b b a a f (x )dx  lim  0  b  a f (x )dx Definite Integral 335 Provided that the limit on right hand side exists finitely. If l denotes the limit on right hand side, then the improper integral  b f (x )dx is said to a converge to l, when l is finite. If l   or l   , then the integral is said to be a divergent integral.  Let f(x) be a bounded function defined on (a, b) such that a and b are only two points of infinite discontinuity of f(x) i.e., f(a) , f(b) . Then the improper integral of f(x) on (a, b) is denoted by  b f (x )dx and is defined as   b  c b  60 a f (x )dx  lim f (x )dx  lim f (x )dx , a  c  b a  0 a   0 a Provided that both the limits on right hand side exist.  Let f(x) be a bounded function defined [a, b]-{c}, c[a, b] and c is the only point of infinite discontinuity of f(x) i.e. f(c). Then the improper  b f (x )dx and is defined as  b a a f (x )dx  lim x 0  cx f (x )dx  lim  0  b f (x )dx E3 integral of f(x) on [a, b] – {c} is denoted by Provided that both the limits on right hand side exist finitely. The improper integral  a c  b f (x )dx is said to be convergent if both the limits on the a right hand side exist finitely. If either of the two or both the limits on RHS are , then the integral is said to be divergent.  Example: 32 e The improper integral x ID  dx is …… and the value is…. 0 (a) Convergent, 1 I  e  x dx  lim Thus, lim  k x k  I   2  dx  lim 2 k  -  - a  x 2 0 k  (b) Convergent and equal to Divergent and equal to  2a  2a dx a2  x 2 U   k  0 x 1 1 1  I  lim  tan 1   lim  tan 1 0  tan 1 k   a a  k k   a a Hence integral is convergent.  1 The integral dx is x  e  ex (a) Convergent and equal to /6 (c) Convergent and equal to /3 I k  e  x dx exists and is finite. Hence the given integral is convergent. 0 ST Solution: (d) k  1 dx , a  0 is 2  a  x  (a) Convergent and equal to a  (c) Divergent and equal to (d) a The integral (d) Divergent, 0 dx  I  lim [e  x ]k0  lim [e k  e 0 ]  I  lim (1  e k )  1  0  1 [ lim e k  e   0] 0 Example: 34 (c) Convergent, 0 D YG k  0 Solution: (b) e k  0 0 Example: 33 (b) Divergent, 1 k U Solution: (a)  1 1     k  I  0  tan 1 ()     a a  2  2a a   1 dx  x e  ex (b) Convergent and equal to /4 (d) Convergent and equal to /2  ex dx 2x  1  e  Put e  t  e dx  dt x  I   0 Example: 35  2 1 x 1 dt  I  [tan 1 t]0  [tan 1   tan 1 0]  I   /2 , which is finite so convergent. 1  t2 x 1 x 1 dx is (a) Convergent and equal to 14 3 (b) Divergent and equal to 3 14 336 Definite Integral (c) Convergent and equal to  2 x  1dx  1  Example: 36 2 1  2 1 dx x 2  5x  4 2 2 2  =  (x  1)3 / 2   [4 x  1 ]12 = 14 /3 which is finite so convergent. 3 1 dx x 1 dx is 1 log 2 3 (a) Convergent and equal to (b) Convergent and equal to 3/log2 (c) Divergent 2 1 1 dx = (x  1)(x  4 ) 3     x  4  x  1  dx 2 1 1 = 1 1 [log 2  ]   3 E3  I Solution: (c) (d) None of these So the given integral is not convergent. 6.11 Some Important results of Definite Integral. (4) If In  (5) If In  (6) If In   /4  0  /4  0 sec n xdx then I n   /4  0  /2  0  /2  0 cot n xdx then In  In  2  1 1n ( 2 )n  2 n  2  In  2 n 1 n 1 cosec xdx then In  n ID (3) If In  0 1 n 1 tan n xdx then In  In  2  U (2) If In   /4  ( 2 )n  2 n  2  In  2 n 1 n 1 D YG (1) If In  60  I Solution: (a) Divergent and equal to  (d) x n sin xdx then In  n(n  1)In  2  n( / 2)n 1 x n cos xdx then In  n(n  1)In  2  ( /2)n  /2  0  /2 dx  a  b cos x dx  a  b cos x U (7) If a  b  0, then  0 ST (8) If 0  a  b then (9) If a  b  0 then  /2  (10) If 0  a  b , then 0 dx  a  b sin x  /2  0 dx  a  b sin x (11) If a  b, a 2  b 2  c 2 , then (12) If a  b, a 2  b 2  c 2 , then   /2 0  /2  0 2 a b 2 2 1 b 2  a2 tan 1 log 2 a2  b 2 b a b a  b a b a  b a tan 1 1 2 2 ab ab log ab ab ba  ba ba  ba dx  a  b cos x  c sin x dx  a  b cos x  c sin x 2 a b c 2 2 2 1 b 2  c 2  a2 tan 1 log ab c a2  b 2  c 2 a  b  c  b 2  c 2  a2 a  b  c  b 2  c 2  a2 Definite Integral 337 (13) If a  b, a 2  b 2  c 2 then  /2  0 1 dx  a  b cos x  c sin x b 2  c2  a2 log b  a  c  b 2  c2  a2 b  a  c  b 2  c2  a2 Important Tips  f (x )dx  f (0)  lim x 0  b 0 x f (x )dx  (b  a) a  f (b  a) t  a dt 1 0 6.12 Integration of Piecewise Continuous Functions. E3  60 x Any function f (x ) which is discontinuous at finite number of points in an interval [a, b] can be made continuous in f (x ) is discontinuous at points x 1 , x 2 , x 3.......... x n in (a, ID sub-intervals by breaking the intervals into these subintervals. If Example: 37  20 10 U b), then we can define subintervals (a, x 1 ), (x 1 , x 2 ).............( x n1 , x n ), (x n , b) such that f (x ) is continuous in each of these subintervals. Such functions are called piecewise continuous functions. For integration of Piecewise continuous function. We integrate f (x ) in these sub-intervals and finally add all the values. [cot 1 x] dx , where [.] denotes greatest integer function Solution: (b) 30  cot 1  cot 2  cot 3 (d) None of these (b) D YG 30  cot 1  cot 3 (c) 8 30  cot 1  cot 2 (a) Let I   20 10 [cot 1 x ] dx , we know cot 1 x (0,  )  x  R 3,  2, x]   1  0 [cot 1 U thus, I Hence,  [x 2 cot 3 10 3 dx   cot 2 2 dx  cot 3  cot 1 1 dx  cot 2  20 0 dx  30  cot 1  cot 2  cot 3 cot 1  x  1] dx , where [.] denotes greatest integer function ST 2 Example: 38  x  (, cot 3) x  (cot 3, cot 2) x  (cot 2, cot 1) x  (cot 1, ) 0 (a) Solution: (a) 7 5 2 Let I   2 0 (b) [x  x  1] dx  2  1 5 2 0 7 5 2 [ x  x  1] dx  2 (c)  2 1 5 2 5 3 2 [ x  x  1] dx  2 (d) None of these  1 5 2 0 1 dx   2 1 5 2 2dx  7 5 2