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If Im,n = ∫ 0 1 x m − 1 ( 1 − x ) n − 1 d x ,     f o r     m ,     n ≥ 1 ,     a n d ∫ 0 1 x m − 1 + x n − 1 ( 1 + x ) m + n d x = α I m , n , α ∈ R , then a equals…....

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    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago

    Given l m , n = ∫ 0 1 x m − 1 ( 1 − x ) n − 1 d x . . . . . . . . . . . . ( i )  

    put 1 - x = t { x = 0 , t = 1 x = 1 , t = 0  

    dx = -dt

    From (i) l m , n = ∫ 1 0 ( 1 − t ) m − 1 . t n − 1 ( − d t )  

    l m , n = ∫ 0 1 t n − 1 ( 1 − t ) m − 1 d t = ∫ 0 1 x n − 1 ( 1 − x ) m − 1 d x . . . . . . . . . . ( i i )                              

    (i)   l m , n = ∫ ∞ 0 1 ( 1 + y ) m − 1 ( 1 − 1 1 + y ) n − 1 ( − d y ( 1 + y ) 2 ) = ∫ 0 ∞ y n − 1 ( 1 + y ) m + n d y . . . . . . . . . . ( i i i )

    Similarly by (ii) l m , n = ∫ 0 ∞ y m − 1 ( y + 1 ) m + n d y . . . . . . . . . . ( i v )  

    Adding (iii) & (iv) 2 l m , n = ∫ 0 ∞ y n − 1 + y m + 1 ( y + 1 ) m + n  

    Putting   1 z { y = 1 , z = 1 y = ∞ , z = 0 ⇒ d y = − 1 z 2 d z  

    Hence  l m , n = ∫ 0 1 x m − 1 + x n − 1 ( 1 + x ) m + n dx = a lm, n

    -> a = 1

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∫ ( s i n x − c o s x ) s i n 2 x s i n x c o s 2 x + t a n x s i n 3 x d x is equal to
A
alok kumar singh

∫ ( s i n x − c o s x ) s i n 2 x t a n x ( s i n 3 x + c o s 3 x ) d x

∫ ( s i n x − c o s x ) s i n x c o s x s i n 3 x + c o s 3 x d x , put sin3x + cos3x = t(3 sin2x×cosx – 3cos2xsinx) dx = dt

-> 1 3 ∫ d t t

= l n t 3 + c

= l n | s i n 3 x + c o s 3 x | 3 + c

             

           

The value of the integral ∫ 1 3 [ x 2 − 2 x − 2 ] d x ,  where [x] denotes the greatest integer less than or equal to x, is
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l = ∫ 1 3 [ x 2 − 2 x + 1 − 3 ] d x = ∫ 1 3 [ ( x − 1 ) 2 − 3 ] d x = ∫ 1 3 [ ( x − 1 ) 2 ] d x − 3 ∫ 1 3 d x ...........(A)

l 1 = ∫ 1 3 [ ( x − 1 ) 2 ] d x P u t ( x − 1 ) 2 = t

l 1 = 1 2 [ 0 ∫ 0 1 d t t ∫ 1 2 d t t + 2 ∫ 2 3 d t t + 3 ∫ 3 4 d t t ] = 1 2 { | t − 1 2 + 1 − 1 2 + 1 | 1 2 | + 2 t − 1 2 + 1 − 1 2 + 1 | 2 3 + 3 t − 1 2 + 1 − 1 2 + 1 ] 3 4 }

Hence from (A)

= 5 − 2 − 3 − 3 − 6 = − 1 − 2 − 3

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∫ 1 3 [ ( x − 1 ) 2 ] d x − 6 . . . . . . . . . . . . . . ( A )

From (A), l = 5 − 2 − 3 − 6 = − 1 − 2 − 3

Let f(x) be a differentiable function defined on [0, 2] such that f’(x) = f’(2 – x) for all x ∈ ( 0 , 2 ) , f(0) = 1 and f(2) = e2. Then the value of ∫ 0 2 f ( x ) d x     i s  
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    l = ∫ 0 2 f ( x ) d x = [ x f ( x ) ] 0 2 − ∫ 0 2 x f ' ( x ) d x = 2 e 2 − ∫ 0 2 x f ' ( x ) d x    ........(A)

Put l 1 = ∫ 0 2 x f ' ( x ) d x               .........(i)

Using properties ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x

l 1 = ∫ 0 2 ( 2 − x ) f ' ( 2 − x ) d x = ∫ 0 2 ( 2 − x ) f ' ( x ) d x ...........(ii)

Adding (i) and (ii) we get

2 l 1 = 2 ∫ 0 2 f ' ( x ) d x ⇒ l 1 = [ f ( x ) ] 0 2

f(2) – f(0) = e2 – 1

From (A) l = 2e2 – e2 + 1 = e2 + 1

The value of the integral 4 8 π 4 ∫ 0 π ( 3 π x 2 2 − x 3 ) s i n x 1 + c o s 2 x d x is equal to………..
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Raj Pandey

l = 4 8 π 4 ∫ 0 π [ ( π 2 − x ) 3 − 3 π 2 4 ( π 2 − x ) + π 3 4 ] s i n x d x 1 + c o s 2 x

Using ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x

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⇒ l = 1 2 π [ − t a n − 1 ( c o s x ) ] 0 π = 1 2 π . π 2 = 6

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∴ ∫ − 6 0 f ( x ) d x = 2 × 1 2 ( 2 + 5 ) × 3 = 2 1

 

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