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Solution

Let the integral be f( x ),

f( x )= 1 1 e x dx

The integration of the function is,

a b f( x )dx =( ba ) lim n 1 n ( f( a )+f( a+h )+f( a+2h )+...+f( a+( n1 )h ) )

Simplify the values,

a=1,b=1 h= 1( 1 ) n = 2 n f( x )= e x

And,

f( 1 )= e 1 f( 1+h )= e ( 1+h ) f( 1+2h )= e ( 1+2h ) f( 1+( n1 )h )= e ( 1+( n1 )h )

Substitute the value in the integral function,

1 1 ( e x )dx =( 1( 1 ) ) lim n 1 n ( f( 1 )+f( 1+h )+f( 1+2h )+...+f( 1+( n1 )h ) ) =2 lim n 1 n ( e 1 + e 1 e h + e 1 e 2h +...+ e 1 e ( n1 )h ) =2 lim n 1 n ( e 1 ( 1+ e h + e 2h +...+ e ( n1 )h ) ) =2 e 1 lim n 1 n ( 1+ e h + e 2h +...+ e ( n1 )h ) (1)

Consider the series,

S=1+ e h + e 2h +...+ e ( n1 )h

It is a geometric progression with common ratio ( r ),

r= e h 1 = e h

Sum of the geometric progression,

S=1( ( e h ) n 1 e h 1 ) = e nh 1 e h 1

Now, substitute the value of series in equation (1),

1 1 ( e x )dx =2 e 1 lim n 1 n ( 1+ e h + e 2h +...+ e ( n1 )h ) = 2 e lim n 1 n ( e nh 1 e h 1 ) = 2 e lim n ( e nh 1 nh e h 1 h ) = 2 e lim n ( e n× 2 n 1 n× 2 n e 2 n 1 2 n )( lim n e t 1 t =1 )

Further, simplify the function,

1 1 ( e x )dx = 2 e lim n e 2 1 2 =e 1 e

Thus, the value of the definite integral function f( x )is ( e 1 e ).


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