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# J -leXdr

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## Let the integral be f( x ), f( x )= ∫ −1 1 e x dx The integration of the function is, ∫ a b f( x )dx =( b−a ) lim n→∞ 1 n ( f( a )+f( a+h )+f( a+2h )+...+f( a+( n−1 )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+( n−1 )h )= e ( 1+( n−1 )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+( n−1 )h ) ) =2 lim n→∞ 1 n ( e −1 + e −1 ⋅ e h + e −1 ⋅ e 2h +...+ e −1 ⋅ e ( n−1 )h ) =2 lim n→∞ 1 n ( e −1 ( 1+ e h + e 2h +...+ e ( n−1 )h ) ) =2 e −1 lim n→∞ 1 n ( 1+ e h + e 2h +...+ e ( n−1 )h ) (1) Consider the series, S=1+ e h + e 2h +...+ e ( n−1 )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 ( n−1 )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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