For: F ( 0 ) = 0 and F ′ ( x ) = f ( x )Euler's method: F ( 0 + h ) = F ( 0 ) + h F ′ ( 0 ) = 0 + h f ( 0 )Continuing the process, F ( 10 h ) = h f ( 0 ) + h f ( h ) + h f ( 2 h ) + . . . . . h f ( 9 h )This resembles the Riemann sum: Σ i = 1 n f ( x i ) ( x i − x i − 1 )Therefore my professor used Euler's method to solve integral problems.Example: ∫ 3 3.09 f ( x ) d x = 0.81. Approximate f ( 3 ) . F ( x ) = h f ( x ) 0.81 = 0.09 f ( x ) f ( x ) = 3My question: How did F ( x + h ) = F ( x ) + h f ( x ) become F ( x ) = h f ( x ) ?

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For:   F  (  0  )  =  0 and       F    ′    (  x  )  =  f  (  x  )Euler&#039;s method:   F  (  0  +  h  )  =  F  (  0  )  +  h      F    ′    (  0  )  =  0  +  h  f  (  0  )Continuing the process,   F  (  10  h  )  =  h  f  (  0  )  +  h  f  (  h  )  +  h  f  (  2  h  )  +  .  .  .  .  .  h  f  (  9  h  )This resembles the Riemann sum:       Σ          i      =      1        n    f  (      x    i    )  (      x    i    −      x          i      −      1        )Therefore my professor used Euler&#039;s method to solve integral problems.Example:       ∫    3          3.09        f  (  x  )  d  x  =  0.81. Approximate   f  (  3  )  .   F  (  x  )  =  h  f  (  x  )  0.81  =  0.09  f  (  x  )  f  (  x  )  =  3My question: How did   F  (  x  +  h  )  =  F  (  x  )  +  h  f  (  x  ) become   F  (  x  )  =  h  f  (  x  )  ?

Sage Guerra · Accepted Answer

I think your confusion is mostly from slightly sloppy notation. You have  F  (  x  +  h  )  −  F  (  x  )  ≈  h  f  (  x  )so  F  (  3.09  )  −  F  (  3  )  ≈  0.09  f  (  3  )and  F  (  3.09  )  −  F  (  3  )  =      ∫    3          3.09        f  (  x  )  d  x

For: F ( 0 ) = 0 and F ′ </msup> ( x ) =

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