Limit as a definite integral: (e^xi/ 3 + xi) * deltaxi

[CalcEqualsUgh]

Express the limit as a definite integral on the given interval:

(i = 1). (e^xi/ 3 + xi) * deltaxi [5,9]

 

[stapel]

(i = 1). (e^xi/ 3 + xi) * deltaxi [5,9]

I will guess that the "[5,9]" is "the closed interval [5, 9}" and is the interval mentioned in the exercise. But what does "i = 1" signify, and what is the limit? Is there a sum anywhere involved in the exercise...?

Thank you!

Eliz.

 

[CalcEqualsUgh]

Yes, there is a sum sign.. Above the sum sign is n and below the sum sign is i = 1. The limit is n approaching infinity.

 

[stapel]

Yes, there is a sum sign.. Above the sum sign is n and below the sum sign is i = 1. The limit is n approaching infinity.

Then compare this limit-summation with the formulation they gave you for Reimann sums, and review how Reimann sums relate to definite integrals. Then convert the limit-summation into the related integral. :wink:

Eliz.

 

[CalcEqualsUgh]

I'm not really sure how to do any of that..

 

[pka]

\(\displaystyle \begin{gathered} \Delta x = \frac{4} {N}\;\& \;x_i = 5 + \left( {\Delta x} \right)i,\;i = 0,1, \cdots ,N \hfill \\ \lim _{N \to \infty } \sum\limits_{k = 1}^N {e^{x_i } + x_i } = \int\limits_5^9 {\left( {e^x + x} \right)dx} \hfill \\ \end{gathered}\)