Limit of a Summation.

1. Oct 10, 2007

azatkgz

1. The problem statement, all variables and given/known data
$$\frac{1}{n}\lim_{n\rightarrow\infty}\sum_{k=1}^{n}f(a+\frac{b-a}{n}k)$$

3. The attempt at a solution
I tried to solve it simply.
$$\frac{1}{n}\lim_{n\rightarrow\infty}\sum_{k=1}^{n}f(a+\frac{b-a}{n}k)=\int_{0}^{1}f(a+(b-a)x)dx$$

$$=f(b)-f(a)$$

Last edited: Oct 10, 2007
2. Oct 10, 2007

azatkgz

Sorry,
$$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}f(a+\frac{b-a}{n}k)=\int_{0}^{1}f(a+(b-a)x)dx$$

3. Oct 10, 2007

HallsofIvy

Staff Emeritus
Very good. Now integrate by letting u= a+ (b-a)x. What is du? Also, be careful about the limits of integration.

4. Oct 10, 2007

azatkgz

$$\int_{a}^{b}\frac{f(u)du}{(b-a)}=\frac{1}{b-a}(f(b)-f(a))$$
Yes?

5. Oct 10, 2007

Dick

Yes, to the left hand side. No, to the right. Unless the 'f' on the right stands for the antiderivative of the 'f' on the left. In which case you should say so and use a different symbol.

6. Oct 10, 2007

azatkgz

$$\frac{1}{b-a}\int_{a}^{b}f(u)du$$