Limit of Summation: Solving with Integration

[azatkgz]

Homework Statement

$$\frac{1}{n}\lim_{n\rightarrow\infty}\sum_{k=1}^{n}f(a+\frac{b-a}{n}k)$$

The Attempt at a Solution

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

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

 

[azatkgz]

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

 

[HallsofIvy]

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

 

[azatkgz]

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

 

[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.

 

[azatkgz]

Than ,actually,answer remains just
$$\frac{1}{b-a}\int_{a}^{b}f(u)du$$

 

[Dick]

Sure the summation is just a definition of a riemann integral.