# Summation and Integral

1. Oct 7, 2007

### azatkgz

Can someone check this solution.
1. The problem statement, all variables and given/known data
$$\lim_{n\rightarrow\infty}\sum^{n}_{i=1}\sqrt{\frac{1}{n^2}+\frac{2i}{n^3}}$$

3. The attempt at a solution
$$=\lim_{n\rightarrow\infty}\frac{1}{n}\sum^{n}_{i=1}\sqrt{1+\frac{2i}{n}}=\int^{1}_{0}\sqrt{1+2x}dx$$
for u=1+2x->du=2dx
$$\int^{1}_{0}\frac{\sqrt{u}du}{2}=\frac{1}{3}$$

2. Oct 7, 2007

### Gib Z

That is correct.

3. Oct 7, 2007

### azatkgz

My prof.says that I forgot to change something,while changing variables.

4. Oct 7, 2007

### Gib Z

O perhaps the bounds on the integral? Sorry im having a bad day..

5. Oct 7, 2007

### azatkgz

yes, I think problem is on Integral.

6. Oct 7, 2007

### Gib Z

I am sure of it now, you forgot the change the bounds. New limits of integration should be 3 and 1.

7. Oct 7, 2007

### azatkgz

and the answer is
$$\int^{3}_{1}\sqrt{1+2x}dx=\frac{3^{\frac{3}{2}}-1}{3}$$
yes?

8. Oct 7, 2007

### Gib Z

Correct, though it may look more pretty as $$\frac{3\sqrt{3}-1}{3}$$ lol.

9. Oct 7, 2007

### azatkgz

Can you explain, please,why we must choose 3 and 1.

10. Oct 7, 2007

### Gib Z

O simple. What the limits on the original integral mean are 'sum for values of x between 1 and 0'. You made the substitution u= 2x+1. So when the original integral says sum for x between 1 and 0, the new integral, where u is the new variable, must say 'sum for x between 1 and 0, and since u=2x+1, sum for u between 3 and 1'.

11. Oct 7, 2007

ok,thanks