# Another method of solving integrals?

1. Aug 14, 2008

### avr10

1. The problem statement, all variables and given/known data

$$\text {Find m such that }\displaystyle\int^m_4 \frac{1}{x\sqrt{x}}\,dx = .9$$

2. Relevant equations

3. The attempt at a solution

$$\displaystyle\int^m_4 \frac{1}{x\sqrt{x}}\,dx = .9 \Rightarrow \displaystyle\int^m_4 x^{-3/2}\,dx = .9 \Rightarrow -2m^{-1/2} +2(4)^{-1/2} = .9 \Rightarrow m = \frac {4}{1.9^{2}} = 1.108$$

If I plug this value back into the original integral, it comes out as $$-.9$$. Should I solve this integral another way? Also, an extention of the problem is

$$\text {Explain why there is no number m such that} \displaystyle\int^m_4 \frac{1}{x\sqrt{x}}\,dx = 1.1$$

It seems like that has to deal with convergence issues, something I'm just beginning to learn. Any hints for the first step?

2. Aug 14, 2008

### nicksauce

Check your numbers again, I get m = 400.

To do the second part, show that the integral is maximal if m=infinity, and show that the value of the integral in that case is less than 1.1.

3. Aug 14, 2008

### rootX

I don't know latex but that bold part is wrong (Addition mistake) I also get 20^2

You added when you should subtract

4. Aug 14, 2008

### avr10

Thanks, in order to show that it is maximal at infinity, does it suffice to say that since x must always be greater than 0, then the integral is maximal at infinity? That doesn't sound very rigorous...how would you phrase it?

5. Aug 14, 2008

### nicksauce

I think it should be enough to show that the function
$$g(t)=\int_4^{t}\frac{dx}{x\sqrt{x}}$$ is always increasing (you could use FTC, with g'(t)>0), and so it follows that g(t) is maximal as t->infinity.