Another method of solving integrals?

[avr10]

Homework Statement

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

Homework Equations

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?

 

[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.

 

[rootX]

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

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

You added when you should subtract

 

[avr10]

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.

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?

 

[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.