MHB Evaluating the Improper Integral

[shamieh]

Hey, I need a little help. I'm a bit stuck.

Evaluate the Integral

$$\int ^{\infty}_2 \frac{1}{x^{1.5}}$$

Here is what I have:

$$\int ^{\infty}_2 = x^{-1.5} = \frac{1}{.5} x^{.5} |^\infty _2$$

Am i doing this correctly or no?

 

[Prove It]

Hey, I need a little help. I'm a bit stuck.

Evaluate the Integral

$$\int ^{\infty}_2 \frac{1}{x^{1.5}}$$

Here is what I have:

$$\int ^{\infty}_2 = x^{-1.5} = \frac{1}{.5} x^{.5} |^\infty _2$$

Am i doing this correctly or no?

For starters, PLEASE use correct integral notation, it should be

$\displaystyle \begin{align*} \int_2^{\infty}{\frac{1}{x^{1.5}}\,dx} \end{align*}$

I agree converting to a negative power is the way to perform the integration. So you need to evaluate

$\displaystyle \begin{align*} \lim_{b \to \infty} \int_2^b{x^{-1.5}\,dx} \end{align*}$

 

[shamieh]

So then:

$$\int ^{\infty}_2 \frac{1}{x^{1.5}}$$

$$\lim_{b\to\infty}$$ $$\int ^{b}_2 = x^{-1.5} = \frac{1}{.5} x^{.5} |^b _2$$

$$= \lim_{b\to\infty}$$ $$[\infty] - [\frac{2}{\sqrt{2}} ]$$ is what I got. Do I need to put + C when evaluating improper integrals?

 

[MarkFL]

You are in essence taking the limit of a definite integral, so any constant of integration would be subtracted away anyway during the application of the FTOC, so there is no need to use a constant of integration. I would write:

$$I=\int_2^{\infty} x^{-\frac{3}{2}}\,dx= \lim_{t\to\infty}\left[\int_2^t x^{-\frac{3}{2}}\,dx \right]$$

Now to finish apply the FTOC and then compute the resulting limit. Be careful to correctly apply the power rule...

 

[shamieh]

$$-2\sqrt{x} | ^t_2 = [\infty] - [2*\sqrt{2}] = -2\sqrt{2}$$ ?

 

[MarkFL]

You are still not applying the power rule correctly:

$$\int x^r\,dx=\frac{x^{r+1}}{r+1}+C$$ where $$r\ne-1$$

What is $$-\frac{3}{2}+1$$ ?

 

[shamieh]

$$-2x^{-1/2} |^t_2$$

$$= 0 - [-2 * \frac{1}{\sqrt{2}}] = \frac{2}{\sqrt{2}} $$

wow that way is so much easier

 

[MarkFL]

Where did the zero come from?

 

[shamieh]

Edited

 

[shamieh]

I guess realizing that $$x^{1.5}$$ is the same as $$x^{1/2}$$ initially probabaly would have helped (Tauri)

 

[MarkFL]

I guess realizing that $$x^{1.5}$$ is the same as $$x^{1/2}$$ initially probabaly would have helped (Tauri)

$$1.5=\frac{3}{2}$$ (Wink)

 

[shamieh]

$$1.5=\frac{3}{2}$$ (Wink)

Oh why of course that's what I meant. I just like to keep you on your feet mark (Headbang)(Drunk)(Toivo)

 

[shamieh]

$$\therefore$$ this would diverge correct?

 

[MarkFL]

You tell me:

$$I=\int_2^{\infty} x^{-\frac{3}{2}}\,dx= \lim_{t\to\infty}\left[\int_2^t x^{-\frac{3}{2}}\,dx \right]=?$$

 

[shamieh]

so we have $$lim_{t\to\infty} -\frac{2}{3} x^{-1/2} |^t_2$$

so: we end up with $$\infty$$ divided by a number which = $$0$$. So it converges but then when we evaluate the other side we end up with $$2/\sqrt{2}$$ which is about $$1.4$$ something which is > 1 so it diverges, and if it diverges here the whole thing diverges right?

 

[Prove It]

so we have $$lim_{t\to\infty} -\frac{2}{3} x^{-1/2} |^t_2$$

so: we end up with $$\infty$$ divided by a number which = $$0$$. So it converges but then when we evaluate the other side we end up with $$2/\sqrt{2}$$ which is about $$1.4$$ something which is > 1 so it diverges, and if it diverges here the whole thing diverges right?

How did you get -2/3?

Surely $\displaystyle \begin{align*} \frac{x^{-\frac{3}{2} + 1}}{-\frac{3}{2} + 1} = \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} = -2x^{-\frac{1}{2}} \end{align*}$.

Now write it as $\displaystyle \begin{align*} -\frac{2}{\sqrt{x}} \end{align*}$. Evaluate this between 2 and t, and see what happens as $\displaystyle \begin{align*} t \to \infty \end{align*}$.

 

[shamieh]

How did you get -2/3?

Surely $\displaystyle \begin{align*} \frac{x^{-\frac{3}{2} + 1}}{-\frac{3}{2} + 1} = \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} = -2x^{-\frac{1}{2}} \end{align*}$.

Now write it as $\displaystyle \begin{align*} -\frac{2}{\sqrt{x}} \end{align*}$. Evaluate this between 2 and t, and see what happens as $\displaystyle \begin{align*} t \to \infty \end{align*}$.

Wow I've screwed up this problem lol.. What I meant to say was:

so we have $$\lim_{t\to\infty} -2x^{-1/2} |^t_2$$

so: we end up with $$-2(\infty)^{-1/2}$$ which would converge. But then when we evaluate the other side we end up with $$\frac{2}{\sqrt{2}}$$ which is about 1.4 but we don't have to worry about that side so would the whole thing converge? See I am wondering if I should just consider t's exponential, which in this case would be $$< 1$$ because $$-1/2 < 1$$ thus: converge? OR should I actually re-write the problem as $$\frac{-2}{\sqrt{\infty}}$$ which is just 0 thus t < 1 , thus converges

Again, thank you for bearing with me on this problem and this subject, I just want to make sure I completely understand what I am finding and what is going on, you guys have been great. Let me know if I am correct in my solution I have just provided.

Thanks

 

[MarkFL]

I would just write:

$$I=\int_2^{\infty} x^{-\frac{3}{2}}\,dx= \lim_{t\to\infty}\left[\int_2^t x^{-\frac{3}{2}}\,dx \right]=-2\lim_{t\to\infty}\left(\left[x^{-\frac{1}{2}} \right]_2^t \right)=2\lim_{t\to\infty}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{t}} \right)=2\left(\frac{1}{\sqrt{2}}-0 \right)=\sqrt{2}$$

 

[shamieh]

Got it. Since this is a finite limit I know it converges. I see now.