MHB Is This Integral Divergent as t Approaches Infinity?

[ozgunozgur]

I have 48 hours and i am bad, i am sorry but i want to understand how it is

 

[MarkFL]

Hello, and welcome to MHB! :)

In the future please make a thread for each problem so the resulting discussion doesn't become convoluted.

1.) I would begin by graphing \(\bf R\):

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I would use the washer method here for the volume of an arbitrary element:

$$dV=\pi(R^2-r^2)\,dx$$

Can you identify the outer radius \(R\) and the inner radius \(r\) of the arbitrary washer above?

 

[Fantini]

Mark, this does look like a test. I'd wait for the 48h before replying.

 

[ozgunozgur]

Thanks, now I have a solution. So iis that true?

https://ibb.co/g7nRs7D

https://i.ibb.co/b3fKp3r/20200602-022640.jpg

 

[MarkFL]

Continuing where I left off, we have:

$$dV=\pi((1+\sin(x))^2-(1)^2)\,dx=\pi(\sin^2(x)+2\sin(x))\dx$$

Using a double-angle identiy for cosine: we may state:

$$\cos(2x)=1-2\sin^2(x)\implies \sin^2(x)=\frac{1-\cos(2x)}{2}$$

And so we may write:

$$dV=\frac{\pi}{2}(4\sin(x)-\cos(2x)+1)\,dx$$

Hence, adding up all the washers, we find:

$$V=\frac{\pi}{2}\int_0^{\pi} 4\sin(x)-\cos(2x)+1\,dx$$

Let's let:

$$u=x-\frac{\pi}{2}\implies du=dx$$

$$V=\frac{\pi}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4\cos(u)+\cos(2x)+1\,dx$$

Using the even-function rule, we have:

$$V=\pi\int_{0}^{\frac{\pi}{2}} 4\cos(u)+\cos(2x)+1\,dx=\pi\left[4\sin(u)+\frac{1}{2}\sin(2u)+u\right]_{0}^{\frac{\pi}{2}}=\pi\left(4+\frac{\pi}{2}\right)=\frac{\pi}{2}(8+\pi)$$

It looks like you were on the right track, but you neglected to distribute \(\pi\) to your additional integral.

For the second problem, it looks like you are using integration by parts, but your work is hard to read. I'll let someone else jump into help with that. :)

 

[ozgunozgur]

Thanks, This is question 5, can you write me?

 

[ozgunozgur]

Is this solution true?

 

[ozgunozgur]

Regarding question 3, ln(1) = 0, so you need to find −∫10ln(x)dx−∫01ln⁡(x)dx.
An indefinite integral is xln(x)−xxln⁡(x)−x. Can you finish the question now?

According to W|A, the integral in question 4 doesn't converge. Did you type it correctly?

 

[ozgunozgur]

In questipn 3, integral x(lnx-1) 1(0-1) - 0(ln0-1) = -1
In question 4, the integral does not converge.
One side is endless when it is solved as a normal integral.

Please help to solve step by step :/

 

[ozgunozgur]

Hello again, is this true?

 

[MarkFL]

3. I would observe that the area in question may be found from:

$$A=\int_{-\infty}^0 e^x\,dx$$

This is an improper integral, and so I would write:

$$A=\lim_{t\to-\infty}\left(\int_t^0 e^x\,dx\right)=\lim_{t\to-\infty}\left(e^0-e^t\right)=1\quad\checkmark$$

 

[ozgunozgur]

3. I would observe that the area in question may be found from:

$$A=\int_{-\infty}^0 e^x\,dx$$

This is an improper integral, and so I would write:

$$A=\lim_{t\to-\infty}\left(\int_t^0 e^x\,dx\right)=\lim_{t\to-\infty}\left(e^0-e^t\right)=1\quad\checkmark$$

Can you help me for question 4? This is my last question, is that true?

 

[MarkFL]

According to W|A the definite integral given in #4 does not converge.

$$I=\int_2^{\infty}\frac{x\sqrt{x}}{x^2-1}\,dx$$

Using your substitution:

$$u=\sqrt{x}\implies du=\frac{1}{2\sqrt{x}}\,dx\implies dx=2u\,du$$

We then have:

$$I=2\int_{\sqrt{2}}^{\infty}\frac{u^4}{u^4-1}\,du=2\int_{\sqrt{2}}^{\infty}1+\frac{1}{u^4-1}\,du$$

And then using partial fractions, we may write:

$$I=\frac{1}{2}\int_{\sqrt{2}}^{\infty} 4+\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{u^2+1}\,du$$

Do you agree so far?

 

[ozgunozgur]

According to W|A the definite integral given in #4 does not converge.

$$I=\int_2^{\infty}\frac{x\sqrt{x}}{x^2-1}\,dx$$

Using your substitution:

$$u=\sqrt{x}\implies du=\frac{1}{2\sqrt{x}}\,dx\implies dx=2u\,du$$

We then have:

$$I=2\int_{\sqrt{2}}^{\infty}\frac{u^4}{u^4-1}\,du=2\int_{\sqrt{2}}^{\infty}1+\frac{1}{u^4-1}\,du$$

And then using partial fractions, we may write:

$$I=\frac{1}{2}\int_{\sqrt{2}}^{\infty} 4+\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{u^2+1}\,du$$

Do you agree so far?

Yes, please write all of steps, sir.

 

[MarkFL]

I'm having trouble following your written work. Can you neatly show how you would proceed?

 

[ozgunozgur]

I'm having trouble following your written work. Can you neatly show how you would proceed?

I did not solve that question, I just put it on paper. I thought I was writing legibly. :/

 

[MarkFL]

My next step would be to write the improper integral as:

$$I=\frac{1}{2}\lim_{t\to\infty}\left(\int_{\sqrt{2}}^{t} 4+\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{u^2+1}\,du\right)$$

Can you give the anti-derivative we would use in our application of the FTOC on the definite integral in the limit?

 

[ozgunozgur]

My next step would be to write the improper integral as:

$$I=\frac{1}{2}\lim_{t\to\infty}\left(\int_{\sqrt{2}}^{t} 4+\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{u^2+1}\,du\right)$$

Can you give the anti-derivative we would use in our application of the FTOC on the definite integral in the limit?

Is 4u + ln|u+1| - ln|u-1| + 2arctan(u) result?
if t --&gt is infinity;
= 4t - 4kok(2) + ln|(t+1)/(kok(2)+1)| - ln|(t-1)/(kok(2)+1)| + 2arctan(t) - 2arctan(kok(2))

is t a divergent integral since it approximates to infinity?

 

[MarkFL]

Is 4u + ln|u+1| - ln|u-1| + 2arctan(u) result?
if t --&gt is infinity;
= 4t - 4kok(2) + ln|(t+1)/(kok(2)+1)| - ln|(t-1)/(kok(2)+1)| + 2arctan(t) - 2arctan(kok(2))

is t a divergent integral since it approximates to infinity?

No, be careful with your signs...