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Six assorted calculus problems

ozgunozgur

New member
Jun 1, 2020
27
I have 48 hours and i am bad, i am sorry but i want to understand how it is

midexam.png
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
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 \(\bf R\):



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

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

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

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Mark, this does look like a test. I'd wait for the 48h before replying.
 

ozgunozgur

New member
Jun 1, 2020
27
Thanks, now I have a solution. So iis that true?


 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
Continuing where I left off, we have:

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

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

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

And so we may write:

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

Hence, adding up all the washers, we find:

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

Let's let:

\(\displaystyle u=x-\frac{\pi}{2}\implies du=dx\)

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

Using the even-function rule, we have:

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

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 in to help with that. :)
 

ozgunozgur

New member
Jun 1, 2020
27

ozgunozgur

New member
Jun 1, 2020
27

ozgunozgur

New member
Jun 1, 2020
27
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

New member
Jun 1, 2020
27
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

New member
Jun 1, 2020
27
ca385868a75a4265a2a2816595471743.png

Hello again, is this true?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
3. I would observe that the area in question may be found from:

\(\displaystyle A=\int_{-\infty}^0 e^x\,dx\)

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

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

ozgunozgur

New member
Jun 1, 2020
27
3. I would observe that the area in question may be found from:

\(\displaystyle A=\int_{-\infty}^0 e^x\,dx\)

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

\(\displaystyle A=\lim_{t\to-\infty}\left(\int_t^0 e^x\,dx\right)=\lim_{t\to-\infty}\left(e^0-e^t\right)=1\quad\checkmark\)
Can you help me for question 4? This is my last question, is that true?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
According to W|A the definite integral given in #4 does not converge.

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

Using your substitution:

\(\displaystyle u=\sqrt{x}\implies du=\frac{1}{2\sqrt{x}}\,dx\implies dx=2u\,du\)

We then have:

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

And then using partial fractions, we may write:

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

Do you agree so far?
 

ozgunozgur

New member
Jun 1, 2020
27
According to W|A the definite integral given in #4 does not converge.

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

Using your substitution:

\(\displaystyle u=\sqrt{x}\implies du=\frac{1}{2\sqrt{x}}\,dx\implies dx=2u\,du\)

We then have:

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

And then using partial fractions, we may write:

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

Do you agree so far?
Yes, please write all of steps, sir.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,774
I'm having trouble following your written work. Can you neatly show how you would proceed?
 

ozgunozgur

New member
Jun 1, 2020
27
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

Administrator
Staff member
Feb 24, 2012
13,774
My next step would be to write the improper integral as:

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

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

ozgunozgur

New member
Jun 1, 2020
27
My next step would be to write the improper integral as:

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

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

Administrator
Staff member
Feb 24, 2012
13,774
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...