Could you help me please to solve my homework

O

ozgunozgur

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


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ozgunozgur said:
Ihave 48 hours and i am bad, i am sorry but i want to understand how it is

UPkBCQ.png

UPkBCQ Fotoğrafı, resimleri, fotoğrafları
hizliresim.com hizliresim.com
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The link doesn't work and you should not post links to your homework. Instead pick one problem, take a screen capture and post the image. And, most importantly, explain where you are stuck.
 
lev888 said:
The link doesn't work and you should not post links to your homework. Instead pick one problem, take a screen capture and post the image. And, most importantly, explain where you are stuck.
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I edited. Thank you.
 
You didn't exactly do what lev888 asked!

Pick one problem to start with, and show your attempt. We can't help if we don't know what you need.
 
This set of questions was also posted at another site on which I participate. I have helped the OP set up the first one, and then showed where his minor error in evaluating the resulting definite integral lies, so that one is done. I have invited others to help with the second one because I found the written work somewhat difficult to read. There's a minor error in there somewhere, but I don't see it among the scribbles.
 
Can you help me to solve question 6 step by step?

< link to imgur's 'Most Viral Images' removed >
 
Subhotosh Khan said:
Your picture is too dark for me to decipher. Post a better lit picture.
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I wrote question 6.

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?
 
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 :/
 
I=∫∞2xx−−√x2−1dxI=∫2∞xxx2−1dx

Using your substitution:

u=x−−√⟹du=12x−−√dx⟹dx=2uduu=x⟹du=12xdx⟹dx=2udu

We then have:

I=2∫∞2√u4u4−1du=2∫∞2√1+1u4−1duI=2∫2∞u4u4−1du=2∫2∞1+1u4−1du

And then using partial fractions, we may write:

I=12∫∞2√4+1u−1−1u+1−2u2+1du

now?
 
ozgunozgur said:
I=∫∞2xx−−√x2−1dxI=∫2∞xxx2−1dx

Using your substitution:

u=x−−√⟹du=12x−−√dx⟹dx=2uduu=x⟹du=12xdx⟹dx=2udu

We then have:

I=2∫∞2√u4u4−1du=2∫∞2√1+1u4−1duI=2∫2∞u4u4−1du=2∫2∞1+1u4−1du

And then using partial fractions, we may write:

I=12∫∞2√4+1u−1−1u+1−2u2+1du

now?
Click to expand...

You're trying to simply copy/paste what I posted on the other site, and it didn't work very well because my post contained LaTeX. This is what I posted:

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

[MATH]I=\int_2^{\infty}\frac{x\sqrt{x}}{x^2-1}\,dx[/MATH]
Using your substitution:

[MATH]u=\sqrt{x}\implies du=\frac{1}{2\sqrt{x}}\,dx\implies dx=2u\,du[/MATH]
We then have:

[MATH]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[/MATH]
And then using partial fractions, we may write:

[MATH]I=\frac{1}{2}\int_{\sqrt{2}}^{\infty} 4+\frac{1}{u-1}-\frac{1}{u+1}-\frac{2}{u^2+1}\,du[/MATH]
Do you agree so far?

And I have asked you to neatly show how you'd proceed, because your hot-linked images are hard to read.
 
Why don't you continue to solve the problem? I said I have no time and I can't do it. The problem is fourth on my first post.
 
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