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Don't know if this is correct (and final step)

  • Thread starter iRaid
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  • #1
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Homework Statement


[tex]\int_0^1 \frac{4ln(3x)}{\sqrt{x}}dx[/tex]


Homework Equations





The Attempt at a Solution


I really dont want to type all this out, but since I hate when people don't, I will. I hope someone takes the time to help me out :)

[tex]\int_0^1 \frac{4ln(3x)}{\sqrt{x}}dx=\lim{t \to 0^{+}} \int_0^{t} \frac{4ln(3x)}{\sqrt{x}}dx[/tex]
[tex]u=4ln3x, v=2\sqrt{x}\\du=\frac{4}{x}dx, dv=\frac{1}{\sqrt{x}}dx[/tex]
[tex](4ln3x)(2\sqrt{x})-\int \frac{8\sqrt{x}}{x}dx=8\sqrt{x}(ln3x)-8\int x^{\frac{-1}{2}}dx=8\sqrt{x}(ln3x)-16\sqrt{x}[/tex]
[tex]\lim_{t \to 0^{+}} (8\sqrt{x}(ln3x)-16\sqrt{x})|_{t}^{1}=\lim_{t \to 0^{+}} 8(ln3)-16-(8\sqrt{t}(ln3t)-16\sqrt{t})[/tex]

Now I think that just equals 8ln3-16, but I'm not sure and wolfram gives me a different answer..

Thanks.
 

Answers and Replies

  • #2
SammyS
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Homework Statement



[itex]\displaystyle \int_0^1 \frac{4\ln(3x)}{\sqrt{x}}dx[/itex]

Homework Equations



The Attempt at a Solution


I really don't want to type all this out, but since I hate when people don't, I will. I hope someone takes the time to help me out :)
[tex]\int_0^1 \frac{4\ln(3x)}{\sqrt{x}}dx=\lim_{t \to 0^{+}} \int_0^{t} \frac{4\ln(3x)}{\sqrt{x}}dx
[/tex][tex]
u=4\ln3x, v=2\sqrt{x}\\du=\frac{4}{x}dx, dv=\frac{1}{\sqrt{x}}dx
[/tex][tex]
(4\ln3x)(2\sqrt{x})-\int \frac{8\sqrt{x}}{x}dx=8\sqrt{x}(\ln3x)-8\int x^{\frac{-1}{2}}dx=8\sqrt{x}(\ln3x)-16\sqrt{x}
[/tex][tex]
\lim_{t \to 0^{+}} (8\sqrt{x}(\ln3x)-16\sqrt{x})|_{t}^{1}=\lim_{t \to 0^{+}} 8(\ln3)-16-(8\sqrt{t}(\ln3t)-16\sqrt{t})[/tex]
Now I think that just equals 8ln3-16, but I'm not sure and wolfram gives me a different answer..

Thanks.
The indefinite integral is correct. (I've checked your work that far.)

What result does Wolfram give you?
 
  • #3
559
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The indefinite integral is correct. (I've checked your work that far.)

What result does Wolfram give you?
Wow I'm stupid I just did it again and I'm not sure why wolfram gives me the answer, but my calculator gives me a different answer?? Log=ln right?
 
  • #4
SammyS
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Wow I'm stupid I just did it again and I'm not sure why wolfram gives me the answer, but my calculator gives me a different answer?? Log=ln right?
That's true for Wolfram.

For you calculator, it's probably log = log10 .
 
  • #5
559
8
That's true for Wolfram.

For you calculator, it's probably log = log10 .
That is so stupid -.-"
 

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