Evaluating a definite integral

[CatchThis2]

I have been trying to solve this problem for a long time and I keep getting stuck maybe someone can help as it would be greatly appreciated.

1/(3x+3) from 0 to 2

So far I get

U=3x+3
DU= 3dx
DX= 1/3 DU

I am stuck at 1/3 ln(3X+3) from 0 to 2

 

[Aladdin]

I have been trying to solve this problem for a long time and I keep getting stuck maybe someone can help as it would be greatly appreciated.

1/(3x+3) from 0 to 2

So far I get

U=3x+3
DU= 3dx
DX= 1/3 DU

I am stuck at 1/3 ln(3X+3) from 0 to 2

Replace :
(1/3)ln(3(2)+3) - (1/3)ln(3(0)+3) = ....

 

[CatchThis2]

That is what I did also. Can you try plugging it in and telling me what you get for an answer. I don't have calculator on me at the moment but It is nice to know someone is doing the same steps as me.

Thanks

 

[galactus]

Factor out the 3:

$\displaystyle \frac{1}{3}\int_{0}^{2}\frac{1}{x+1}dx$

Now, it's just a matter of an ln

 

[BigGlenntheHeavy]

$\displaystyle \int_{0}^{2}\frac{dx}{3x+3}, \ Let \ u \ = \ 3x+3, \ then \ du \ = \ 3dx, \ hence;$

$\displaystyle \frac{1}{3}\int_{3}^{9}u^{-1}du \ = \ \frac{1}{3}ln|u|\bigg]_{3}^{9} \ = \ \frac{1}{3}[ln(9)-ln(3)] \ = \ ln(3^{1/3}).$

 

[CatchThis2]

Thank you so much for your help. That was the correct answer. If you don't mind could you take a look at another question I have.

 

[CatchThis2]

I need to find the integral of (z+1)e^4z DZ

I have:

U=4z
DU=4dz
Dz=(1/4)DU
DU=(1/4)e^4z

Not sure if all of this is right. I have much more work on paper but it is to much to actually type. If anyone would try to solve this it would be greatly appreciated as I have attempted it numerous times with no luck.

 

[Aladdin]

Try expanding ....

Change of variable didn't work with me ..

 

[galactus]

Did you try parts?.

$\displaystyle \int (z+1)e^{4z}dz$

Let $\displaystyle u=z+1, \;\ dv=e^{4z}dz, \;\ du=dz, \;\ v=\frac{1}{4}e^{4z}$

Then, $\displaystyle \frac{(z+1)e^{4z}}{4}-\frac{1}{4}\int e^{4z}dz$

Finish up?.

 

[CatchThis2]

It is definitely an integration by parts problem, I think that is a two step problem though. I looked at my notes and I actually attempted it the way you did and stopped where you were at because I was unsure how to expand upon the work you have. Can you possibly finish the problem showing me step by step how you got your answer? I would really appreciate it. For some reason I am having trouble wit this integration by parts topic.

 

[galactus]

It's pretty much done except for integrating $\displaystyle \frac{1}{4}\int e^{4z}dz$ and that's straightfroward.

We should get $\displaystyle \frac{z}{4}e^{4z}+\frac{1}{4}e^{4z}-\frac{1}{16}e^{4z}=\frac{z}{4}e^{4z}+\frac{3}{16}e^{4z}$

 

[CatchThis2]

For some reason that answer is not working. I have a computer program which instantly tells you if it is right or wrong but it won't accept that answer. I really appreciate your help on that problem though as your work looks right to me.

 

[galactus]

I don't know what the computer is expecting, but that is correct.

I assume the z is just a variable. This is not complex, is it?. I wouldn't think so.

 

[CatchThis2]

I have one more definite integral problem which I am unable to solve. It is evaluate the definite integral of -7x^2e^4x dx from 0 to 1

 

[BigGlenntheHeavy]

One way,reduced: Integration by parts twice.

 

[galactus]

$\displaystyle 7\int x^{2}e^{4x}dx$

Try it this way instead of parts:

$\displaystyle \frac{d}{dx}p(x)e^{4x}=x^{2}e^{4x}$

$\displaystyle p(x)=ax^{2}+bx+c$

$\displaystyle \frac{d}{dx}(ax^{2}+bx+c)e^{4x}=x^{2}e^{4x}$..........[1]

$\displaystyle (2ax+b)e^{4x}+(ax^{2}+bx+c)4e^{4x}=x^{2}e^{4x}$

Expand and equate coefficients:

$\displaystyle (4ax^{2}+2ax+4bx+b+4c)e^{4x}=x^{2}e^{4x}$

$\displaystyle 4a=1$

$\displaystyle 2a+4b=0$

$\displaystyle b+4c=0$

$\displaystyle a=\frac{1}{4}, \;\ b=\frac{-1}{8}, \;\ c=\frac{1}{32}$

Sub back into [1]:

$\displaystyle \frac{d}{dx}(\frac{1}{4}x^{2}-\frac{1}{8}x+\frac{1}{32})e^{4x}=x^{2}e^{4x}$

Integrate boths sides and the derivative cancels and we have:

$\displaystyle \frac{1}{4}x^{2}e^{4x}-\frac{1}{8}xe^{4x}+\frac{1}{32}e^{4x}=\int x^{2}e^{4x}$

Just multiply by 7 now

 

[CatchThis2]

Your going to have to show me what to multiply by 7.

 

[galactus]

What?. It's part of your problem

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ \ -7\int_{0}^{1}x^{2}e^{4x}dx, \ I \ by \ P \ \implies \ \int udv \ = \ uv-\int vdu$

$\displaystyle Let \ u \ = \ x^{2}, \ du \ = \ 2xdx \ and \ dv \ = \ e^{4x}dx, \ v \ = \ \frac{e^{4x}}{4}.$

$\displaystyle Hence, \ f(x) \ = \ -7[\frac{x^{2}e^{4x}}{4}\bigg]_{0}^{1} \ -\frac{1}{2} \int_{0}^{1}xe^{4x}dx]$

$\displaystyle u \ = \ x, \ du \ = \ dx, \ dv \ = \ e^{4x}dx, \ v \ = \ \frac{e^{4x}}{4}$

$\displaystyle = \ -7[\frac{e^{4}}{4}-\frac{1}{2}[\frac{xe^{4x}}{4}\bigg]_{0}^{1}-\frac{1}{4}\int_{0}^{1}e^{4x}dx]]$

$\displaystyle = \ -7[\frac{e^{4}}{4}-\frac{e^{4}}{8}+\frac{1}{8}\bigg(\frac{e^{4}}{4}-\frac{1}{4}\bigg)]$

$\displaystyle = \ -7[\frac{e^{4}}{4}-\frac{e^{4}}{8}+\frac{e^{4}}{32}-\frac{1}{32}] \ = \ -7\bigg[\frac{5e^{4}-1}{32}\bigg]$

 

[CatchThis2]

Thanks so much for the help. I could follow the work you posted very well. I really appreciate your time.

GARRETT