Integration of (z + 2) sqroot (1 - z) dz by Parts
- Thread starter ajtnewmex
- Start date
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
Re: Integration by Parts
Hi AJ:
Well, just continue around and around like you're doing.
u = 1-z
z+u=1
1-u = z
u = 1-z
z+u=1
1-u = z
u = 1-z
z+u=1
1-u = z
u = 1-z
.
.
.
in a downward spiral until you crash and burn.
Just kidding. :twisted:
If your exercise calls for using Integration by Parts, then here's some info for you. (Double-click images to expand them.)
It looks like you tried to pick (1-z)^(1/2) for u.
[attachment=2:18u6fn2e]intparts23.JPG[/attachment:18u6fn2e]
With THAT choice, you need to pick dv to be Z + 2 dz.
Notice how the LEFT side of the Integration by Parts formula turns into your given problem when you make the substitutions. (This is always a good check.)
Now, to continue, you need to find v and du. I'll do it for you.
[attachment=0:18u6fn2e]intparts24.JPG[/attachment:18u6fn2e]
With the initial choice of u = (1-z)^(1/2), look what we end up with for the RIGHT side of the Integration by Parts formula.
Not very nice looking.
When something like this happens with Integration by Parts, then it's worthwhile to start over by picking the other factor for u.
[attachment=1:18u6fn2e]intparts25.JPG[/attachment:18u6fn2e]
If you are doing Integration by Parts, then I would begin as above.
Cheers,
~ Mark
ajtnewmex said:
Hi AJ:
Well, just continue around and around like you're doing.
u = 1-z
z+u=1
1-u = z
u = 1-z
z+u=1
1-u = z
u = 1-z
z+u=1
1-u = z
u = 1-z
.
.
.
in a downward spiral until you crash and burn.
Just kidding. :twisted:
If your exercise calls for using Integration by Parts, then here's some info for you. (Double-click images to expand them.)
It looks like you tried to pick (1-z)^(1/2) for u.
[attachment=2:18u6fn2e]intparts23.JPG[/attachment:18u6fn2e]
With THAT choice, you need to pick dv to be Z + 2 dz.
Notice how the LEFT side of the Integration by Parts formula turns into your given problem when you make the substitutions. (This is always a good check.)
Now, to continue, you need to find v and du. I'll do it for you.
[attachment=0:18u6fn2e]intparts24.JPG[/attachment:18u6fn2e]
With the initial choice of u = (1-z)^(1/2), look what we end up with for the RIGHT side of the Integration by Parts formula.
Not very nice looking.
When something like this happens with Integration by Parts, then it's worthwhile to start over by picking the other factor for u.
[attachment=1:18u6fn2e]intparts25.JPG[/attachment:18u6fn2e]
If you are doing Integration by Parts, then I would begin as above.
Cheers,
~ Mark
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pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
Re: Integration by Parts
\(\displaystyle \begin{gathered} \int {\left( {{\text{z + 2}}} \right)\sqrt {1 - z} dz} \hfill \\ u = 1 - z,\;\;z = 1 - u\;\& \;dz = - du \hfill \\ \int {\left( {3 - u} \right)\left( {u^{1/2} } \right)\left( { - du} \right)} = \int {\left( {u^{3/2} - 3u^{1/2} } \right)du} \hfill \\ \end{gathered}\)
\(\displaystyle \begin{gathered} \int {\left( {{\text{z + 2}}} \right)\sqrt {1 - z} dz} \hfill \\ u = 1 - z,\;\;z = 1 - u\;\& \;dz = - du \hfill \\ \int {\left( {3 - u} \right)\left( {u^{1/2} } \right)\left( { - du} \right)} = \int {\left( {u^{3/2} - 3u^{1/2} } \right)du} \hfill \\ \end{gathered}\)
Re: Integration by Parts
Hello, ajtnewmex!
Another approach . . .
\(\displaystyle \text{Let }\,u \:=\:\sqrt{1-z}\quad\Rightarrow\quad u^2 \:=\:1-z \quad\Rightarrow\quad z \:=\:1-u^2 \quad\Rightarrow\quad dz \:=\:-2u\,du\)
\(\displaystyle \text{Substitute: }\:\int \left(3-u^2\right)\cdot u \cdot (-2u\,du) \;=\;-2\int \left(3u^2-u^4\right)\,du\)
\(\displaystyle \text{Integrate: }\;-2\left(u^3 - \frac{u^5}{5}\right) + C \;=\;-\frac{2}{5}u^3\left(5 - u^2\right) + C\)
\(\displaystyle \text{Back-substitute: }\;-\frac{2}{5}(1-z)^{\frac{3}{2}}\bigg[5 - (1 - z)\bigg] + C \quad\Rightarrow\quad -\frac{2}{5}(1-z)^{\frac{3}{2}}(4 + z) + C\)
Hello, ajtnewmex!
Another approach . . .
\(\displaystyle \text{Let }\,u \:=\:\sqrt{1-z}\quad\Rightarrow\quad u^2 \:=\:1-z \quad\Rightarrow\quad z \:=\:1-u^2 \quad\Rightarrow\quad dz \:=\:-2u\,du\)
\(\displaystyle \text{Substitute: }\:\int \left(3-u^2\right)\cdot u \cdot (-2u\,du) \;=\;-2\int \left(3u^2-u^4\right)\,du\)
\(\displaystyle \text{Integrate: }\;-2\left(u^3 - \frac{u^5}{5}\right) + C \;=\;-\frac{2}{5}u^3\left(5 - u^2\right) + C\)
\(\displaystyle \text{Back-substitute: }\;-\frac{2}{5}(1-z)^{\frac{3}{2}}\bigg[5 - (1 - z)\bigg] + C \quad\Rightarrow\quad -\frac{2}{5}(1-z)^{\frac{3}{2}}(4 + z) + C\)