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Trouble getting solution to this U sub integral
- Thread starter PalmOlive
- Start date
D
Deleted member 4993
Guest
[FONT=MathJax_Size2]Been wrapping my head around it, but can't figure it out. Not sure where to begin.. I tried using e^ln7 to sub out the 7.... but lost myself in a dead end... Any ideas??
∫ (3-x)7^((3-x)^2)dx[/FONT]
\(\displaystyle \displaystyle{\int (3-x)^{7^{(3-x)^2}}dx}\)
first substitute u = 3-x then substitute u2 = v and continue... it is messy ...
What are your thoughts?
Please share your work with us ...even if you know it is wrong
If you are stuck at the beginning tell us and we'll start with the definitions.
You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:
http://www.freemathhelp.com/forum/th...Before-Posting
Last edited by a moderator:
\(\displaystyle \displaystyle{\int (3-x)^{7^{(3-x)^2}}dx}\)
first substitute u = 3-x then substitute u2 = v and continue... it is messy ...
What are your thoughts?
Please share your work with us ...even if you know it is wrong
If you are stuck at the beginning tell us and we'll start with the definitions.
You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:
http://www.freemathhelp.com/forum/th...Before-Posting
It's actually \(\displaystyle \displaystyle \begin{align*} \int{ \left( 3 - x \right) \cdot 7^{ \left( 3 - x \right) ^2 } \,\mathrm{d}x } \end{align*}\), not \(\displaystyle \displaystyle \begin{align*} \int{ \left( 3 - x \right) ^{ 7^{ \left( 3 - x \right) ^2 } } \,\mathrm{d}x } \end{align*}\). Anyway
\(\displaystyle \displaystyle \begin{align*} \int{ \left( 3 - x \right) \cdot 7^{ \left( 3 - x \right) ^2 } \,\mathrm{d}x } &= -\int{ -\left( 3 - x \right) \cdot 7^{ \left( 3 - x \right) ^2 } \,\mathrm{d}x } \\ &= -\int{ u\cdot 7^{ u^2 } \,\mathrm{d}u } \textrm{ after substituting } u = 3 - x \implies \mathrm{d}u = -\mathrm{d}x \\ &= -\frac{1}{2} \int{ 2\,u \cdot 7^{ u^2 } \,\mathrm{d}u } \\ &= -\frac{1}{2} \int{ 7^v \,\mathrm{d}v } \textrm{ after substituting } v = u^2 \implies \mathrm{d}v = 2\,u\,\mathrm{d}u \end{align*}\)
Can you continue? It might also help to remember that \(\displaystyle \displaystyle \begin{align*} 7^v = \mathrm{e}^{ \ln{ \left( 7^v \right) } } = \mathrm{e}^{ v\ln{(7)} } \end{align*}\)...
D
Deleted member 4993
Guest
It's actually \(\displaystyle \displaystyle \begin{align*} \int{ \left( 3 - x \right) \cdot 7^{ \left( 3 - x \right) ^2 } \,\mathrm{d}x } \end{align*}\), not \(\displaystyle \displaystyle \begin{align*} \int{ \left( 3 - x \right) ^{ 7^{ \left( 3 - x \right) ^2 } } \,\mathrm{d}x } \end{align*}\). Anyway ...
I suspected that! However, I wanted to determine OP's interest in the problem - so that s/he can come back and correct my interpretation.