Integration Example

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Jason76

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Look right? Did I do the back substitution too late? :confused:

\(\displaystyle \int(x^{3} + 9)^{2} x^{2} dx\).

\(\displaystyle u = x^{3} + 9\).

\(\displaystyle du = 3x^{2} dx\)

\(\displaystyle \dfrac{1}{3} du = dx\).

\(\displaystyle \dfrac{1}{3}\int (u)^{2} du\)

\(\displaystyle (\dfrac{1}{3})\dfrac{(u)^{3}}{3} + C\).

\(\displaystyle \dfrac{(u)^{3}}{9} + C\).

\(\displaystyle \dfrac{(x^{3} + 9)^{3}}{9} + C\). - Back substitute.


OR


\(\displaystyle \int(x^{3} + 9)^{2} x^{2} dx\).

\(\displaystyle u = x^{3} + 9\).

\(\displaystyle du = 3x^{2} dx\)

\(\displaystyle \dfrac{1}{3} du = dx\).

\(\displaystyle \dfrac{1}{3}\int (u)^{2} du\)

\(\displaystyle (\dfrac{1}{3})\dfrac{(u)^{3}}{3} du\).

\(\displaystyle \dfrac{(u)^{3}}{9} du\).

\(\displaystyle \dfrac{(x^{3} + 9)^{3}}{9} + C\). - Back substitute.
 
Last edited:
Jason76 said:
Look right? Did I do the back substitution too late? :confused:

\(\displaystyle \int(x^{3} + 9)^{2} x^{2} dx\).

\(\displaystyle u = x^{3} + 9\).

\(\displaystyle du = 3x^{2} dx\)

\(\displaystyle \dfrac{1}{3} du = dx\). X where is the x^2?

\(\displaystyle \dfrac{1}{3}\int (u)^{2} du\)

\(\displaystyle (\dfrac{1}{3})\dfrac{(u)^{3}}{3} + C\).

\(\displaystyle \dfrac{(u)^{3}}{9} + C\).

\(\displaystyle \dfrac{(x^{3} + 9)^{3}}{9} + C\). - Back substitute.


OR ...
Click to expand...
Good :)
Your 2nd version looks identical to the 1st one to me.
 
Jason76 said:
First version is correct. For the second version... which somewhat identical to the first version (except for the mistake as indicated)


\(\displaystyle \int(x^{3} + 9)^{2} x^{2} dx\).

\(\displaystyle u = x^{3} + 9\).

\(\displaystyle du = 3x^{2} dx\)

\(\displaystyle \dfrac{1}{3} du = dx\).

\(\displaystyle \dfrac{1}{3}\int (u)^{2} du\)

\(\displaystyle (\dfrac{1}{3})\dfrac{(u)^{3}}{3} du\) ..............incorrect ...... It should be \(\displaystyle (\dfrac{1}{3})\dfrac{(u)^{3}}{3}\) + C

\(\displaystyle \dfrac{(u)^{3}}{9} du\)...............incorrect ...... It should be \(\displaystyle \dfrac{(u)^{3}}{9}\) + C

\(\displaystyle \dfrac{(x^{3} + 9)^{3}}{9} + C\). - Back substitute.
Click to expand...
.
 
Last edited by a moderator:
du

Jason76 said:
Look right? Did I do the back substitution too late? :confused:

\(\displaystyle \int(x^{3} + 9)^{2} x^{2} dx\).

\(\displaystyle u = x^{3} + 9\).

\(\displaystyle du = 3x^{2} dx\)

\(\displaystyle \dfrac{1}{3} du = dx\).

\(\displaystyle \dfrac{1}{3}\int (u)^{2} du\)

\(\displaystyle (\dfrac{1}{3})\dfrac{(u)^{3}}{3} + C\).

\(\displaystyle \dfrac{(u)^{3}}{9} + C\).

\(\displaystyle \dfrac{(x^{3} + 9)^{3}}{9} + C\). - Back substitute.


OR


\(\displaystyle \int(x^{3} + 9)^{2} x^{2} dx\).

\(\displaystyle u = x^{3} + 9\).

\(\displaystyle du = 3x^{2} dx\)

\(\displaystyle \dfrac{1}{3} du = dx\).

\(\displaystyle \dfrac{1}{3}\int (u)^{2} du\)

\(\displaystyle (\dfrac{1}{3})\dfrac{(u)^{3}}{3} du\).

\(\displaystyle \dfrac{(u)^{3}}{9} du\).

\(\displaystyle \dfrac{(x^{3} + 9)^{3}}{9} + C\). - Back substitute.
Click to expand...

du refers to the variable with which you are integrating, which is u; once you take the integrand, you can't write du afterwards, it would be incorrect to do so.
 
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