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Mathematics 7 Online
OpenStudy (anonymous):

U-Substitution Integration

OpenStudy (anonymous):

\[\int\limits_{0}^{1}x(x^2+1)^5 \]

OpenStudy (anonymous):

\[u=x^2+1\\ du=2x~dx\\ \frac{1}{2}du=x~dx\] \[\frac{1}{2}\int_1^2u^5~du\]

OpenStudy (anonymous):

ok thats exactly what I did.. maybe I just did my math wrong. I then took the anti and got u^6 / 6 so it turned into \[(1/12) \int\limits_{1}^{2}(x^2+1)^6 \]

OpenStudy (anonymous):

and then 5^6 *1/12 - 2^6 * 1/12 =/= 21/4

OpenStudy (anonymous):

The new limits (the 1 and 2) only apply to \(u\). If you change back to a function of \(x\), you must use the given limits (0 and 1).

OpenStudy (anonymous):

ohhhh! wow, super noob move. Thank you.

OpenStudy (anonymous):

only to the 6th power, not 7th, I got it.

OpenStudy (anonymous):

Sorry, made a mistake there. You were right

OpenStudy (anonymous):

Thank you! just a little mistake.

OpenStudy (anonymous):

\[\frac{1}{12}\bigg[u^6\bigg]_1^2\] etc

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