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find the following integral
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\[\int\limits_{?}^{?} (3-x)^10 dx\]
the bracket is raised to 10
Make a substitution. Let u = 3 - x, du = -dx, dx = -du
\[u = 3 - x, -du = dx\]\[\int\limits (3-x)^{10}dx \rightarrow - \int\limits u^{10} du = - \frac {u^{11}}{11} + C \rightarrow - \frac {(3-x)^{11}}{11} + C\]
If you wanna make that positive, you can switch the 3 and the x.\[\int\limits\limits (3-x)^{10}dx = \frac {(x-3)^{11}}{11} + C\]
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