anonymous
  • anonymous
What about x/(2x-1)^(1/2) with an upper of 5 and a lower of 1. My calculator is giving me a correct answer of 5.333333. But I can't work the problem out to be the same
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
You have to use u substitution and then back substitute x in terms of u: \[\int\limits_{1}^{5} \frac{x}{(2x-1)^\frac{1}{2}}dx \rightarrow u=2x-1, du=2dx, \frac{1}{2}du=dx\] so you end up with..... \[\frac{1}{2} \int\limits \frac{x}{u^\frac{1}{2}}du \rightarrow since: u=2x-1, u+1=2x, \frac{1}{2}(u+1)=x\] \[\frac{1}{2} \int\limits\limits \frac{\frac{1}{2}(u+1)}{u^\frac{1}{2}}du \rightarrow \frac{1}{4} \int\limits\limits \frac{u+1}{u^\frac{1}{2}}du\] Now change the limits of integration to u... \[Upper:x=5, u=2x-1, so... u=2(5)-1, u=9\] \[Lower:x=1, u=2x-1, so... u=2(1)-1, u=1\] Thus... \[\frac{1}{4} \int\limits_{u=1}^{u=9} \frac{u+1}{u^\frac{1}{2}}du \rightarrow \frac{1}{4} \int\limits_{1}^{9} \frac {u}{u^{\frac{1}{2}}} du + \frac{1}{4} \int\limits_{1}^{9} \frac{1}{u^\frac{1}{2}}du\] so.... \[\frac{1}{4}[\frac{2}{3}u^\frac{3}{2}+2u^\frac{1}{2}]_{1}^{9}\rightarrow \frac{1}{4}(\frac{72}{3}-\frac{8}{3})=\frac{16}{3}=5.3333\]
anonymous
  • anonymous
You are a lifesaver. Thank you very much!
anonymous
  • anonymous
you're welcome

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anonymous
  • anonymous
Raise the quantity in parentheses to the indicated exponent, and simplify the resulting expression. (-2x^9y^4)^ 4

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