## anonymous 5 years ago 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

1. 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.... $\frac{1}{4}[\frac{2}{3}u^\frac{3}{2}+2u^\frac{1}{2}]_{1}^{9}\rightarrow \frac{1}{4}(\frac{72}{3}-\frac{8}{3})=\frac{16}{3}=5.3333$

2. anonymous

You are a lifesaver. Thank you very much!

3. anonymous

you're welcome

4. anonymous

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

Find more explanations on OpenStudy