Question

I'm having trouble with an improper integral question. when calculating the integral between 4 and 69 of abs(x-5)^(-1/3) I'm getting an answer of 45/2 when the answer should be 51/2. I've pinpointed my error to one integral equating -3/2 when it should be equal to 3/2. I don't understand how that should be a positive value though. help please?

Answer 1

\[\int\limits_{4}^{69}\frac{ 1 }{ \sqrt[3]{\left| x-5 \right|} } \] just incase anyone got confused

Answer 2

my work so far: integral 1 = \[\lim_{R \rightarrow 5-}\int\limits_{4}^{R}(5-x)^{-1/3} dx = -3/2\]

Answer 3

you should show your integral, because you are missing a sign.

you can integrate
\[ \int u^{-\frac{1}{3}} \ du = \frac{3}{2}u^{\frac{2}{3}}\]
let u= 5-x
du = - dx (notice the minus sign)

we need a -dx so, multiply by -1 twice:

\[-\int\limits_{4}^{R}(5-x)^{-1/3} (-dx) = - \frac{3}{2}(5-x)^{\frac{2}{3}} \mid_4^{5}=0 - - \frac{3}{2}= + \frac{3}{2}\]

Answer 4

thank you so much!!