49
S dx/ (27+2x)^(2/3)
0
I'm pretty confused on what to do, help me figure it out please.
$\displaystyle \int_0^{49}\frac{1}{(27+2x)^{\frac{2}{3}}}dx$
So u = 27 + 2x
du = 2dx
dx = du/2
change the limits, 27 to 125
$\displaystyle \int_{27}^{125}\frac{1}{2(u)^{\frac{2}{3}}}du$
$\displaystyle \int_{27}^{125}\frac{1}{2}*u^{\frac{-2}{3}}du$
power rule:
$\displaystyle \frac{3}{2}*u^{\frac{1}{3}}$ from 27 to 125, can you take it from there?
I'm still not sure where you get this 277 from but the point of the last question was to check that you followed why I simply substituted 125 and 27 in for u instead of replacing it with the 27+2x, because i'm lazy and don't feel like substituting back in if I already know what it's going to be.
But here is the solution with substituting the 27+2x back in:
$\displaystyle \frac{3}{2}*u^{\frac{1}{3}}
$ u goes from 27 to 125
$\displaystyle \frac{3}{2}*(27+2x)^{\frac{1}{3}}$ x goes from 0 to 49
$\displaystyle =\frac{3}{2}*(27+2(49))^{\frac{1}{3}}-\frac{3}{2}*(27+2(0))^{\frac{1}{3}}$
$\displaystyle =\frac{3}{2}*(27+98)^{\frac{1}{3}}-\frac{3}{2}*(27)^{\frac{1}{3}}$
$\displaystyle =\frac{3}{2}*(125)^{\frac{1}{3}}-\frac{3}{2}*(27)^{\frac{1}{3}}$
$\displaystyle =\frac{3}{2}*5-\frac{3}{2}*3$
does that help?