Calculus 8th Edition

$36$
We simplify the fraction by changing the fraction to the constants multiplied by some power of $u$ and get $\int_1^4 4u^{-\frac{1}{2}} + 6u^{\frac {1}{2}} du$ This makes it easier to see the powers of $u$ that are being changed int eh integrand. Evaluating the integral results in $\frac{4u^{\frac{1}{2}}} {\frac {1}{2}} + \frac{6u^{\frac{3}{2}}} {\frac {3}{2}}$ $= 8u^{\frac{1}{2}} + 4u^{\frac{3}{2}}$ We now substitute $4$ and $1$ into $u$ and get $(8(4)^{\frac {1}{2}} + 4(4)^{\frac{3}{2}}) - (8(1)^{\frac {1}{2}} + 4(1)^{\frac{3}{2}})= (8(2) + 4(8)) - (8 + 4) = (16 + 32) - (12) = 36$ *Recall that $4^{\frac {1}{2} } = \sqrt 4 = 2$ and $4^{\frac {3}{2} } = 2^3 =8$