Question

Evaluate the Intregral posted below

Answer 1

\[\int\limits_{0}^{1} \frac{ 6x dx }{\sqrt{16+3x^2} }\]

Answer 2

try \(u=16+3x^2, du=6xdx,u(0)=16, u(1)=19\)

Answer 3

then
\[\int_{16}^{19}\frac{du}{\sqrt{u}}\] should work

Answer 4

would 19 be U and 16 be du?

Answer 5

i know the answer is \[2\sqrt{19}-8 \] but i dont know how to get there