Question

help on integral please ??

Answer 1

express it as
\[
I=\sqrt{2}\int_1^8x^{1/2}dx
\]

Answer 2

\[\sqrt{\dfrac{2}{x}} = \dfrac{\sqrt2}{\sqrt{x}}\]
So \(\displaystyle\int _1 ^8 \sqrt{\dfrac{2}{x}}dx = \sqrt{2}\int_1^8x^{-1/2}dx\)

Answer 3

Remember: \[\int x^ndx = \dfrac{x^{n+1}}{n+1} + C\]

Answer 4

@geerky42 thank you ! (:

Answer 5

No problem.