Question

Integration help...

Answer 1

\[\int\limits_{0}^{8} \frac{ dx }{ \sqrt{1+x} } =\]

Answer 2

do you know how you're suppose to write this or do you need to solve for x?

Answer 3

i think u=1+x , du/dx=1 , dx=du
im not sure what to do after this.
\[\int\limits_{0}^{8} \frac{ dx }{\sqrt{u} }\] ?

Answer 4

i think ur supos to use sin

Answer 5

you are correct change the value of limits
when x=0,u=?
when x=8,u=?

Answer 6

\[\int\limits\frac{ du }{ \sqrt{a^{2} - u^{2}} } = \sin^{-1} \frac{ u }{ a } +C\] ?

Answer 7

\[\int\limits_{0}^{8}\frac{ dx }{ \sqrt{1+x} }=\int\limits_{0}^{8}\left( 1+x \right)^{\frac{- 1 }{ 2 }}dx\]
solve it

Answer 8

\[\int\limits x^n~dx=\frac{ x ^{n+1} }{ n+1 }\]

Answer 9

would that be 2sqrt(1+x) for [0,8]

Answer 10

correct ,now you use the values.

Answer 11

i got 4.
thanks

Answer 12

correct.yw.