Question

evaluating an integral:

Answer 1

\[\int\limits_{0}^{7}\frac{1}{\sqrt{x+9}}dx\]
u = x+9
du = dx
\[=\int\limits_{0}^{7}\frac{1}{\sqrt{u}}du\]
\[=2\sqrt{u} = 2\sqrt{x+9}\]
Then Fundamental Theorem of Calculus.

Answer 2

\[\int\limits_{0}^{7}\frac{ dx }{\sqrt{x+9} }\]
\[=\int\limits_{0}^{7}\left( x+9 \right)^{\frac{ -1 }{2 }}dx\]
use the formula
\[\int\limits f ^{n}\left( x \right)f \prime \left( x \right) dx=\frac{ f ^{n+1}\left( x \right) }{n+1 }\]

Answer 3

thanks guys i got it...i get 2.

Answer 4

Woo, correct!!

Answer 5

just to note that u = x+9 so limits of integral in u where both perfect squares upper 16 and lower 9 easy to square root