Question

$\int\limits^0_1 { \frac{1}{ \sqrt{x+1} } } \, dx$

Answer 1

x+1 =y dx=dy  Scimbii   limitele de  integrare
x=1 y=2  x=0  y=1 Integrala devine
I=[tex] \int\limits^2_1 { \frac{1}{ \sqrt{y} } } \, dy= \int\limits^1_2 { y^{-2} } \, dy= - \frac{1}{y} =-( \frac{1}{2} - \frac{1}{1} =-(- \frac{1}{2} )= \frac{1}{2} [/tex]
S-a folosit formula
$\int\limits { t^{n} } \, dt= \frac{1}{a+1} * t^{a+1} unde a=-1/2$