` int_(0)^(lambda) (ydy)/(sqrt(y+lambda))` is equal to

Options

A. `(2)/(3)(2-sqrt2)lambda sqrt(lambda)`
B. `(2)/(3)(2+sqrt2)lambda sqrt(lambda)`
C. `(1)/(3)(2-sqrt2)lambdasqrt(lambda)`
D. `(1)/(3)(2+sqrt2)lambdasqrt(lambda)`

Answer

A

Explanation

Let ` l = int_(0)^(lambda) (ydy)/(sqrt(y+lambda))` ,
` = int_(0)^(lambda) [ sqrt(y+lambda) - lambda/(sqrt(y+lambda))]dy` ,
` = [ (y+lambda)^(3//2)/(3//2)]_(0)^(lambda) - [ (lambda sqrt(y+lambda))/(1//2) ] _(0)^(lambda)`
` = 2/3 [ (2 lambda )^(3//2) - lambda ^(3//2) ] - 2lambda[(2 lambda )^(1//2) - (lambda)^(1//2)] `
` = 2lambda sqrt(lambda) [ (2 sqrt(2)-1)/3 - (sqrt(2)-1) ] = 2/3 lambda sqrt(lambda) (2 - sqrt(2))` ,