Asked by Ja
integrate (x+6)/(sqrt(x+2)) with range 2 , -2 ; with u = sqrt(x+2)
Answers
Answered by
mathhelper
Let u = √(x+2)
u^2 = x + 2
x = u^2 - 2
dx = 2u du
then,
∫ (x+6)/(√(x+2)) dx
= ∫ (u^2 -2 + 6)/u (2u du)
= 2∫ (u^2 + 4) du
= 2[ (1/3)u^3 + 4u]
= 2[(1/3)(x+2)^(3/2) + 4√(x+2) ] for x = -2 to +2
= 2 [(1/3) (4)^(3/2) + 4√4 - ((1/3)(0) + 4√0) ]
= 2 [ 8/3 + 8)
= 64/3
u^2 = x + 2
x = u^2 - 2
dx = 2u du
then,
∫ (x+6)/(√(x+2)) dx
= ∫ (u^2 -2 + 6)/u (2u du)
= 2∫ (u^2 + 4) du
= 2[ (1/3)u^3 + 4u]
= 2[(1/3)(x+2)^(3/2) + 4√(x+2) ] for x = -2 to +2
= 2 [(1/3) (4)^(3/2) + 4√4 - ((1/3)(0) + 4√0) ]
= 2 [ 8/3 + 8)
= 64/3