Expert: Ahmed Salami - 8/3/2010

Question
Find the area of the region that lies beneath the curve  y=sqrt (2x+2) ; 0<x<1

Answer
Hi Ken,
The area beneath a curve f(x) between the x-limits a and b is given by
∫f(x) dx from a to b
Now,
f(x) = √(2x+2)
    = √2.√(x+1)
A = ∫√2.√(x+1) dx    from 0 to 1
 = √2∫(x+1)^(1/2) dx    from 0 to 1
 = √2[(2/3)(x+1)^(3/2)]    from 0 to 1
 = √2[(2/3)(1+1)^(3/2)] - √2[(2/3)(0+1)^(3/2)]
 = √2[(2/3)(2)^(3/2)] - √2[(2/3)(1)^(3/2)]
 = 8/3 - 2√2/3
 = (8 - 2√2)/3
 = 1.724

Regards