# What is area of region between the function y=x*square root(2x-1), x axis and lines x=1 and x=2?

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The area of the region can be determined evaluating the definite integral of the given function.

`int` xsqrt(2x-1)dx

2x - 1 = t => 2dx = 2dt => dx = dt/2

2x = t+1

x = (t+1)/2

`int` xsqrt(2x-1)dx =(1/4) `int` (t+1)*sqrt t dt = (1/4)`int` t sqrt (t)dt + (1/4) `int` sqrt(t) dt

`int` xsqrt(2x-1)dx = (1/4) [2(2x-1)^(5/2)/5 + 2(2x-1)^(3/2)/2]

Now, we'll apply Leibniz Newton formula to determine the definite integral:

`int` xsqrt(2x-1)dx = F(2) - F(1)

F(2) - F(1) = (1/4) [2*3^(5/2)/5 + 2*3^(3/2)/2 - 2/5 - 1]

F(2) - F(1) = (1/4) [18*sqrt3/5 + 6sqrt3/2 - 2/5 - 1]

F(2) - F(1) = (1/4) (36*sqrt3/10 + 30sqrt3/10 - 14/10)

F(2) - F(1) = (1/4)[(33sqrt3 - 7)/5]

**The requested area of the region bounded by the given function, x axis and the lines x = 1 and x = 2, is A = (33sqrt3 - 7)/20 square units.**