evaluate integral?

Question

integral {2 to 10} of ( 1/ sqrt(x-1) )dx

Answer 1

I = ∫ (x - 1)^(-1/2) dx
I = 2 (x - 1)^(1/2) between limits 2 and 10
I = 2(9)^(1/2) - 2(1)^(1/2)
I = 6 - 2
I = 4

Answer 2

du U substitution. u=x-1 thus du=dx

you have the integral of 1/ sqrt(u) which is equal to u^(-1/2) which is 2u^(1/2) after integrating. Put u in terms of x again and you have 2(x-1)^(1/2).

from 2 to 10 >> 2(10-1)^(1/2)- 2(2-1)^(1/2)=6-2=4

Answer 3

∫dx/sqrt(x-1)

=>∫(x-1)^(-1/2) dx

=>2sqrt(x-1)

applying limits 2 to 10

2[sqrt(10-1) - sqrt(2-1)]

=>2[sqrt(9) - sqrt(1)]

=>2[3 - 1] = 4

Answer 4

=4

f(x)=1/sqrt(x-1) = (sqrt(x-1))^(-1) = ((x-1)^(1/2))^(-1) = (x-1)^(-1/2)

F(x)= 2 * (x-1)^(1/2) = 2 * sqrt(x-1)

When the borders 2 and 10 are inserted you get:
integral= 2 * sqrt(10-1) - 2 * sqrt(2-1) = 2 * 3 - 2 * 1 = 6 - 2 = 4

Answer 5

integ [2 to 10] of ( 1/sqrt(x-1) ) dx
Let x-1=t
dx=dt
integ [1 to 9] dt/sqrt(t)
integ [1 to 9] t^(-1/2)dt
integ [1 to 9] t^(-1/2+1)/(-1/2+1)
integ [1 to 9] 2/t^(1/2)
x-1 = t
when x=2, t=1
when x=10, t = 9
integ [1 to 9] 2 t^(1/2)
2 9^(1/2) -2 1^(1/2)
6-2=4

Answer 6

Let u = x-1
Then du = dx
So:

int (2,10) (1/sqrt (x-1)) dx
= int (1,9) (1/sqrt (u)) du
= 2sqrt(u) | (1,9)
= 2sqrt(9) - 2sqrt(1)
= 2*3 - 2
= 6-2
= 4