How do you evaluate $int 1/sqrt(2-x)dx$ from 1 to 2?

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Answer 1 · 2017-08-02T06:26:25

$int_1^2 1/sqrt(2-x)dx = -int_1^2 (d(2-x))/sqrt(2-x)$

$int_1^2 1/sqrt(2-x)dx = -int_1^2 (2-x)^(-1/2)d(2-x)$

$int_1^2 1/sqrt(2-x)dx = - [(2-x)^(1/2)/(1/2)]_1^2$

$int_1^2 1/sqrt(2-x)dx = - [2sqrt(2-x)]_1^2 = -2sqrt(2-2)+2sqrt(2-1) = 2$

Answer 2 · 2017-08-02T06:27:02

The answer is $=2$

We need

$intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

We are going to perform a substitution

Let $u=2-x$ , $=>$ , $du=-dx$

Therefore,

$int(dx)/sqrt(2-x)=int(-du)/sqrtu$

$=-intu^(-1/2du)=(-u^(1/2))/(1/2)$

$=-2sqrtu=-2sqrt(2-x)+C$

So,

$int_1^2(dx)/sqrt(2-x)=[-2sqrt(2-x)]_1^2$

$=(0)-(-2*1)$

$=2$

How do you evaluate int 1/sqrt(2-x)dxint 1/sqrt(2-x)dx from 1 to 2?