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#