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

Answer 1

Emilia Aguilar

#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#

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Answer 2

Charles Arocha

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#

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Answer 3

Adam Adkins

To evaluate ∫(1/√(2-x))dx from 1 to 2, you can use the substitution method. Let u = 2 - x, then du = -dx.

When x = 1, u = 2 - 1 = 1. When x = 2, u = 2 - 2 = 0.

So, the integral becomes ∫(1/√u)(-du) from 1 to 0.

This equals -∫(1/√u)du from 0 to 1.

The antiderivative of 1/√u is 2√u.

Now, evaluate this from 0 to 1: 2√1 - 2√0.

This simplifies to 2 - 0 = 2.

Therefore, the value of the integral ∫(1/√(2-x))dx from 1 to 2 is 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.