# How do you integrate #int (x-1)sqrt(2-x)dx# from [1,2]?

-4/15

You don't need to u-substitute back and you'll get the same answer if you do.

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To integrate ( \int_{1}^{2} (x-1) \sqrt{2-x} , dx ) from ([1,2]), we can use the substitution method. Let's set ( u = 2-x ), then ( du = -dx ). Also, when ( x = 1 ), ( u = 2 - 1 = 1 ), and when ( x = 2 ), ( u = 2 - 2 = 0 ).

Substitute these into the integral:

[ \int_{1}^{2} (x-1) \sqrt{2-x} , dx = -\int_{1}^{2} u \sqrt{u} , du ]

Now we integrate ( -u \sqrt{u} ) with respect to ( u ):

[ -\int u^{3/2} , du = -\frac{2}{5}u^{5/2} + C ]

Substitute back ( u = 2-x ) and the limits:

[ -\frac{2}{5}(2-x)^{5/2} \Bigg|_{1}^{2} ]

Now plug in the upper limit:

[ -\frac{2}{5}(2-2)^{5/2} = 0 ]

And subtract the result when plugging in the lower limit:

[ 0 - \left(-\frac{2}{5}(2-1)^{5/2}\right) = -\frac{2}{5} ]

So, ( \int_{1}^{2} (x-1) \sqrt{2-x} , dx = -\frac{2}{5} ).

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