Find the area given by the curves.
y=sqrt(x-1), x-y=1

Question

Find the area given by the curves.
y=sqrt(x-1), x-y=1

anonymous · Answer

ok so to find the area between curves you need to determine which curve is on top of teh other right?

anonymous · Answer

Yeah, I graphed them and the y=sqrt(x-1) is on top. I also set up the intergral so far, but I'm not sure if it is correct. I got $$A=\int\limits_{1}^{2}[(\sqrt{x-1})-(x-1)]dx$$

anonymous · Answer

yeah it is right good job

anonymous · Answer

Okay. The part that is confusing me is when you start to simplify it. How would the antiderivate work for (x-1)^1/2?

anonymous · Answer

kk so (x-1)^1/2 can be thought of as x^1/2

anonymous · Answer

you know the antiderivative of x^1/2 is 2/3(x^3/2)

anonymous · Answer

so the int(x-1)^1/2 is 2/3(x-1)^3/2

anonymous · Answer

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anonymous · Answer

So there is no rule like a chain rule involved in it?

anonymous · Answer

I got 3/2 as the answer.

anonymous · Answer

$$\int\limits f^n(x) f'(x) dx=\frac{ f ^{n+1}(x) }{ n+1 }$$

anonymous · Answer

nah..chain rule is for deriviatives...I used a u-sub with x-1=u

anonymous · Answer

Oh alright! Thank you!