Question

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

Answer 1

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

Answer 2

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 \[A=\int\limits_{1}^{2}[(\sqrt{x-1})-(x-1)]dx\]

Answer 3

yeah it is right good job

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

I got 3/2 as the answer.

Answer 10

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

Answer 11

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

Answer 12

Oh alright! Thank you!