Question

\[\int_0^2 \frac{dx}{\sqrt{2x - x^2}}\] i got a 6 out of 12 points here i dont know why =_=

btw my answer is \(2\pi\)

Answer 1

\[2x-x^2=-x^2+2x=-(x^2-2x)=-(x^2-2x+1)+1 =-(x-1)^2+1\]
Was this the way you wrote your bottom inside that square root thingy?

Answer 2

uhmm yep \(1 - (x-1)^2\)

Answer 3

\[\large \lim_{t \rightarrow 0} \int_t^1 \frac{dx}{\sqrt{1 - (x-1)^2}}\]

Answer 4

now just do a trig sub and you got it

Answer 5

\[\large \lim_{t \rightarrow 2} \int_1^t \frac{dx}{\sqrt{1 - (x-1)^2}}\]

Answer 6

those are my limits

Answer 7

where did i go wrong @myininaya

Answer 8

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


I chose the labeling that way so that when I find the other side it comes out to be the bottom :)

\[\text{ Let } \cos(\theta)=\frac{x-1}{1} (=\frac{adj}{hyp})\]
\[\cos(\theta) =x-1 => -\sin(\theta) d \theta =dx\]
\[\int\limits_{}^{}\frac{-\sin(\theta) d \theta}{\sqrt{1-\cos^2(\theta)}}=\int\limits_{}^{}-1 d \theta=-\theta+C\]
\[=-\cos^{-1}(x-1)+C\]
So we need to look back at the limits right ?