Question

Calculus question. Been on it all day. Still not solved

Answer 1

\[\int\limits y \sqrt{10+4y-4y^2} dy\]

Answer 2

set u=10+4y-4y^2

Answer 3

I feel `completing the square` and then using a `trig substitution` works out smoothly..

Answer 4

I am always up for most elementary methods as possible. u=... is easier.

Answer 5

oh wait, to complete the square and then use u sub.... that would be even better looking work.

Answer 6

but I mean, either way... I will leave it to the mathematicians here. Won't interrupt ya'll...

Answer 7

solomon you are right. final answer is what i have wrong

Answer 8

\[I=\int\limits y \sqrt{10+4y-4y^2}dy\]
\[=-\frac{ 1 }{ 8 } \int\limits \left( -8y \right)\sqrt{10+4y-4y^2}dy\]
\[=-\frac{ 1 }{ 8 }\int\limits \left( -8y+4-4 \right)\sqrt{10+4y-4y^2}dy\]
\[=-\frac{ 1 }{ 8 }\int\limits \left( 10+4y-4y^2 \right)^{\frac{ 1 }{ 2 }}\left( -8y+4 \right)dy+\frac{ 4 }{ 8 }\int\limits \sqrt{10+4y-4y^2}dy\]

Answer 9

\[=-\frac{ 1 }{ 8 }\frac{ \left( 10+4y-4y^2 \right)^{\frac{ 3 }{ 2 }} }{ \frac{ 3 }{ 2 } }+\frac{ 1 }{ 2 }I _{1}+c\]

Answer 10

\[I _{1}=\int\limits \sqrt{10-4\left( y^2-y+\frac{ 1 }{ 4 }-\frac{ 1 }{ 4 } \right)}dy=2 \int\limits \sqrt{\frac{ 11 }{ 4 }-\left( y-\frac{ 1 }{ 2 } \right) ^2}dy\]
put\[y-\frac{ 1 }{ 2 }=\frac{ \sqrt{11} }{ 2 }\sin t\]
\[dy=\frac{ \sqrt{11} }{ 2 }\cos t~dt\]
?

Answer 11

this is where i get stuck to...at the end

Answer 12

the answer has sin-1 in it...i know that

Answer 13

number 30 is the form we are using for this question

Answer 14

up until the form...i had about what you do to. what to do with the form?

Answer 15

\[I _{1}=2*\frac{ 11 }{ 4 }\int\limits \cos ^2~t~dt=\frac{ 11 }{ 4 }\int\limits \left( 1+\cos 2t \right)dt\]
\[=\frac{ 11 }{ 4 }\left( t+\frac{ \sin 2t }{ 2 } \right)=\frac{ 11 }{ 4 }\left( t+\sin t \cos t \right)\]
\[=\frac{ 11 }{ 4 }\left( t+\sin t \sqrt{1-\sin ^2t} \right)\]
replace the value of t