Question

Integral with complete the square and trig subsitution

Answer 1

give me a minute to post the question

Answer 2

1st of all... I am told to use #30 on the reference page i attached

Answer 3

the integral to evaluate is \[\int\limits y \sqrt{10+4y-4y^2} dy\]

Answer 4

the final answer has \[\sin^{-1} \] in it

Answer 5

I cannot get the final right! I have submitted so many answers.. they are all coming up wrong.

Answer 6

those are the clues the professor gave me

Answer 7

Hi!!

Answer 8

factor out the \(-4\)first

Answer 9

\[-4(y^2-y)+10\] the complete the square

Answer 10

you should get \(11-4 (x-1/2)^2\)

Answer 11

i got 11 - (2y-1)^2

Answer 12

\[11- (2y-1)^2\]

Answer 13

hi Freckles!

Answer 14

glad to see you

Answer 15

gotta run to class
@freckles got this i am sure

Answer 16

\[y \sqrt{10+4y-4y^2} \\ y \sqrt{10-(4y^2-4y)} \\ y \sqrt{10-(4y^2-4y+1)+1} \\ y \sqrt{11-(2y-1)^2} \\ \]
\[y \sqrt{11 (1-\frac{(2y-1)^2}{11}) }\\ \sqrt{11} y \sqrt{1-\frac{(2y-1)^2}{11}} \\ \sqrt{11} y \sqrt{1-(\frac{2y-1}{\sqrt{11}})^2}\]

Answer 17

so recall
sin^2(x)+cos^2(x)=1
cos^2(x)=1-sin^2(x)
so use the sub sin(x)=(2y-1)/sqrt(11)

Answer 18

i have \[\int\limits 1/2 (u+1) \sqrt{a^2-b^2}1/2 du\]

Answer 19

then

Answer 20

\[(\frac{ 1 }{ 4}) \int\limits u \sqrt{a^2-b^2} du + \frac{ 1 }{ 4 }\int\limits \sqrt{a^2-b^2} du\]

Answer 21

what is a and b and u?

Answer 22

u= 2y-1 a= sqrt 11

Answer 23

\[w= \sqrt{a^2-b^2}\]

Answer 24

\[-1/8 \int\limits \sqrt w dw + (u/8) \sqrt (a^2-b^2) + (a^2/8) \sin^{-1} (u/a)\]

Answer 25

this is where i get lost

Answer 26

i think \[\sqrt{a^2- b^2}\] = \[\sqrt{(11-(2y-1)^2)}\]

Answer 27

but i got lost and confused myself

Answer 28

ok one sec
I guess you are trying to use some kinda formula...

Answer 29

there is a pic of what i did...idk where i went wrong. but you can see what i did. that equation thing is hard for me to get in quick

Answer 30

\[\int\limits \sqrt{11} y \sqrt{1-(\frac{2y-1}{\sqrt{11}})^2} dy \\ \text{ Let } \sin(t)=\frac{2y-1}{\sqrt{11} } \\ \cos(t) dt=\frac{2}{ \sqrt{11}} dy \\ \int\limits \sqrt{11} (\frac{\sqrt{11} \sin(t)+1}{2}) \sqrt{1-\sin^2(t)} \frac{\sqrt{11}}{2} \cos(t) dt \\ \frac{11}{4}\int\limits (\sqrt{11} \sin(t)+1) \sqrt{1-\sin^2(t)} \cos(t) dt \\ \frac{11}{4}\int\limits_{}^{} \sqrt{11} \sin(t) \sqrt{1-\sin^2(t) } \cos(t) dt+\frac{11}{4} \int\limits_{}^{} 1 \sqrt{1-\sin^2(t)} \cos(t) dt \]
\[\frac{11 \sqrt{11}}{4} \int\limits_{}^{}\sin(t) \sqrt{\cos^2(t)} \cos(t) dt +\frac{11}{4}\int\limits_{}^{} \sqrt{\cos^2(t)} \cos(t) dt \\ \frac{11 \sqrt{11}}{4} \int\limits_{}^{}\sin(t) \cos^2(t) dt+\frac{11}{4} \int\limits \cos^2(t) dt \\ \text{ Let } u=\cos(t) \\ du=-\sin(t) dt \\ \frac{11 \sqrt{11}}{4} \int\limits_{}^{} u^2 (-du)+\frac{11}{4} \int\limits_{}^{} \frac{1}{2}(1+\cos(2t)) dt \]
\[\frac{11 \sqrt{11}}{4} \frac{-u^3}{3}+\frac{11}{8}(t +\frac{1}{2}\sin(2t))+C \\ \frac{-11 \sqrt{11}}{12} u^3+\frac{11}{8} t+\frac{11}{16} \sin(2t)+C \\ \frac{-11 \sqrt{11}}{12} \cos^3(t)+\frac{11}{8}t+\frac{11}{16} 2 \sin(t)\cos(t)+C \\ \]
Now to get it back in terms of y we refer to our substitution and its corresponding reference right triangle

Answer 31

now I suppose you are using a formula ... so let me look at your work

Answer 32

yea..i do substitutions a lot... thats the way the professor makes us

Answer 33

all i know is my final answer came up as wrong when i plugged it in. and i do not know where im going wrong with the question. do not forget... he told us to use #30 of the reference page i posted above

Answer 34

formulas are boring :(
and you have to memorize more stuff

Answer 35

i agree... i feel very lost in all the substitutions but idk how to do anything else because it hasn't been taught to me. I feel like this question is too long and too hard lol

Answer 36

Ok i have found a formula in my book:
\[\int\limits u \sqrt{2au-u^2} du =\frac{2u^2-au-3a^2}{6} \sqrt{2au-u^2}+\frac{a^3}{2} \cos^{-1}(\frac{a-u}{a})+C\]
i see this formula

Answer 37

is that the one you are trying to use?

Answer 38

professor said to use # 30

Answer 39

on the paper i posted... right after i split up the integral u +1 and right before the w substitution started... the formula is used on the integral that doesnt have the u

Answer 40

the second integration

Answer 41

ok we are going to do this...
\[I= \int\limits_{}^{}y \sqrt{10+4y-4y^2} dy \\ \text{ Let } u=10+4y-4y^2 \text{ So } du=(4-8y ) dy \\ I=\frac{1}{-8} \int\limits -8y \sqrt{10+4y-4y^2} dy \\ I=-\frac{1}{8} \int\limits (-8y+4-4) \sqrt{10+4y-4y^2} dy \\ I=\frac{-1}{8} \int\limits [(4-8y) \sqrt{10+4y-4y^2}-4 \sqrt{10+4y-4y^2} ]dy ) \\ I=\frac{-1}{8} \int\limits_{}^{} (4-8y) \sqrt{10+4y-4y^2} dy -\frac{1}{8} \int\limits -4 \sqrt{10+4y-4y^2} dy\]


\[\text{ Let } u=10+4y-4y^2 \\ du=(4-8y) dy \\ I=\frac{-1}{8} \int\limits u^\frac{1}{2} du+\frac{1}{2} \int\limits \sqrt{11-(2y-1)^2}\]
\[I=\frac{-1}{8} \frac{u^\frac{3}{2}}{\frac{3}{2}}+\frac{1}{2} \int\limits \sqrt{11-(2y-1)^2} dy \\ \text{ Let } b=2y-1 \\ db=2 dy \\ \frac{1}{2} db=dy \\ I=\frac{-1}{8} \frac{2}{3} u^\frac{3}{2}+\frac{1}{2} \int\limits \sqrt{11-b^2} \frac{1}{2} db \\ I=\frac{-1}{12} u^\frac{3}{2}+\frac{1}{4} \int\limits \sqrt{11-b^2}\]

Answer 42

not that one integral should be in the exact form you need

Answer 43

where a^2=11
so a =sqrt(11)

Answer 44

ok.. im trying to review real quick

Answer 45

but I think you got to this part right?

Answer 46

yea... my final answer came out wrong tho

Answer 47

oops one sec

Answer 48

\[I=\frac{-1}{12}(10+4y-4y^2)^\frac{3}{2}+\frac{1}{4} \int\limits\limits \sqrt{11-b^2} \\ I=\frac{-1}{12}(10+4y-4y^2)^\frac{3}{2}+\frac{1}{4} [\frac{ b}{2} \sqrt{11-b^2}+\frac{11}{2}\sin^{-1}(\frac{b}{\sqrt{11}})]+C \\ I=\frac{-1}{12}(10+4y-4y^2)^\frac{3}{2}+\frac{b}{8} \sqrt{11-b^2}+\frac{11}{8} \sin^{-1}(\frac{b}{\sqrt{11}})+C \\ I=\frac{-1}{12}(10+4y-4y^2)^\frac{3}{2}+\frac{2y-1}{8} \sqrt{11-(2y-1)^2}+\frac{11}{8} \sin^{-1}(\frac{2y-1}{\sqrt{11}})+C \]

Answer 49

1/2+2/2=3/2
I see one mistake so far

Answer 50

you didn't integrate sqrt(w) correctly

Answer 51

well you half like did

Answer 52

just the exponent is wrong there

Answer 53

looking at your other stuff now

Answer 54

oooohhh!

Answer 55

and the only other complaint I would make is that inside the arcsin I would put (2y-1)/sqrt(11)
just so my teach knows I know that just the 1 isn't be divided by sqrt(11)

Answer 56

that is the only thing I see wrong with the final answer though :)

Answer 57

Sorry I had to do it myself first to see if I got what you got

Answer 58

dont be sorry...i WANT to see how you did it!

Answer 59

I thank you kindly... once again! I am still confused lol but I feel that If i review the way you did a few times... i will get it!

Answer 60

well how did you get the final solution being confused?
i mean you basically got it
just a little tiny error is all

Answer 61

it took me 8 hrs tho lol

Answer 62

and i got on line with a chat tutor that my school provides... and watch 3 videos of similar questions

Answer 63

its been a "nightmare"

Answer 64

well I be around for maybe another hour if you decide you need more help
or if I need to clarify anything

Answer 65

ok! I am actually writing down what you did... I like your way much better. I will then study this over and over until I feel I have a grasp on it!

Answer 66

the first way or the second way? lol
the first way is without formulas
second way is your teacher preferred way from what I understood (Formula)

Answer 67

the second way lol...i have to use what the professor tole me for the form or i would try your form choice