Question

Can some pls help, integrals

Answer 1

Hello!

Answer 2

F=F(x) is the anti-derivative of
\[f(x)=(2-x)^2-\frac{ 5 }{ (2-x)^2 }\]
which satisfies F(0)=0, find F(3)

Answer 3

Hi! :)

Answer 4

I set u=2-x

Answer 5

So basically
\(\displaystyle f(x)=(2-x)^2-\frac{ 5 }{ (2-x)^2 }\)

is the derivative of some function (will call it F), and
this function F(x) satisfies F(0)=0 (we will use this to solve
for integration constant C). Then as we find the function,
we need to find F(3).

Answer 6

If I read that correctly. Alright?

Answer 7

Yup! Sounds good to me

Answer 8

Ok, so we need to find (indefinite) integral of this.

Yes, u=2-x, is correct.

Answer 9

Okay cool. Maybe I'm making a really small mistake. Should I show my work first?

Answer 10

Sure

Answer 11

Okay gimme a min

Answer 12

\[\int\limits_{}^{} (u)^2-\frac{ 5 }{ (u)^2 } (\frac{ du }{ -1 })\]

Answer 13

\[-1 \int\limits_{}^{} (u)^2-\frac{ 5 }{ (u)^2 }du\]

Answer 14

\[-1[\frac{ 1 }{ 3}(u)^3+5(u)^{-1}]\]

Answer 15

\[-1[\frac{ 1 }{ 3 }(2-x)^6+5(2-x)+c]\]

Answer 16

-1 power.

Answer 17

and 3 power for the first one.

Answer 18

ahhh

Answer 19

was that your mistake:)

Answer 20

right. Idk y i kept the ^2. Thank you!!!

Answer 21

No problem ... write it correctly, and if you wish to verify your solution, go on to solve the problem. (If you have any questions, just say that you do.)

Answer 22

Just in case....
-----------------------------------
\(\color{black}{\displaystyle \int f(x)~dx=\int (2-x)^2-\frac{ 5 }{ (2-x)^2 }~dx}\)

\(\color{gray}{u=2-x}\)
\(\color{gray}{-du=dx}\)

\(\color{black}{\displaystyle \int -u^2+\frac{ 5 }{ u^2 }~du=-\frac{1}{3}u^3-\frac{ 5 }{ u }+C}\)

\(\color{black}{\displaystyle =-\frac{1}{3}(2-x)^3-\frac{ 5 }{ 2-x }+C}\)

Answer 23

So, your antiderivative function is:

\(\color{black}{\displaystyle F(x)=-\frac{1}{3}(2-x)^3-\frac{ 5 }{ 2-x }+C}\)

Answer 24

thanks for writing it out lol my work is so messy xD

Answer 25

You know that \(F(0)=0\). (Given by the problem.)
So, you plug this in, directly.

\(\color{black}{\displaystyle F(\color{red}{0})=-\frac{1}{3}(2-\color{red}{0})^3-\frac{ 5 }{ 2-\color{red}{0}}+C=0}\)

and then go on to solve for C.

Answer 26

would c be -c ?

Answer 27

or does it stay positive?

Answer 28

It doesn't matter if you adjust for it.
I will try to explain with an example.

Answer 29

Okay, thanks so much

Answer 30

btw i got c=31/6 and F(3)=21/2.

Answer 31

Suppose you have \(f(x)=\int g(x) dx\),
\(g(x)=(3-x)^2+2\), you are given that
\(g(0)=3\), and you are ASKED TO
FIND g(2).

Well, your first step is to integrate:
\(f(x)=\int g(x) dx=\int [(3-x)^2+2] dx\)

then, substitute,
\(u=3-x\); \(du=-dx\)

\(f(x)=-\int [u^2+2]du\)

\(\color{gray}{\rm not~I~will~post~what~I~mean~by~adjusting}\)
(sorry for going off track ...)

Answer 32

No problem, I appreciate it!

Answer 33

Algebraically,
\(f(x)=-(-x+3)^3-2(-x+3)+C\)
\(f(x)=(x-3)^3+2(x-3)+C\)

If you say your C is positive (which is what I did now), then
\(f(x)=(x-3)^3+2(x-3)+C\)

Since \(g(0)=3\), therefore.
\(f(0)=3=(-3)^3+2(-3)+C\)
\(3=-27-6+C\)
\(36=C\)

\(f(x)=(x-3)^3+2(x-3)+36\)
\(f(2)=(2-3)^3+2(2-3)+36=1-2+36=35\)
------------------------------------------
If you say your C is negative, then
\(f(x)=(x-3)^3+2(x-3)-C\)

Since \(g(0)=3\), therefore.
\(f(0)=3=(-3)^3+2(-3)-C\)
\(3=-27-6-C\)
\(-36=C\)

\(f(x)=(x-3)^3+2(x-3)-(-36)\)
\(f(x)=(x-3)^3+2(x-3)+36\) (nothing changes)
\(f(2)=(2-3)^3+2(2-3)+36=1-2+36=35\)

Answer 34

I made a mistake at the last line of each case,

= - 1, so in both cases it is 33.

Answer 35

BUT, YOU SEE

Answer 36

Ohhh I see, that clears it up!! Thanks soo much!!!! :D

Answer 37

As a matter of fact, in this case,
\(f(x)=(x-3)^3+2(x-3)+C\)
\(f(x)=(x-3)^3+2x-6+C\)
\(f(x)=(x-3)^3+2x+C\) (arbitrary constant - 6 = arbitrary constant)

then,
\(f(0)=3=(-3)^3+C\)
\(C=30\)

and then,
\(f(x)=(x-3)^3+2x+30\)
\(f(2)=(2-3)^3+2(2)+30\)
\(f(2)=-1+4+30=33\) (still 33)

Answer 38

because it's subtracting a negative so it'll be positive anyways, or add a positive. Thx :)

Answer 39

So, if you don't plug in the solution from C if you assumed C is positive, into the equation with the negative C, (or vica versa) you are good to go.

Answer 40

Sweet, thank you thank you thank you!!

Answer 41

or see what I did above just now, you can even get rid of this "-6", and still get g(2)=33, as long as you plug in the C correctly.

Answer 42

Ok, anyway, now with your problem.

Answer 43

You might want to assume it's always positive.
That's easier, and I think is certainly "grammatically" correct.

Answer 44

lol okay I will do that

Answer 45

Sure:)

Answer 46

If you want you may definitely post it here.
If you don't, then I won't insist:)) (your choice)

Answer 47

The answer or the work? :D

Answer 48

What do you mean? (Excuse me)

Answer 49

Oh ... whatever you want to post.

Answer 50

You said you can post "it". What do u mean by it? loll

Answer 51

You can post your work or your answer for verification, if you wish.

Answer 52

(Translating what I said_

Answer 53

Oh it's actually multiple choice and I submitted it online and got it correct :) Thanks to you!

Answer 54

Nice to hear that:)
Good luck with your maths!

Answer 55

Thank you!!! Good luck with all your stuff too :P