Question

can someone help me simplify the integral from x to 1 of (t-1) f(t) dt if x = 0? I'm still stuck on how this =0??

Answer 1

\[\int\limits_{x}^{1}(t-1) f(t) dt \] if x = 0

Answer 2

What would this look like when I sub in the 0 for the x? How does it work out?

Answer 3

I don't understand the question
It would look like this:
\[\int\limits_{0}^{1}(t-1)f(t) dt \]

Answer 4

and then how would you evaluate it??

Answer 5

what is f?

Answer 6

no clue ..is never given....

Answer 7

this is the right half of the problem y(x) = (x-1)\[\int\limits_{0}^{x} t * f(t) dt + x \int\limits_{x}^{1}(t-1) * f(t) dt\]

Answer 8

laurie214 where is this problem located? what book? I don't understand right half of the problem? y(x)=x-1?
is y(x)=f(x)?

Answer 9

and I know how to get the value if x=0 for y(x)....but then

Answer 10

no book...prof problem...let me type the entire thing again...

Answer 11

Are you showing this side equals the other side you mentioned?

Answer 12

\[I first wrote the y(x) original problem....should have written y('x)=\int\limits_{0}^{x} t * f(t) dt + \int\limits_{x}^{1}\]

Answer 13

i'm sleepy...i will look at this later...
@foolformath
you want to look at this?

Answer 14

cut me off again....\[y'(x)=\int\limits_{0}^{x}t *f(t) + \int\limits_{x}^{1}(t-1) f(t) dt\]

Answer 15

Ok so what are we trying to do?
Find y(x)?

Answer 16

and I need to evaluate if x = 0

Answer 17

y'(0)
or
y(0)?

Answer 18

I know the first part of the problem would result in a 0 since .....this is what I got when I found y"(x)...now I need to evaluate this if x=0 and x=1

Answer 19

y'(x) , I mean....so looking for y'(0)

Answer 20

the first part is \[\int\limits_{0}^{0}t * f(t) d\] and would =0 because of the integral from 0 to 0...

Answer 21

but the second half has me stuck....the \[+ \int\limits_{0}^{1}(t-1) f(t) dt\]

Answer 22

would this half also = 0?

Answer 23

bmp had helped check over my work yesterday when I was finding the first and second derivatives of the original y(x) expression, but he hasn't been available today when I;ve checked in....he told me to send a message, but not sure how to without his name to click on...

Answer 24

\[y'(x)=\int\limits_{0}^{x}t f(t) dt+\int\limits_{x}^{1}(t-1) f(t) dt\]
So you did get the following:
\[y''(x)=xf(x)-(x-1)f(x)=f(x)\]

Answer 25

\[y''(x)=f(x)\]

Answer 26

\[y'(x)=\int\limits_{}^{}f(x) dx +C\]

Answer 27

yes :-) but now I have to evaluate y(0), y(1), y'(0) and y'(1)

Answer 28

\[\int\limits_{}^{}f(x)dx+C=\int\limits_{0}^{x}t f(t) dt+\int\limits_{x}^{1}(t-1)f(t) dt\]

Answer 29

Nothing else is given?

Answer 30

No conditions?

Answer 31

nope

Answer 32

just f is integrable on [0,1}

Answer 33

how do get the value of an integral that has x=0 yet the terms to be integrated are in terms of t?

Answer 34

\[\int\limits_{0}^{x}f(t) dt+C=\int\limits_{0}^{x}tf(t)+\int\limits_{x}^{1}(t-1)f(t) dt \]

Answer 35

the limits contain x

Answer 36

This is a function of x not t

Answer 37

ok

Answer 38

You want to see an example?

Answer 39

please! :-)

Answer 40

Pretend like
\[f(t)=t^2\]
\[y'(x)=\int\limits_0^x t \cdot t^2 dt+\int\limits_{x}^{1}(t-1)t^2 dt \]
\[y'(x)=\frac{x^4}{4}+\frac{1}{4}-\frac{x^4}{4}-\frac{1}{3}+\frac{x^3}{3}=\frac{x^3}{3}+\frac{-1}{12}=\frac{x^3}{3}+C\]

But see for unknown function f(t) I'm trying to figure out that constant :(

Answer 41

But see how it is a function of x not t?

Answer 42

\[y'(x)=\int\limits_{}^{}f(x) dx+C\]

Answer 43

\[y'(x)=\int\limits_{0}^{x}f(t) dt +C\]

Answer 44

yes

Answer 45

so the value of this would be......?

Answer 46

if x=0 then this = 0 because the integral is from 0 to 0, or like a to a....

Answer 47

I get that part, but it's the last part \[\int\limits_{0}^{1}(t-1)f(t) dt\] I'm not sure of

Answer 48

\[\int\limits_{0}^{x}f(t) dt+C=\int\limits_{0}^{x}tf(t) dt+\int\limits_{x}^{1}(1-t)f(t) dt\]
So if x=0
we have
\[\int\limits_{0}^{0}f(t) dt +C=\int\limits_{0}^{0}tf(t) dt+\int\limits_{0}^{1}(1-t)f(t) dt \]
\[C=\int\limits_{0}^{1}(1-t)f(t) dt \]
But I don't see how this helps :(

Answer 49

\[\int\limits_{0}^{x}f(t)dx+\int\limits_{0}^{1}(1-t) f(t) dt=\int\limits_{0}^{x}tf(t)+\int\limits_{x}^{1}(1-t)f(t) dt \]

Answer 50

I know...I can't get any further...I think he said the value of y' when x =1 was a 0, but that would make the last part a 0 also, and I'm not sure how that happens...

Answer 51

that's a (t-1) in the parentheses

Answer 52

So if x=1 we have
\[\int\limits_{0}^{1}f(t) dt+\int\limits_{0}^{1}(1-t) f(t) dt=\int\limits_{0}^{1}tf(t) dt +\int\limits_{1}^{1}(1-t)f(t) dt \]
So this gives us
\[\int\limits_{0}^{1}f(t) dt +\int\limits_{0}^{1}f(t) dt -\int\limits_{0}^{1}tf(t) dt =\int\limits_{0}^{1}t f(t) dt +0\]
this means we have
\[2\int\limits_{0}^{1}f(t) dt-\int\limits_{0}^{1}t d(t) dt=\int\limits_{0}^{1}tf(t) dt \]
did I do something wrong?

Answer 53

Because this means we have
\[2\int\limits_{0}^{1} f(t) dt=2 \int\limits_{0}^{1}t f(t) dt \]

Answer 54

which means
\[\int\limits_{0}^{1}f(t) dt=\int\limits_{0}^{1}t f(t) dt \]

Answer 55

I have what you have down to the next to last line...

Answer 56

now how did you get the 2 integral etc.....on the last line ?

Answer 57

I wonder if this is true for our little example
lets see what happens for \[f(t)=t^2\]

Answer 58

i added on both side int(0 to 1 of tf(t) dt)

Answer 59

\[\int\limits_{0}^{1}t^2 dt =? \int\limits_{0}^{1}t\cdot t^2 dt \]
I'm pretty sure this is not true :(

Answer 60

well actually i know it isn't true lol
1/3 does not equal 1/4

Answer 61

@bmp
I will mention bmp here but you can also do it the message thing laurie
like act as if you are going to message me but instead delete my name and put bmp
ok?

Answer 62

i need sleep laurie
I'm sorry
It is way passed my bed time

Answer 63

ok, thanks a bunch! I need sleep, too...tried but couldn't ..this was still on my mind! It's 5:30 am here....awake all night.. :-(

Answer 64

omg lol
its 4:35 AM here

Answer 65

we both need sleep! :-) thanks again!

Answer 66

i think with a fresher brain we both can work better at this problem
well hopefully lol

Answer 67

i will see you later today maybe

Answer 68

@estudier

Answer 69

I sent bmp a message...thanks!

Answer 70

hey can you take a pic of the problem and put it up
it will be easier

Answer 71

This question is not clear at all.

Answer 72

lol one more question did you say y(x)=x-1
if so then y'(x)=1 and y''(x)=0
but we had y''(x)=f(x)
So f(x)=0
?

Answer 73

ok now i go to sleep

Answer 74

@myininaya you made a mistake when you chose the limits
its true that y''(x)=f(x)
but the limits you chose for integrating should be x to 1 and not 0 to x.\[\int\limits_{0}^{x} tf(t)dt+\int\limits_{x}^{1} tf(t)dt+\int\limits_{x}^{1}f(t)dt =\int\limits_{0}^{1}tf(t)dt+\int\limits_{x}^{1}f(t)dt\]

Answer 75

so \[C=\int\limits_{0}^{1}tf(t)dt. \]
which gets cancelled off during the second differentiation.

Answer 76

so we definately need the function f(t) to go any further
we are back at where we started