# Thread: solving an intergal without knowing the funtion

1. ## solving an intergal without knowing the funtion

$\displaystyle \int_{0}^{^{X}}(x-t)y(t)=2x+\int_{0}^{^{X}}y(t)$
iknow that i should differentiate by x

$\displaystyle \frac{d\int_{0}^{^{X}}(x-t)y(t)}{dx}=\frac{2x+\int_{0}^{^{X}}y(t)}{dx}$
but i dont know if it the correct way.
i know that differentiation cancels integration but here i have intervals with variables on the integrals
and i dont know what is y(x)

2. Assuming your variable of integration is t, you would do this:

$\displaystyle \displaystyle\int_{0}^{x}(x-t)y(t)\,dt=2x+\int_{0}^{x}y(t)\,dt$

$\displaystyle \displaystyle x\int_{0}^{x}y(t)\,dt-\int_{0}^{x}t y(t)\,dt=2x+\int_{0}^{x}y(t)\,dt.$

Differentiating yields

$\displaystyle \displaystyle \int_{0}^{x}y(t)\,dt+xy(x)-x y(x)=2+y(x).$

Can you go from here?

3. why you dissregarded the zero in the lower interval when you differentiate the integral.
and why the integra of ty(t) turns into xy(x)-xy(x) and not xy(x)-xy(0)
?
about the going further ,i tried is it ok

$\displaystyle \int_{0}^{x}y(t)\,dt+xy(x)-x y(x)=2+y(x)$
turns into
$\displaystyle \int_{0}^{x}y(t)\,dt+0=2+y(x)$
turns to
$\displaystyle \int_{0}^{x}y(t)\,dt+y(x)=2$
after another differentiation
turns to
$\displaystyle y'(x)+y(x)=0$
correct?

4. why you disregarded the zero in the lower interval when you differentiate the integral
Because, when you plug in the zero, the result is a constant. The derivative of a constant is zero.

and why the integra of ty(t) turns into xy(x)-xy(x) and not xy(x)-xy(0)
.

It didn't come from that. You have to use the product rule on

$\displaystyle \displaystyle\frac{d}{dx}\left[x\int_{0}^{x}y(t)\,dt\right]=\int_{0}^{x}y(t)\,dt+xy(x).$

$\displaystyle y'(x)+y(x)=0$
You have a sign error there, as well as a losing of the constant. It should be

$\displaystyle y'(x)-y(x)=2.$

[EDIT]: See below for a correction.

5. regarding the first clame:
when we do a deriavtive to an integral we dont do the derivative of the inside function we just put one interval function input minus other interval function input

6. Ah, but what you're talking about is only evaluating the integral in the first place. The next step in our problem is to differentiate. So yes, you start with

$\displaystyle \displaystyle\int_{0}^{x}f(t)\,dt=F(x)-F(0),$ by the fundamental theorem of calculus, part 2. But then you differentiate thus:

$\displaystyle \displaystyle\frac{d}{dx}\int_{0}^{x}f(t)\,dt=\fra c{d}{dx}F(x)-0=f(x),$ by the fundamental theorem of calculus, part 1.

7. ahhh thanks now i get it

8. You're welcome. All set?

9. how did you differentiatied $\displaystyle \int_{0}^{x}t y(t)$
?

10. It's just the fundamental theorem of the calculus. You get

$\displaystyle \displaystyle\frac{d}{dx}\int_{0}^{x}ty(t)\,dt=xy( x).$

11. i got it

12. Ok, great. Have a good one!

13. Sorry from post #2 I get

$\displaystyle y'(x)-y(x)=0.$

14. Oh, right. You have to differentiate once more. My mistake.

15. Originally Posted by transgalactic
$\displaystyle \int_{0}^{^{X}}(x-t)y(t)=2x+\int_{0}^{^{X}}y(t)$
iknow that i should differentiate by x

$\displaystyle \frac{d\int_{0}^{^{X}}(x-t)y(t)}{dx}=\frac{2x+\int_{0}^{^{X}}y(t)}{dx}$
but i dont know if it the correct way.
i know that differentiation cancels integration but here i have intervals with variables on the integrals
and i dont know what is y(x)
I know it is a bit late but there is a different approch using the LaPlace transform.

The left hand side is the convolution of two functions so its transform is the product of the 2.

Taking the transform gives

$\displaystyle \frac{1}{s^2}Y(s)=\frac{2}{s^2}+\frac{1}{s}Y(s)$

Solving this for $\displaystyle Y(s)$ gives

$\displaystyle Y(s)=\frac{2}{s+1}$ and inverting the transform gives

$\displaystyle y(x)=2e^{-x}$