# Thread: definite integrals, upper limit variable

1. ## definite integrals, upper limit variable

Hello

it is correct to write:

$\int^x_0 f(x)\,dx$

or i need write ?

$\int^x_0 f(t)\,dt$

why ?

The question is :
if I need a substitution of f(x) in a formula :

$f(x)=sin(x)$

$y=f(x) + \int^x_0 f(x)\,dx$

I can directly replace f(x)

otherwise :

$y=f(x) + \int^x_0 f(t)\,dt$

don't seem me elegant imho..

Thanks for the possible answers !

2. Originally Posted by Gustavis
Hello

it is correct to write:

$\int^x_0 f(x)\,dx$

or i need write ?

$\int^x_0 f(t)\,dt$ Mr F says: This is the one you want.

why ? Mr F says: Let me re-write your two choices and then you tell me which one doesn't make sense:

${\color{red}\int^{x = x}_{x = 0} f(x)\,dx}$

${\color{red}\int^{t = x}_{t = 0} f(t)\,dt}$

The question is :
if I need a substitution of f(x) in a formula :

$f(x)=sin(x)$

$y=f(x) + \int^x_0 f(x)\,dx$

I can directly replace f(x)

otherwise :

$y=f(x) + \int^x_0 f(t)\,dt$

don't seem me elegant imho..

Thanks for the possible answers !
..

3. Originally Posted by Gustavis
Hello

it is correct to write:

$\int^x_0 f(x)\,dx$

or i need write ?

$\int^x_0 f(t)\,dt$
As Mr. Fantastic indicatses, the second is far less confusing. In the first, you would have to recognize that you are using "x" to mean two different things- and that's never a good idea.

why ?

The question is :
if I need a substitution of f(x) in a formula :

$f(x)=sin(x)$

$y=f(x) + \int^x_0 f(x)\,dx$

I can directly replace f(x)

otherwise :

$y=f(x) + \int^x_0 f(t)\,dt$

don't seem me elegant imho..

Thanks for the possible answers !
Would you say that,, when x= 1, $y(1)= f(1)+ \int_0^1 f(1)d1$? That doesn't even make sense: there is no such thing as "d1"! If you write instead $y(1)= f(1)+\int_0^1 f(x)dx$ then you are recognizing that the "x" outside the integral and in the upper limit is NOT the same as the "x" inside the integral- and so you shouldn't use the same symbol!
Far better to write $y(x)= f(x)+ \int_0^x f(t)dt$ so that when x= 1, $y(1)= f(1)+ \int_0^1 f(t)dt$