1. ## indefinite/definite integral

Hello all,

My first post(question) here.

what is indefinite integral,is integral without limits called as indefintie integral?

and my teacher said that for definite integral the variable of integration could be anything .I understand this since it will read the same result.

But is this not possible for indefinite integral?

2. An indefinite integral yields an equation. A definite integral yeilds a number.

$\int_{a}^{b}f(x)dx = F(b) - F(a)$

When dealing with a definite integral, we don't care about the + C. It's completely irrelevant:

$\int2xdx = F(x) = x^2 + C$

So

$F(b) - F(a) = (b^2 + C) - (a^2 + C) = b^2 - a^2$

3. Thank you derfleurer.That cleard things up a bit.
But can't we change the variable of integration indefinite integral?
It would just yield the same equation with the variable changed.So we can get back to the original variable whenever we want.correct?

4. If you change C, you change the function. $x + 4$ and $x + 3$ may have the same derivative, but for all x, the two functions are completely different. Your constant of integration is just that; a constant. Constants in no way affect your derivative and so your +C can be anything from $ln5$ to $\pi$.

We don't always take the integral from b to a. That's only when we're finding accumulated area, in which case the +C is negligable. But sometimes we just want F(a) or F(b), in which case it's very important to know.

Say we know the antiderivative of $f(x)$ is $x^3 + 4x + C$. And say we know that $F(0) = 1$. This gives us a point on the function with which to calculate C. We know that $(0)^3 + 4(0) + C = 1$, so C must equal 1.

5. Originally Posted by essex
Thank you derfleurer.That cleard things up a bit.
But can't we change the variable of integration indefinite integral?
It would just yield the same equation with the variable changed.So we can get back to the original variable whenever we want.correct?
Well, we have to know what variable to use in the result! (There is no variable in the result of a definite integral so that doesn't come up.)

It is common to write $\int^x f(t)dt$ where the variable is shown in the upper limit so that the "dummy variable" inside the integral really is "dummy".

$2\int^x t dt= 2\int^x udu= 2\int^x ydy= x^2+ C$

6. Originally Posted by derfleurer
If you change C, you change the function. $x + 4$ and $x + 3$ may have the same derivative, but for all x, the two functions are completely different. Your constant of integration is just that; a constant. Constants in no way affect your derivative and so your +C can be anything from $ln5$ to $\pi$.

We don't always take the integral from b to a. That's only when we're finding accumulated area, in which case the +C is negligable. But sometimes we just want F(a) or F(b), in which case it's very important to know.

Say we know the antiderivative of $f(x)$ is $x^3 + 4x + C$. And say we know that $F(0) = 1$. This gives us a point on the function with which to calculate C. We know that $(0)^3 + 4(0) + C = 1$, so C must equal 1.
So we cannot change the variable of integration in Indefinite integral(as we do in Definite integral ) because the constant will not be the same .correct?

Actually what i wanted to ask (say) in your example,how would anything change if i change the variable to u and integrate and subtitute again the original variable (say x) as HallsofIvy has done.
Thank you vary much once again

7. So you're asking if this statement would be true?

Where u = x + 1: $\int (x + 1)dx = \int udu = \frac{u^2}2 + C$

If so, it certainly is. Only change would be that our point F(1) = 5 would be converted to F(2) = 5. All the substitutions in the world won't change your + C.

8. I get it,thank you very much .