# Definite Integral and Constant of Integration

I understand that when we are doing indefinite integrals on the real line, we have $\int f(x) dx = g(x) + C$, where $C$ is some constant of integration.

If I do an integral from $\int f(x) dx$ on $[0,x]$, then is this considered a definite integral? Can I just leave out the constant of integration now? I am skeptical of the fact that this is a definite integral, because our value $x$ is still a variable.

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$\int f(x)\,\mathrm{d}x$ is an antiderivative. It represents any function whose derivative with respect to $x$ is $f(x)$.

$\int_0^af(x)\,\mathrm{d}x$ is a definite integral, and for any of the antiderivatives $g(x)=\int f(x)\,\mathrm{d}x$ (which incorporate a constant of integration), $$\int_0^af(x)\,\mathrm{d}x=g(a)-g(0)$$ For example, $$\int x^3\,\mathrm{d}x=\frac14x^4+C$$ for some constant $C$, and \begin{align} \int_0^ax^3\,\mathrm{d}x &=\left(\frac14a^4+C\right)-\left(\frac140^4+C\right)\\ &=\frac14a^4 \end{align} no matter which $C$ is chosen.

In the case above, $\int_0^xf(x)\,\mathrm{d}x$, there is confusion because the same variable is used inside the integration as in the bounds. The bound variable $x$ inside the integral is not the same as the free variable $x$ in the limit. To reduce the confusion, your integral can also be written as $\int_0^xf(t)\,\mathrm{d}t$ by renaming the bound variable. In any case, this is a definite integral.

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Would this be the case even for functional integration? –  Ravi Thiagarajan Jun 10 '12 at 9:24
In higher dimensions (and to my understanding functional integration), most integrals are definite. Perhaps someone who understands functional integration better can illuminate further. –  robjohn Jun 10 '12 at 13:12