Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

$\int f(x)\,\mathrm{d}x$ 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.

share|cite|improve this answer
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

Yes, your function is a definite integral, because it is evaluated over a certain interval. Although the constant is strictly not necessary, because it will be subtracted when the integral is evaluated, it is good practice to keep the constant of integration. If you want to be consistent, rename the variable in the function that you are integrating to avoid confusion, if you are using x as the value of a point. It seems your reasoning is ok.

share|cite|improve this answer

A definite integral is nothing different from an indefinite integral but the constant, that was eliminated during the differentiation, has some definite value. For instance in indefinite integrals we have to write a C that represents all constants after the integration has been done. So in definite integrals this C has some value and can be determined by some function value for instance if (int)f(x)=g(x)+C in 0 to a limits then for calculating the C there would be a function value i.e. f(n) (n belongs to Z) would be given or some details would be given to calculate it. And for integrating definite integrals, just integrate them normally like indefinite and then put the values of limits in the answer (upper limits - lower limits) to get the answer.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.