Maybe it's a really basic question as i just started learning calculus but If integration represents the area under a curve, why indefinite integrals gives a function as area and how it is related to the area under that curve? Like definite integrals which gives a 'number' as the area.

also why integration of sin is -cos and not '0(zero)' as if we see sin graph it has equal no of crests as valleys of equal areas so why they don't cancel each other out and give '0'?

thank you.

  • $\begingroup$ The integral is only equal to the area under the curve if the given function is nonnegative in the given interval. Aditionally, this area can as well depend on the limits of the interval , so it makes sense that the area is a function of one or more variables. $\endgroup$
    – Peter
    Jan 16 at 11:27
  • $\begingroup$ One cal also finf excellent ressources on the site OpenStax ( Rice University, free open textbooks) , in particular : <openstax.org/details/books/calculus-volume-1>, with Gilbert Strang as senior contributor. $\endgroup$ Jan 16 at 12:32

2 Answers 2


If you're talking about between a graph of a function $f(x)$ and the x-axis, you need actually specify two lines parallel to $y$ axis to say 'which' area you are talking about. The choice of these two lines is basically determined by the boundary of the subset of you integrate over. For example, if you integrate over the set [3,4], then the lines between which you area is bounded is $x=3$ and $x=4$.

The question indefinite integral is, to ask about a function, which can directly take in these sets and spit out area? For instance, let's say I want to know the integral of $x^2$ over different sets $[0,1]$ , $[2,3]$, then I can first find the indefinite integral:

$$ \int x^2 dx = \frac{x^3}{3}+C$$

And find the area between each set by just plugging in the numbers, explicitly:

$$ \int_{[0,1]} x^2 dx = \frac{1^3}{3} - \frac{0^3}{3}$$

$$ \int_{[2,3]} x^2 dx = \frac{3^3}{3} - \frac{2^3}{3}$$

Note that the constant $C$ cancels out under subtraction of the function evaluated at the two boundary point.

  • $\begingroup$ that make a lot of sense, thank you $\endgroup$
    – Dev Partap
    Jan 16 at 11:50
  • $\begingroup$ You may have a look at Fowler's " Calculus, a mooculus" . $\endgroup$ Jan 16 at 12:05
  • I think it is useful to make the following distinctions:

(1) $\LARGE {\int}_a^b f(x)dx $ is a number ( which corresponds to the area under the graph of function $f$ from $x=a$ to $x=b$ in case this graph is above the $X-$ axis.)

(2) $\LARGE g(x)={\int}_a^x f(t)dt $ is a function that gives as output the number called " definite integral " ( as defined in (1) above) for a varying limit of integration $x$. So, the input of this function , $x$ , is the varying upper limit of integration. Maybe one could call $g(x)$ the definite integral function; it is sometimes called the " accumulation function".

Example : https://www.desmos.com/calculator/todsdsoeaf

(3) $ \LARGE h(x) = \int f(x)dx $ is the indefinite integral , which is defined as a generic function representing all the antiderivatives or primitives of function $f$. So $h(x)= \LARGE \int f(x)dx $ reads as " function $h$ is any primitive of $f$ " .

  • The link between all these objects is provided by the Fundamental Theorem Of Calculus.

With $f, g, h$ as defined above , we have :

(1) $\LARGE g'(x) = f(x)$ , i.e. the derivative of $g(x)$ - the accumulation function of $f(x)$ - is functon $f$ itself ; meaning that integration and derivation are inverse processes.

(2) $\LARGE {\int}_a^b f(x)dx = h(b) - h(a) $ .

Since the most dffcult concept is the idea of an " accumulation function" , I add this picture . In this picture, I have arbitrarily chosen $a=0$ as the lower limit of integration.

enter image description here

  • 1
    $\begingroup$ so indefinite integral of f(x) will give you a function g(x) which represents the set of areas under f(x)'s graph for limits 0 and x where g(x) for a value of x will give you area under the graph of f(x) till between 0 and x. and if your lower limit is not 0 but a variable y(let's say) then the area is g(x)(the area under 0 to x ) - g(y)(the area under 0 to y). right? $\endgroup$
    – Dev Partap
    Jan 16 at 12:43
  • $\begingroup$ I think common usage is to reserve " indefinite integral" to the generic primitive of $f$ ( function $h$ in my post). The accumulation function is just one specific prmitive ( identified by number $a$, its lower limit of integration): that is, if one chooses another number for $a$, one gets another primitive. You're right as to the fact that the set of all g(x) is the set of all areas with limits $a$ and $x$. $\endgroup$ Jan 16 at 13:02
  • $\begingroup$ Here a Desmos graph exhibiting different accumulation functions ( with different values for $a$, the lower limit of integration). You can ask Desmos for the derivatives of $ g_1, g_2, g_3$ and you'll see that the anwer is $f$ for all three : <desmos.com/calculator/todsdsoeaf> $\endgroup$ Jan 16 at 13:05
  • $\begingroup$ If the lower limit is not $0$ , but say $2$ , the accumulation function gives as output all the areas from $2$ to $x$ , with $x$ a variable upper limit. But note that the " area" is a signed area , and can be a negative number. $\endgroup$ Jan 16 at 13:14
  • 1
    $\begingroup$ got it. thank you :) $\endgroup$
    – Dev Partap
    Jan 16 at 13:26

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