If integration represents the area under a curve, why indefinite integrals gives a function.

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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.

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  • 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

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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.