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Indefinite integral is the family of all its primitives or antiderivatives. It represents geometrically a family of curves having parallel tangents at their points of intersection with the lines orthogonal to the axis representing the variable of integration.

This is the geometrical interprdtation of indefinite integral. Also the book introduces integration as

Integration really means to sum up unlike differentiation which divides things infinitesimally.

Really?? But in Thomas Calculus it is written

...indefinite integral is a function plus constant.

So, what does the book want to say by saying integration is summation. Yes, it is right in case of definite integral. But what about indefinite integral? It just represents family of curves ie. a function. How can it be related to summation?

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Integration really means to sum up unlike differentiation which divides things infinitesimally.

I'm quite sure this is just the explanation of there the word 'integration' came from, nothing more.

This is a part of the usual confusion between antiderivative and an integral, which was adressed a lot on this site. 'Indefinite integral' seems like a very unfortunate choice of words.


On the other hand, if we choose a particular value for the integration constant, then indefinite and definite integrals look very much alike.

For example, the definition of the logarithm is:

$$\ln x=\int_1^x \frac{dt}{t}, \qquad x>0$$

The indefinite integral, or the antiderivative of a hyperbole would be:

$$\int \frac{dt}{t}=\ln |t|+C$$

For $C=0$ and $t>0$ there is not much difference between the two expressions, is it?


Not let's talk a little about the geometric interpretation:

$\displaystyle \int_1^x \frac{dt}{t}$ represents the area under a family of curves which all begin at the same point $(1,1)$ and go through the same set of points, but have different endpoints $x$.

In the picture below are area representations of $\ln 2, \ln 3, \ln 4$:

enter image description here

On the other hand, we can transform this integral:

$$\int_1^x \frac{dt}{t}=\int_0^{x-1} \frac{du}{1+u}=(x-1)\int_0^1 \frac{dv}{1+(x-1)v}$$

Now we have the area under a family of curves which all have the same $x-$ coordinates of the beginning and end points, but go through different sets of points in-between. Basically, this is a definite integral with a parameter.

In the picture below we represent the same area values of $\ln 2, \ln 3, \ln 4$, but in another form:

enter image description here


In other words, once we fix the constant, the indefinite integral becomes something like definite and can be definitely talked about in terms of area. But we still shouldn't confuse the two objects.

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