For the readers of “A Case of Pre-FTC Definite Integral“, this post illustrates yet another way of evaluating the following definite integral without FTC:

Let us divide into *unequal* subintervals (see Fig. 1) such that

,

i.e.,

in general and

so that the subintervals become finer as increases.

Fig. 1

We have, for the sum of the rectangles *above *the curve ,

see “Beer Theorems and Their Proofs“

Since as . Consequently,

.

And so,

It follows that

i.e.,

For

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