Prove that the area between the $x$-axis and the function $y=e^x$ in the interval $0 < x < c$ is $e^c - 1$. You're not allowed to use integrals.
I have started to calculate the area by using the right and left Riemann sums. I got the following expressions:
Right Riemann sum:
$$\frac{c}{n} \cdot \frac{e^{\frac{c}{n}}}{e^{\frac{c}{n}}-1}\cdot(e^c - 1)\tag{1}$$
and by using L'Hospital's rule it is easy to show that the sum is $e^c -1$ as $n \rightarrow \infty$
Similarly the left Riemann sum:
$$\frac{c}{n} \cdot \frac{1}{e^{\frac{c}{n}}-1}\cdot(e^{\frac{c}{n}} - 1)\tag{2}$$ again I can show that the limit as n approches infinity is $e^c - 1$.
Now I would like to use the concept of supremum and infinimum to show that the expression (1) and (2) lead to the same conclusion that is the area is: $e^c - 1$.
As for the infimum I have also concluded that $$\frac{c}{n}\cdot\frac{1}{e^{\frac{c}{n}}-1}\cdot(e^{\frac{c}{n}} - 1) < \left(1-\frac{c}{n}\right)(e^{\frac{c}{n}} - 1)$$ but how do I apply that
$$\inf \left(\frac{c}{n}\cdot\frac{1}{e^{\frac{c}{n}}-1}(e^{\frac{c}{n}} - 1)\right) = e^{\frac{c}{n}} - 1?$$
The second question is: how on earth can I use the sup to get to the same conclusion?
Can anybody give me a step by step solution for this problem using the sup and inf concept?
I would use the fact that $y=e^x$ is increasing to show that one of the two Riemann sums is always greater than the area between the $x$-axis and the curve (regardless of $n$), while the other is always less than the area between the $x$-axis and the curve. Consequently, the area between the $x$-axis and the curve is is at most the infimum of the greater sums, and at least the supremum of the lesser sums. Show that both the infimum of the greater sums and the supremum of the lesser sums are equal to $e^c-1,$ and you're done!