$\int_a^b e^x dx$ using limits.

Question

$$\int_{a}^{b}{e^{-x}}dx$$

by definition $$b-a=nh$$

$$\int_{a}^{b}{e^{-x}} dx = \lim_{h \to 0}{he^{-(a)}+he^{-(a+h)}+he^{-(a+2h)}+...+he^{-(a+(n-1)h)}}$$

which can further be simplified to

$$\int_{a}^{b}{e^{-x}} dx= \lim_{h \to 0}{h\left(\frac{1}{e^a}+\frac{1}{e^ae^{2h}}+\frac{1}{e^ae^{3h}}+...+\frac{1}{e^ae^{(n-1)h}}\right)}$$

From here $\frac{1}{e^a}$ can be taken common but after that I m confused on how to eliminate $h$

Answer 1

Now you need to use the geometric sum formula:-

$$\lim_{n\to\infty}he^{-a}\frac{e^{-nh}-1}{e^{-h}-1}=e^{-a}\cdot (e^{-(b-a)}-1)\lim_{h\to 0}\frac{h}{e^{-h}-1}=e^{-a}-e^{-b}$$.

As $\lim_{h\to 0}\frac{h}{e^{-h}-1}=-1$. Prove it using L'Hospital or any other way.

But this requires proper justification. This happens as if $P_{n}$ is a partition of $[a,b]$ and $||P_{n}||\to 0$ as $n\to\infty$. Then

$\lim_{n\to\infty}U(P_{n},f)=\int_{a}^{\bar{b}}f$ where $\int_{a}^{\bar{b}}f$ denotes the upper sum = $\inf U(P,f)$ where the infimum is taken over all partitions.

And $\lim_{n\to\infty}L(P_{n},f)=\int_{\bar{a}}^{b}f$ . Where $\int_{\bar{a}}^{b}f$ denotes the lower sum = $\sup L(P,f)$ where supremum is taken over all partitions.

Then their equality proves the Darboux Integrability of f

Answer 2

Into the Riemann Sum formula:

$$\int_a^b e^x dx=\lim_{n\to\infty}\frac{b-a}n\sum_{k=0}^n e^{a+k\frac{b-a}n}=\lim_{n\to\infty}\frac{b-a}n \frac{e^{b+\frac bn}-e^{a+\frac an}}{e^\frac b n-e^\frac an}$$

Now for the limit properties:

$$\lim_{n\to\infty}\frac{b-a}n \frac{e^{b+\frac bn}-e^{a+\frac an}}{e^\frac b n-e^\frac an}= \lim_{n\to\infty}\frac{b-a}n \frac{e^{b+\frac b\infty}-e^{a+\frac a\infty}}{e^\frac b n-e^\frac an}\to \lim_{n\to\infty}\frac{b-a}n \frac{e^b-e^a}{e^\frac b n-e^\frac an}=\frac{b-a}{b-a} \frac{e^b-e^a}{e^\frac b n-e^\frac an}=e^b-e^a $$

Note the product rule for the limit and therefore:

$$\int_a^b e^x dx=e^b-e^a\implies \int e^x dx=e^x+c$$

Please correct me and give me feedback!