Q.Evaluate as a limit of sums: $\int_{0}^{2} e^{x}\,dx$

The key idea is to interpret the definite integral as the limit of a Riemann sum, then evaluate that sum using the formula for a geometric series.

Step 1: Set up the Riemann sum.

Divide $[0,2]$ into $n$ equal subintervals, each of width $\Delta x = \frac{2}{n}$. Using right endpoints, $x_i^* = \frac{2i}{n}$ for $i=1,2,\dots,n$. The Riemann sum is:

Step 2: Sum the geometric series.

Factor out $\frac{2}{n}$; the sum $\sum_{i=1}^{n} (e^{2/n})^i$ is a geometric series with first term $e^{2/n}$ and ratio $e^{2/n}$: …

The core idea here is that a definite integral is defined as the limit of a sum of rectangles. For $\int_{0}^{2} e^{x}\,dx$, we are finding the area under $e^{x}$ from $x=0$ to $x=2$. Instead of using the Fundamental Theorem of Calculus directly (which would give $e^{2} - e^{0} = e^{2} - 1$), we are asked to evaluate it as a limit of sums — this means we reconstruct the Riemann sum and take the limit as the number of rectangles goes to infinity.

Why does this work? Because $e^{x}$ is a smooth, continuous function, the Riemann sum converges to the exact area. The trick is to set up the sum so that it becomes a geometric series, which we can sum in closed form.

Let’s walk through it step by step.

1. Partition the interval We divide $[0, 2]$ into $n$ equal subintervals. Each subinterval has width:

The partition points are $x_i = 0 + i \cdot \Delta x = \frac{2i}{n}$ for $i = 0, 1, 2, \dots, n$.

2. Choose sample points

   For the Riemann sum, we need a point in each subinterval. A common choice (and the one that leads to a clean sum here) is the right endpoint. So for the $i$-th subinterval $[x_{i-1}, x_i]$, we take $x_i^* = x_i = \frac{2i}{n}$.

3. Write the Riemann sum

   The function value at the right endpoint is $e^{x_i^*} = e^{2i/n}$. The area of the $i$-th rectangle is $f(x_i^*) \Delta x = e^{2i/n} \cdot \frac{2}{n}$. Summing over all $n$ rectangles gives:

4. Factor out constants The factor $\frac{2}{n}$ is common to every term, so:

5. Recognize the geometric series The sum $\sum_{i=1}^{n} e^{2i/n}$ is a geometric series with first term $a = e^{2/n}$ and common ratio $r = e^{2/n}$. Why? Because $e^{2i/n} = (e^{2/n})^i$. So:

Note that $(e^{2/n})^n = e^{2}$. So:

6. Substitute back into $S_n$

7. Take the limit as $n \to \infty$ …

Method: Limit of sums for an exponential integrand

Use this for $\int_a^b e^{x}\,dx$ "as a limit of sums", where the Riemann sum becomes a geometric series that you sum in closed form before taking the limit.

Steps

Step 1: Partition and write the sum.

With $h=\dfrac{b-a}{n}$, $x_i=a+ih$:

Step 2: Sum the geometric series.

Step 3: Insert and simplify. …

Common Mistakes

Mistake 1: Not recognising the geometric series.

Why it's wrong: $\sum e^{ih}=\sum(e^h)^i$ is geometric with ratio $e^h$; treating it as a generic sum stalls. Correct approach: use $\sum_{i=0}^{n-1}r^i=\tfrac{r^n-1}{r-1}$ with $r=e^h$.

Mistake 2: Mishandling the limit $\tfrac{h}{e^h-1}$.

Why it's wrong: the key limit is $\tfrac{e^h-1}{h}\to1$ as $h\to0$; skipping it leaves an unresolved $\tfrac00$ form. Correct approach: factor out $(e^b-e^a)$ and apply the standard limit. …