Q.Evaluate as a limit of sums: $\int_{0}^{2} (x^2+3)\,dx$

The key idea is to interpret the definite integral as the limit of a Riemann sum, then evaluate that limit using standard summation formulas.

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, the sample points are $x_i = \frac{2i}{n}$ for $i=1,2,\dots,n$.

Step 2: Write the sum.

The Riemann sum is

Step 3: Simplify and take the limit.

$S_n = \frac{2}{n} \sum_{i=1}^{n} \left( \frac{4i^2}{n^2} + 3 \right) = \frac{8}{n^3} \sum_{i=1}^{n} i^2 + \frac{6}{n} \sum_{i=1}^{n} 1.$ …

The core idea here is that a definite integral is defined as the limit of a sum of areas of thin rectangles. For $\int_a^b f(x)\,dx$, we slice the interval $[a,b]$ into $n$ equal parts, each of width $\Delta x = \frac{b-a}{n}$. Then we pick a sample point $x_i^*$ in each subinterval (often the right endpoint) and form the sum $\sum_{i=1}^n f(x_i^*) \Delta x$. As $n \to \infty$, this sum approaches the exact area under the curve.

Why does this work? Because the integral measures the accumulated area, and the Riemann sum approximates it with rectangles. The limit removes the approximation error. For a polynomial like $x^2+3$, the sum can be evaluated exactly using summation formulas, and then the limit is straightforward.

Let’s apply this to $\int_{0}^{2} (x^2+3)\,dx$.

1. Set up the partition.

   Here $a=0$, $b=2$, so $\Delta x = \frac{2-0}{n} = \frac{2}{n}$.

   The right endpoint of the $i$-th subinterval is $x_i = a + i\Delta x = 0 + i\cdot\frac{2}{n} = \frac{2i}{n}$.

2. Write the Riemann sum.

   Using right endpoints, the sum is

Since $f(x) = x^2 + 3$, we have

3. Simplify the sum.

Use the standard formulas:

So

4. Simplify algebraically. First term inside: $\frac{4}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{4(n+1)(2n+1)}{6n} = \frac{2(n+1)(2n+1)}{3n}$. So

The second term simplifies: $\frac{2}{n} \cdot 3n = 6$.

The first term: $\frac{2}{n} \cdot \frac{2(n+1)(2n+1)}{3n} = \frac{4(n+1)(2n+1)}{3n^2}$.

Thus …

Method: Evaluating a definite integral as a limit of sums

Use this when a question explicitly asks for $\int_a^b f(x)\,dx$ "as a limit of sums" (first principles). You build Riemann sums and take the limit as the number of strips $\to\infty$.

Steps

Step 1: Set up the partition.

Divide $[a,b]$ into $n$ equal strips of width $h=\dfrac{b-a}{n}$; sample points $x_i=a+ih$.

Step 2: Write the Riemann sum.

Step 3: Substitute $f$ and use standard summation formulas.

Expand $f(a+ih)$ and apply $\sum i=\tfrac{n(n-1)}{2}$ and $\sum i^2=\tfrac{(n-1)n(2n-1)}{6}$, keeping $h=\tfrac{b-a}{n}$. …

Common Mistakes

Mistake 1: Using the antiderivative instead of a limit of sums.

Why it's wrong: the question demands the first-principles method; jumping to $[\tfrac{x^3}{3}+3x]_0^2$ ignores the instruction. Correct approach: build the Riemann sum and take $n\to\infty$ (you may check with the antiderivative).

Mistake 2: Wrong summation formulas.

Why it's wrong: errors in $\sum i$ or $\sum i^2$ propagate into the final value. Correct approach: use $\sum_{i=0}^{n-1}i=\tfrac{n(n-1)}{2}$ and $\sum_{i=0}^{n-1}i^2=\tfrac{(n-1)n(2n-1)}{6}$. …