# Definite integral as limit of a sum

### Definite Integral as Limit of a Sum:

1. Purpose of Definite Integral:

• Represents the accumulation of quantities or area under a curve.
• Denoted as ${\int }_{a}^{b}f\left(x\right)\text{\hspace{0.17em}}dx$ for a function $f\left(x\right)$ over the interval $\left[a,b\right]$.
2. Riemann Sum Approximation:

• Divides the interval $\left[a,b\right]$ into $n$ subintervals.
• Width of each subinterval: $\mathrm{\Delta }x=\frac{b-a}{n}$.
3. Summation over Subintervals:

• For a partition $P$ with $n$ subintervals: $S={\sum }_{i=1}^{n}f\left({c}_{i}\right)\cdot \mathrm{\Delta }x$ is a sample point within the $i$th subinterval.
4. Definite Integral as a Limit:

• The definite integral of $f\left(x\right)$ over $\left[a,b\right]$ is expressed as a limit of Riemann sums: ${\int }_{a}^{b}f\left(x\right)\text{\hspace{0.17em}}dx={\mathrm{lim}}_{n\to \mathrm{\infty }}{\sum }_{i=1}^{n}f\left({c}_{i}\right)\cdot \mathrm{\Delta }x$ as the number of subintervals approaches infinity ($n\to \mathrm{\infty }$).
5. Interpretation and Geometric Meaning:

• The definite integral represents the exact area under the curve $f\left(x\right)$ from $a$ to $b$ on the $x$-axis.
• As $n$ increases (subintervals become infinitesimally small), the Riemann sum approaches the precise area.
6. Types of Riemann Sums:

• Left Riemann Sum: Uses left endpoints of subintervals.
• Right Riemann Sum: Uses right endpoints of subintervals.
• Midpoint Riemann Sum: Uses midpoints of subintervals.
7. Properties and Applications:

• Crucial in calculus for solving problems related to areas, volumes, work done, and various physical quantities.
• Basis for understanding the Fundamental Theorem of Calculus and techniques like integration by substitution.
8. Integration Techniques:

• Helps in evaluating definite integrals of functions that might be challenging to integrate directly.

### Example:

Consider the function $f\left(x\right)=2x$ over the interval $\left[0,2\right]$. We want to find the definite integral of $f\left(x\right)$ from $0$ to $2$ using the limit of a sum.

1. Partition the Interval:

• Divide the interval $\left[0,2\right]$ into $n$ subintervals.
• Width of each subinterval: $\mathrm{\Delta }x=\frac{2-0}{n}=\frac{2}{n}$.
2. Riemann Sum:

• For the $i$th subinterval, choose ${c}_{i}$ as the right endpoint ${x}_{i}=0+i\cdot \mathrm{\Delta }x=\frac{2i}{n}$.
• The Riemann sum for the function $f\left(x\right)=2x$ is: $S={\sum }_{i=1}^{n}f\left({c}_{i}\right)\cdot \mathrm{\Delta }x={\sum }_{i=1}^{n}2\cdot \left(\frac{2i}{n}\right)\cdot \frac{2}{n}$
3. Definite Integral as Limit:

• Express the definite integral of $f\left(x\right)=2x$ from $0$ to $2$ as a limit of the Riemann sum: ${\int }_{0}^{2}2x\text{\hspace{0.17em}}dx={\mathrm{lim}}_{n\to \mathrm{\infty }}{\sum }_{i=1}^{n}2\cdot \left(\frac{2i}{n}\right)\cdot \frac{2}{n}$
4. Simplify the Sum:

• Simplify the sum to solve for the definite integral: $S={\mathrm{lim}}_{n\to \mathrm{\infty }}{\sum }_{i=1}^{n}\frac{8i}{{n}^{2}}={\mathrm{lim}}_{n\to \mathrm{\infty }}\frac{8}{{n}^{2}}\cdot \frac{n\left(n+1\right)}{2}$ $S={\mathrm{lim}}_{n\to \mathrm{\infty }}\frac{4\left(n+1\right)}{n}={\mathrm{lim}}_{n\to \mathrm{\infty }}\left(4+\frac{4}{n}\right)=4$
5. Result:

• Therefore, by using the definite integral as the limit of a sum, we found that: ${\int }_{0}^{2}2x\text{\hspace{0.17em}}dx=4$