# Proof of Definite Integral

Let’s assume that $f$ is continuous and positive on the interval $[a, b]$. Then the definite integral $\int^b_a f(x) dx$ represents the area of the region bounded by the graph of $f$ and the x-axis, from x = a to x = b. First, we partition the interval $[a, b]$ into n subintervals, each of width $\Delta x = (b - a)/n$ such that $a = x_0 < x_1 < x_2 < . . . < x_n = b$ Then we can form a trapezoid for each subinterval and the area of the ith trapezoid = $[\frac{f(x_{i-1}) + f(x_i)}{2}](\frac{b-a}{n})$. This implies that the sum of the areas of the n trapezoids is Area = $\frac{b - a}{2n}[f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)] = \frac{b - a}{2n}[f(x_0) + f(x_n) + 2\displaystyle\sum\limits_{i=1}^{n-1} f(x_i)] = \frac{b - a}{2n}(f(x_0) + f(x_n)) + \displaystyle\sum\limits_{i=1}^{n-1} f(x_i)(\frac{b - a}{n}) = \frac{b - a}{2n}(f(x_0) + f(x_n) - 2f(x_n)) + \displaystyle\sum\limits_{i=1}^{n} f(x_i)(\frac{b - a}{n}) - 2f(x_n)(\frac{b - a}{2n}) = \frac{b - a}{2n}(f(x_0) - f(x_n)) + \displaystyle\sum\limits_{i=1}^{n} f(x_i)\Delta x = \lim_{n\to\infty}\frac{b - a}{2n}(f(x_0) - f(x_n)) + \lim_{n\to\infty}\displaystyle\sum\limits_{i=1}^{n} f(x_i)\Delta x = 0 + \lim_{n\to\infty}\displaystyle\sum\limits_{i=1}^{n}f(x_i)\Delta x = \int^b_a f(x) dx$