How do you prove integration gives the area under a curve?

The idea is approximating the area under a curve using very small rectangles. The more rectangles we use and the narrower they become, the more accurate our approximation becomes. This concept is known as a Riemann sum.

To compute the area under a curve, we first need to establish an interval over which we want to calculate the area, say [a,b]. We then divide this interval into smaller subintervals by placing vertical lines at specific points within the interval. For each subinterval, we draw a rectangle with width being the width of the subinterval, and height being determined by the function's value at a specific point within the subinterval. The sum of the areas of these rectangles gives us an approximation of the area under the curve.

As we make the subintervals narrower and narrower, and the number of rectangles approaches infinity, the approximation becomes more accurate. In mathematical terms, this is expressed by taking the limit of the Riemann sum as the width of the rectangles approaches zero. So, basically we calculate the limit of a sum (sigma)

For example, consider a continuous and non-negative function f(x) defined on an interval [a, b]. The graph of this function lies above the x-axis within this interval. The area under the curve of this function and between the vertical lines x = a and x = b can be approximated by dividing the interval [a, b] into small subintervals, each of width Δx. The area under the curve within each subinterval can be approximated using rectangles, where the height of each rectangle is the value of the function f(x) at some point within that subinterval. As we make the subintervals narrower and more numerous, the approximation gets better (many many rectangles, with very narrow width). The total area under the curve can be approximated as the sum of the areas of these rectangles:

Approximate Total Area=$\sum_{i=1}^{n} f(x_{i} )\triangle x$

where $x_{i}$ is a point within the i-th subinterval.

As you make the subintervals infinitely small (Δx → 0), the sum approaches the integral:

Actual Total Area=$\lim_{\triangle x\rightarrow 0} \sum_{i=1}^{n} f(x_{i} )\triangle x$ =$\int_{a}^{b}f(x)\;\mathrm{d}x$

This integral represents the exact area under the curve of the function f(x) from x=a to x=b.