Area Under a Curve

In the previous chapter, a definite integral $\displaystyle\int_{a}^{b} f(x)\,dx$ was evaluated as a single number using the Fundamental Theorem of Integral Calculus, $\displaystyle\int_{a}^{b} f(x)\,dx = F(b)-F(a)$. This Std XII Commerce Mathematics and Statistics chapter gives that number its most important geometric meaning: it measures area. The same definite-integration principles used here are the standard treatment of area found in commerce- and general-mathematics curricula across India.

The area formula

Let $y=f(x)$ be a continuous function that is non-negative (its curve lies on or above the x-axis) on the interval $a \le x \le b$. Then the area $A$ of the region bounded by:

• the curve $y=f(x)$ (above),
• the x-axis (below),
• the ordinate (vertical line) $x=a$ (on the left), and
• the ordinate $x=b$ (on the right)

is given by

Why the integral gives the area — the idea of accumulation

Imagine slicing the shaded region into a large number of very thin vertical strips, each of width $dx$. A single strip standing at position $x$ has height $y=f(x)$ and width $dx$, so its area is approximately $y\,dx$ (the area of a thin rectangle). Adding up the areas of all such strips from $x=a$ to $x=b$ — which is exactly what the integral sign $\displaystyle\int_{a}^{b}$ does — gives the total area of the region. The integral is a limit of this sum as the strips become infinitely thin, so the approximation becomes exact.

The three-step method

1. Set up the integral with the correct function and limits: $A=\displaystyle\int_{a}^{b} f(x)\,dx$.
2. Integrate $f(x)$ to get an antiderivative $F(x)$ — no constant of integration is needed for a definite integral.
3. Substitute the limits and subtract: $A = F(b)-F(a)$. Since this is an area, the result must be reported as a positive quantity in square units.