Q.Find the area bounded by the curve $y=x^2$, the x-axis, and the lines $x=1$ and $x=3$.

The area under a curve $y = f(x)$, bounded by the X-axis and two vertical lines $x = a$ and $x = b$, is found by slicing the region into a very large number of extremely thin vertical strips, each of width $dx$ and height $y = f(x)$, and adding up their areas. This sum, in the limit as the strips become infinitesimally thin, is exactly the definite integral $\int_a^b f(x)\,dx$ -- so the same tool built in the previous chapter to evaluate integrals now directly measures plane area.

If the curve lies on or above the X-axis throughout $[a, b]$, the integral itself gives the area. But an area can never be negative, so if the curve dips below the X-axis, the raw integral comes out negative and its absolute value must be taken to get the true area. If the curve changes sign somewhere inside the interval -- crossing the X-axis at an interior point -- the interval has to be split at that crossing point, the area of each piece found and made positive on its own, and the two positive areas added; a single integral taken across the whole interval would let the positive and negative parts cancel and understate the actual area.

The same idea works the other way round: when a curve is more naturally written as $x = g(y)$, the area between the curve, the Y-axis, and two horizontal lines $y = c$ and $y = d$ is found by summing horizontal strips instead, giving $\int_c^d x\,dy$. Deciding whether to integrate with respect to $x$ or $y$ is simply a matter of which form keeps the calculation simplest for the particular curve in the problem. This idea also underlies standard results such as the area of a circle or an ellipse, both of which can be derived by integrating one quadrant and using the curve's symmetry, and the area of a sector cut off from a circle by a straight line.