Q.Explain why the area of a region lying below the x-axis is found by taking the absolute value of the definite integral, and illustrate with the region bounded by $y=x-3$, the x-axis, and the ordinates $x=0$ and $x=2$.

A definite integral measures NET SIGNED area, counting area below the x-axis as negative; since a geometric area cannot be negative, its absolute value is taken.

The reason. A definite integral is built from thin vertical strips of signed height $y=f(x)$ and width $dx$. When the curve lies below the x-axis, $f(x)$ is negative, so each strip contributes a NEGATIVE amount and the integral totals to a negative number. This negative number is the net signed area — it correctly records that the region sits below the axis, but a real geometric area (a physical size) can never be negative. Taking the absolute value strips off the sign and leaves the true size of the region.

Illustration. For $0\le x\le 2$, the line $y=x-3$ ranges from $-3$ (at $x=0$) up to $-1$ (at $x=2$) — negative throughout, so the whole region is below the x-axis.

$\int_{0}^{2}(x-3)\,dx = \left[\frac{x^{2}}{2}-3x\right]_{0}^{2} = \left(\frac{4}{2}-6\right)-(0) = (2-6) = -4$ …

Dual-check by geometry. The region is a trapezium below the axis with vertical sides of length $3$ (at $x=0$) and $1$ (at $x=2$) and width $2$: its area is $\tfrac{1}{2}(3+1)(2)=\t …

Reporting the area as $-4$. The negative sign only tells you the region is below the axis; the area itself is $4$. A related error is failing to check the sign of $f(x)$ on …