Q.Find the area of the region bounded by the curve $y=x^{2}-4$, the x-axis, and the ordinates $x=0$ and $x=2$.

On $0\le x\le 2$, $x^{2}-4\le 0$, so the curve lies below the x-axis; the definite integral is negative and the area is its absolute value.

Step 1 — check the sign of the curve on the interval. For $0\le x\le 2$, $x^{2}\le 4$, so $y=x^{2}-4\le 0$: the whole region lies BELOW the x-axis. The integral will therefore be negative, and the area is its absolute value.

Step 2 — set up and integrate.

Step 3 — apply the limits.

$= \left(\frac{2^{3}}{3}-4\cdot2\right)-\left(0-0\right) = \left(\frac{8}{3}-8\right) = \frac{8-24}{3} = -\frac{16}{3}$ …

Dual-check by reflecting the curve. Reflecting a below-axis region across the x-axis turns $y=x^{2}-4$ into $y=4-x^{2}$, which is positive on $[0,2]$ and encloses the SAME area. Then $\int_{0}^{2}(4-x^{2}),dx=\left[4x-\tfrac{x^{3}}{3}\right]_{0}^{2}=\lef …

Reporting the raw integral $-\dfrac{16}{3}$ as the area. A geometric area is never negative; when the region lies below the x-axis …