Q.Sketch the graph $y = |x + 3|$ and evaluate $\int_{-6}^{0} |x + 3|\, dx$.

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. …