Q.Father of a family wishes to divide his square field bounded by $x=0$, $x=4$, $y=4$ and $y=0$ along the curve $y^2=4x$ and $x^2=4y$ into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.

Building on the geometric meaning of $\int_a^b f(x)\,dx$ (Remarks under the limit-of-a-sum definition), this topic packages the vertical/horizontal-strip argument into ready-to-use area formulas.

Bounded by a curve, the $x$-axis, and $x=a,\,x=b$ (vertical strips, height $=|y|$, width $\Delta x$):

• If $y=f(x)\ge0$ throughout $[a,b]$ (curve above the $x$-axis): $A=\displaystyle\int_a^b y\,dx$.
• If $y=f(x)\le0$ throughout $[a,b]$ (curve below the $x$-axis): $A=\displaystyle-\int_a^b y\,dx=\left|\int_a^b y\,dx\right|$.
• If $f$ changes sign on $[a,b]$: split at the zeros $c_1,c_2,\ldots$ into subintervals of constant sign, apply the two rules above on each piece, and add the absolute values — $A=\left|\int_a^{c_1}f\right|+\left|\int_{c_1}^{c_2}f\right|+\cdots$ (never just $\int_a^b f\,dx$, which can cancel positive against negative area).

Bounded by a curve, the $y$-axis, and $y=c,\,y=d$ (horizontal strips, mirror image): $A=\displaystyle\int_c^d x\,dy$ if the curve lies to the right of the $y$-axis ($x\ge0$); $A=-\displaystyle\int_c^d x\,dy$ if to the left; split-and-add-absolute-values if it crosses.

Bounded between two curves. If $f(x)\ge g(x)$ on $[a,b]$ (an "upper" curve $U$ and "lower" curve $L$), the region between them and the ordinates $x=a,x=b$ has area

The $y$-axis mirror (curves $x=f(y)\ge x=g(y)$, a "right" curve $R$ and "left" curve $L$) gives $A=\int_c^d(x_R-x_L)\,dy$. …