Q.Find the area of the region bounded by $2x-y+1=0$, $y=-1$, $y=3$ and $y$-axis.

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 $A=\int_c^d(x_R-x_L)\,dy$.

General working rule (no need to identify upper/lower by name): draw an arbitrary vertical line cutting the region; call the $y$-value where it enters the region $y_{\text{ENTRY}}$ and where it exits $y_{\text{EXIT}}$ (both read off the bounding curves' equations); then

(and the horizontal-strip mirror $A=\int_c^d[x_{\text{EXIT}}-x_{\text{ENTRY}}]\,dy$). This is the most robust approach when a region's boundary switches which curve is "on top" partway through, or when it is more natural to sketch and read off entry/exit points than to name $f,g$ globally.