Q.The area of the region bounded by the curve $x = 2y + 3$ and the lines $y = 1$ and $y = -1$ is
(A) $4$ sq units
(B) $\frac{3}{2}$ sq units
(C) $6$ sq units
(D) $8$ sq units

Idea: The boundary is given as $x=2y+3$, so integrate with respect to $y$ between the horizontal lines $y=-1$ and $y=1$ (the region between this line and the $y$-axis). On this interval $x=2y+3$ stays positive. …

Read the boundaries

The curve is written as $x=2y+3$ — $x$ in terms of $y$ — and the limits given are horizontal lines $y=-1$ and $y=1$. That is the signal to integrate with respect to $y$, measuring the strip out to the line from the $y$-axis. On $-1\le y\le 1$ the value $x=2y+3$ ranges from $1$ to $5$, always positive, so no absolute value is needed.

Set up the integral

Evaluate …

Method: Area bounded by $x = f(y)$ between two horizontal lines

Use this when the boundary is given as $x$ in terms of $y$ and the limits are horizontal lines $y = c$ and $y = d$ — the signal to integrate with respect to $y$.

Steps

Step 1: Recognise the natural variable.

When the curve is written as $x = f(y)$ and the bounds are $y$-values, integrate in $y$ — the horizontal strip runs from the $y$-axis out to the curve, of width $x = f(y)$.

Step 2: Check the sign of $f(y)$ on $[c,d]$. …

Common Mistakes

Mistake 1: Trying to integrate with respect to $x$.

Why it's wrong: the boundary is given as $x = 2y+3$ with horizontal limits $y = \pm 1$; converting to $y = \tfrac{x-3}{2}$ and integrating in $x$ needlessly complicates a clean horizontal strip and invites limit errors. Correct approach: integrate in $y$ directly, $\int_{-1}^{1}(2y+3)\,dy$.

Mistake 2: Mishandling the negative lower limit. …