Q.The area of the region bounded by the curve $y = x + 1$ and the lines $x = 2$ and $x = 3$ is
(A) $\frac{7}{2}$ sq units
(B) $\frac{9}{2}$ sq units
(C) $\frac{11}{2}$ sq units
(D) $\frac{13}{2}$ sq units

The region is a trapezoid under the line $y = x+1$ from $x=2$ to $x=3$. The area is the definite integral of the curve between these limits.

Step 1: Write the area as

Step 2: Integrate term by term:

Step 3: Evaluate from $2$ to $3$: …

The problem asks for the area bounded by a straight line and two vertical lines. This is a classic application of the Area Under the Curve (AUC) concept. When a curve $y = f(x)$ lies above the $x$-axis over an interval $[a, b]$, the area between the curve, the $x$-axis, and the lines $x = a$, $x = b$ is given by the definite integral $\int_a^b f(x) \, dx$.

Here, $y = x + 1$ is a straight line with slope 1 and intercept 1. Over $x = 2$ to $x = 3$, the function is positive (since $x+1 > 0$ for all $x \geq 2$). So the region is simply a trapezoid (or a rectangle plus a triangle) under the line. The integral will give us the exact area.

Let’s work through it step by step.

1. Set up the integral. The area $A$ is the definite integral of $y = x + 1$ from $x = 2$ to $x = 3$:

2. Find the antiderivative. The antiderivative of $x$ is $\frac{x^2}{2}$, and the antiderivative of $1$ is $x$. So:

3. Evaluate the definite integral using the Fundamental Theorem of Calculus. Plug in the upper limit $x = 3$ and the lower limit $x = 2$, then subtract:

4. Simplify each term. At $x = 3$: $\frac{9}{2} + 3 = \frac{9}{2} + \frac{6}{2} = \frac{15}{2}$ At $x = 2$: $\frac{4}{2} + 2 = 2 + 2 = 4$ So: …

Method: Area under a straight line between two vertical ordinates

Use this when the boundary is a line $y = mx + c$ that stays above the $x$-axis between $x = a$ and $x = b$.

Steps

Step 1: Check the line is positive across $[a,b]$.

If $y > 0$ throughout, the integral gives the area directly. (If it crosses the axis inside the interval, you would instead split at the crossing — not needed here.)

Step 2: Set up and evaluate the definite integral.

$\text{Area} = \int_a^b (mx + c)\, dx = \left[\frac{mx^2}{2} + cx\right]_a^b$ …

Common Mistakes

Mistake 1: Subtracting the limits in the wrong order.

Why it's wrong: computing $F(2) - F(3)$ instead of $F(3) - F(2)$ flips the sign and gives $-\tfrac72$. Correct approach: definite integrals are always upper minus lower, $F(b) - F(a)$.

Mistake 2: Integrating $x$ instead of the full $x+1$. …