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

The key idea is that the area between a curve and the x-axis is the integral of $|y|$ between the given limits, because area below the axis counts as positive.

Step 1: Find where the line crosses the x-axis.

Set $y = 0$: $3x + 2 = 0 \Rightarrow x = -\frac{2}{3}$. This lies between $x = -1$ and $x = 1$, so the region has parts above and below the axis.

Step 2: Split the integral at $x = -\frac{2}{3}$.

For $x \in [-1, -\frac{2}{3}]$, $y \le 0$, so area = $-\int_{-1}^{-2/3} (3x+2)\,dx$.

For $x \in [-\frac{2}{3}, 1]$, $y \ge 0$, so area = $\int_{-2/3}^{1} (3x+2)\,dx$.

Step 3: Evaluate.

First integral: $-\left[ \frac{3x^2}{2} + 2x \right]_{-1}^{-2/3} = -\left( \left(\frac{3\cdot\frac{4}{9}}{2} - \frac{4}{3}\right) - \left(\frac{3}{2} - 2\right) \right) = -\left( \left(\frac{2}{3} - \frac{4}{3}\right) - \left(-\frac{1}{2}\right) \right) = -\left( -\frac{2}{3} + \frac{1}{2} \right) = \frac{1}{6}$.

Second integral: $\left[ \frac{3x^2}{2} + 2x \right]_{-2/3}^{1} = \left(\frac{3}{2} + 2\right) - \left(\frac{2}{3} - \frac{4}{3}\right) = \frac{7}{2} - \left(-\frac{2}{3}\right) = \frac{7}{2} + \frac{2}{3} = \frac{25}{6}$.

Step 4: Total area = $\frac{1}{6} + \frac{25}{6} = \frac{26}{6} = \frac{13}{3}$.

We are finding the area between the curve $y = 3x + 2$, the x‑axis, and the vertical lines $x = -1$ and $x = 1$. The key point: area is always positive. If the curve dips below the x‑axis, the definite integral gives a negative value for that portion, so we must split the region and take absolute values.

The line $y = 3x + 2$ crosses the x‑axis where $3x + 2 = 0$, i.e. at $x = -\frac{2}{3}$. Between $x = -1$ and $x = -\frac{2}{3}$, the line is below the axis; between $x = -\frac{2}{3}$ and $x = 1$, it is above. So the total area is the sum of the absolute areas of these two parts.

1. Find the x‑intercept Set $y = 0$:

This is the point where the sign of $y$ changes.

2. Area below the axis (from $x = -1$ to $x = -\frac{2}{3}$) Here $y$ is negative, so the definite integral gives a negative number. The area is the absolute value:

Compute the integral:

Evaluate from $-1$ to $-\frac{2}{3}$:

Simplify term by term:

$\left(-\frac{2}{3}\right)^2 = \frac{4}{9}$, so $\frac{3 \cdot \frac{4}{9}}{2} = \frac{4/3}{2} = \frac{2}{3}$.

Then $2\left(-\frac{2}{3}\right) = -\frac{4}{3}$.

So the upper limit value is $\frac{2}{3} - \frac{4}{3} = -\frac{2}{3}$.

Lower limit: $\frac{3(1)}{2} = \frac{3}{2}$, and $2(-1) = -2$, so $\frac{3}{2} - 2 = -\frac{1}{2}$.

Hence the integral equals:

The area is the absolute value: $\frac{1}{6}$.

3. Area above the axis (from $x = -\frac{2}{3}$ to $x = 1$) Here $y$ is positive, so the integral directly gives the area:

Using the same antiderivative:

Upper limit: $\frac{3}{2} + 2 = \frac{3}{2} + \frac{4}{2} = \frac{7}{2}$.

Lower limit (we already computed this as $-\frac{2}{3}$ above).

So the integral is:

4. Total area Add the two parts:

Method: Area of a line that crosses the x-axis inside the interval

Use this for the area bounded by a line $y = mx + c$, the $x$-axis, and two ordinates, when the line crosses the axis between the limits so part of the region is below the axis.

Steps

Step 1: Find the x-intercept.

Solve $mx + c = 0$ to get $x = -\dfrac{c}{m}$. If this lies inside $[a,b]$, the region has both a below-axis and an above-axis part.

Step 2: Split at the intercept and take magnitudes.

where $x_0 = -c/m$. On the below-axis piece the integral is negative, so take its absolute value.

Step 3: Evaluate each piece and add the positive amounts.

Integrating straight from $a$ to $b$ without splitting lets the negative (below-axis) part cancel the positive part, understating the true area — always test whether the line changes sign first.

Common Mistakes

Mistake 1: Integrating straight from $-1$ to $1$ without splitting.

Why it's wrong: the line $y=3x+2$ is below the axis on $[-1,-\tfrac23]$, so $\int_{-1}^{1}(3x+2)\,dx = 4$ lets the negative part cancel and understates the true area. Correct approach: split at the intercept $x=-\tfrac23$ and add magnitudes, $\tfrac16 + \tfrac{25}{6} = \tfrac{13}{3}$.

Mistake 2: Missing the x-intercept inside the interval.

Why it's wrong: not solving $3x+2=0$ hides that the curve changes sign at $x=-\tfrac23$. Correct approach: always find where the line meets the axis and check whether it lies between the limits before integrating.