Q.Using integration, find the area of the region bounded by the line $2y = 5x + 7$, x-axis and the lines $x = 2$ and $x = 8$.

Concept: Area under a curve — the definite integral of the function between the given vertical lines.

Step 1: Rewrite the line in slope-intercept form:

Step 2: The required area is bounded by this line, the $x$-axis, and $x = 2$, $x = 8$. Since the line is above the $x$-axis for all $x$ in $[2, 8]$, the area is:

Step 3: Integrate:

Step 4: Evaluate: …

Why Integration Works Here

The problem asks for the area bounded by a straight line, the x-axis, and two vertical lines. This is a classic application of definite integration: the area between a curve $y = f(x)$ and the x-axis from $x = a$ to $x = b$ is $\int_a^b f(x) \, dx$, provided $f(x) \geq 0$ on that interval.

Here, the "curve" is a straight line — but integration still works perfectly. In fact, integrating a linear function over an interval gives the area of a trapezoid (or a triangle if one endpoint is at the x-axis). The power of integration is that it handles any continuous curve, including straight lines, without needing separate geometry formulas.

Let's check: the line is $2y = 5x + 7$, so $y = \frac{5x+7}{2}$. For $x$ between 2 and 8, is this always above the x-axis? At $x=2$, $y = \frac{5(2)+7}{2} = \frac{17}{2} = 8.5 > 0$. At $x=8$, $y = \frac{5(8)+7}{2} = \frac{47}{2} = 23.5 > 0$. Since the line is increasing, it stays positive throughout. So we can integrate directly.

Step-by-Step Solution

1. Express the line in the form $y = f(x)$.

The given equation is $2y = 5x + 7$. Divide both sides by 2:

This is the function we'll integrate.

2. Set up the definite integral.

The region is bounded by:

• The line $y = \frac{5x+7}{2}$ (top boundary)
• The x-axis, i.e., $y = 0$ (bottom boundary)
• The vertical lines $x = 2$ (left boundary) and $x = 8$ (right boundary)

So the area $A$ is:

3. Simplify the integrand.

Factor out the constant $\frac{1}{2}$:

4. Integrate term by term.

The antiderivative of $5x$ is $\frac{5x^2}{2}$, and the antiderivative of $7$ is $7x$. So: …

Method: Area under a straight line between two vertical lines

This technique finds the area trapped between a line, the $x$-axis, and two vertical boundaries $x = a$ and $x = b$ — a trapezium that integration handles automatically.

Steps

Step 1: Solve the line for $y$.

A line given as $px + qy = r$ is not ready to integrate. Rearrange it into the form $y = f(x)$ first, so the integrand is an explicit function of $x$.

Step 2: Check the line lies above the $x$-axis on $[a, b]$.

Evaluate $y$ at both endpoints. If both are positive (and the line is monotonic), the whole strip sits above the axis and the plain integral gives the geometric area.

Step 3: Integrate the linear function.

$A = \int_a^b f(x)\, dx,$ …

Common Mistakes

Mistake 1: Integrating $2y = 5x + 7$ without solving for $y$ first

Why it's wrong: the height of a vertical strip is $y = \dfrac{5x+7}{2}$, so integrating $(5x+7)$ directly ignores the division by $2$ and doubles the area to $192$. Correct approach: rewrite as $y = \dfrac{5x+7}{2}$ before integrating, giving $96$.

Mistake 2: Not checking the line stays above the $x$-axis on $[2,8]$ …