Question

The area of the region bounded by the curve \(y = x\) and the x-axis, between the boundaries \(x = 0\) and \(x = 2\) is:

• A. 2 sq. units
• B. \(\frac{1}{2}\) sq. unit
• C. 1 sq. unit
• D. 4 sq. units

Hint:

When finding the area under simple linear graphs like \(y = x\) or \(y = mx + c\), verifying your integration result using basic geometric formulas for triangles or trapezoids is a quick way to catch arithmetic slips.

Answer 1

The Correct Option is A

Concept: The definite integral can be used to compute the total area under a curve \(y = f(x)\) bounded by the vertical lines \(x = a\) and \(x = b\) and the horizontal x-axis line (\(y=0\)). The formula is: \[ \text{Area} = \int_{a}^{b} |f(x)| \, dx \] For the interval \(x \in [0, 2]\), the function \(y = x\) remains non-negative throughout, allowing us to drop the absolute value bars.

Step 1: Set up the area integration formula
Given values: \[ f(x) = x, \quad a = 0, \quad b = 2 \] Setting up the definite integral expression: \[ \text{Area} = \int_{0}^{2} x \, dx \]

Step 2: Compute the antiderivative and apply limits
Using the power rule of integration, \(\int x^n dx = \frac{x^{n+1}}{n+1}\): \[ \int x \, dx = \frac{x^2}{2} \] Applying the lower limit $0$ and upper limit $2$: \[ \text{Area} = \left[ \frac{x^2}{2} \right]_{0}^{2} \] \[ \text{Area} = \left( \frac{2^2}{2} \right) - \left( \frac{0^2}{2} \right) \] \[ \text{Area} = \left( \frac{4}{2} \right) - 0 = 2 \text{ square units} \]

Step 3: Geometric validation check
The region forms a right-angled triangle with a base along the x-axis from \(x=0\) to \(x=2\) (length $= 2$ units) and a height at \(x=2\) where \(y=2\) (height $= 2$ units). \[ \text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2 \text{ sq. units} \] Both analytical pathways match exactly, confirming option (A).