Question

An ant is observed crawling on a sheet of paper along a straight line given by equation \( y = 2x - 4 \). Area of the surface covered by the ant bounded by y-axis, x-axis and \( x = 1 \) is :

• A. \( 1 \) sq. unit
• B. \( 3 \) sq. units
• C. \( 2 \) sq. units
• D. \( 4 \) sq. units

Hint:

Sketching the region helps identify if it's a simple geometric shape. This region is a trapezoid with height 1 and parallel sides of lengths 4 and 2.
Area of Trapezoid = \( \frac{1}{2}(4 + 2) \times 1 = 3 \).

Answer 1

The Correct Option is B

Concept:
The area bounded by a curve \(y=f(x)\), the x-axis, and the lines \(x=a\) and \(x=b\) is: \[ \text{Area}=\int_a^b |f(x)|\,dx \] If the curve lies below the x-axis, we take the negative of the function while calculating the area. 
Step 1: Determine the boundaries
The region is bounded by: 
• The line \(y=2x-4\) 
• The y-axis \(x=0\) 
• The x-axis \(y=0\) 
• The line \(x=1\) For \(x=0\): \[ y=2(0)-4=-4 \] For \(x=1\): \[ y=2(1)-4=-2 \] Thus, the line lies below the x-axis in the interval \([0,1]\). 
Step 2: Set up the definite integral
Since the line lies below the x-axis: \[ A=\int_0^1 -(2x-4)\,dx \] Therefore, \[ A=\int_0^1 (4-2x)\,dx \] 
Step 3: Evaluate the integral
\[ A=\left[4x-x^2\right]_0^1 \] \[ A=(4(1)-1^2)-(4(0)-0^2) \] \[ A=4-1 \] \[ A=3 \] 
Final Answer:
Therefore, the area of the bounded region is: \[ \boxed{3\text{ sq. units}} \] Hence, the correct answer is Option (B).