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:

Areas represent real physical regions and can never be negative. If your integral yields a negative value, take its absolute value.

Answer 1

The Correct Option is B

Concept: The area bounded by a curve \( y = f(x) \), the x-axis, and vertical lines \( x = a \), \( x = b \) is evaluated using definite integration: \( \text{Area} = \int_{a}^{b} |y| \, dx \). If the curve falls below the x-axis within the region, the integral will yield a negative value, so we take its absolute magnitude.

Step 1: Identify bounds and find critical intercepts. The boundaries are the y-axis (\( x = 0 \)), the vertical line \( x = 1 \), and the x-axis (\( y = 0 \)). Let's check where the line \( y = 2x - 4 \) intersects the x-axis by setting \( y = 0 \): \[ 2x - 4 = 0 \implies 2x = 4 \implies x = 2 \] Since our region lies strictly within the interval \( x \in [0, 1] \), the line does not cross the x-axis inside the integration window. For any value in \( [0, 1] \), \( y = 2x - 4 \) is negative.

Step 2: Set up the area definite integral. \[ \text{Area} = \int_{0}^{1} |2x - 4| \, dx = \int_{0}^{1} -(2x - 4) \, dx = \int_{0}^{1} (4 - 2x) \, dx \]

Step 3: Compute the integral value. \[ \text{Area} = \left[ 4x - \frac{2x^2}{2} \right]_{0}^{1} = \left[ 4x - x^2 \right]_{0}^{1} \] Substitute the upper boundary \( x = 1 \): \[ \text{Area}_{\text{upper}} = 4(1) - (1)^2 = 4 - 1 = 3 \] Substitute the lower boundary \( x = 0 \): \[ \text{Area}_{\text{lower}} = 4(0) - (0)^2 = 0 \] Subtract to find total area: \[ \text{Area} = 3 - 0 = 3 \text{ sq. units} \]