Question

The area bounded by $y=x-1$, $1\le x\le 2$, $y=0$ (in sq.units) is

• A. 2
• B. 1
• C. $\frac{1}{2}$
• D. 4
• E. $\frac{1}{4}$

Hint:

This area is a triangle with base 1 ($2-1$) and height 1 ($f(2)$). Area $= 1/2 \times 1 \times 1 = 1/2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the area of the region enclosed by a straight line, the x-axis (y=0), and two vertical lines. This area can be calculated using a definite integral.
Step 2: Key Formula or Approach:
The area (A) under a curve \( y=f(x) \) from \( x=a \) to \( x=b \), bounded below by the x-axis, is given by the definite integral:
\[ A = \int_a^b f(x) dx \] This formula is valid if \( f(x) \geq 0 \) on the interval \( [a, b] \). We must first check this condition.
Step 3: Detailed Explanation:
The given boundaries are:
- The function: \( f(x) = x - 1 \)
- The interval: \( [a, b] = [1, 2] \)
- The lower bound: The x-axis, \( y=0 \)
First, let's check if \( f(x) \geq 0 \) for \( x \in [1, 2] \).
For \( x=1 \), \( y=1-1=0 \).
For any \( x>1 \), \( x-1>0 \).
So, the function is non-negative on the interval \( [1, 2] \).
Now we can set up the definite integral for the area:
\[ A = \int_1^2 (x-1) dx \] Find the antiderivative of \( (x-1) \):
\[ \int (x-1) dx = \frac{x^2}{2} - x \] Now, evaluate the definite integral using the Fundamental Theorem of Calculus:
\[ A = \left[ \frac{x^2}{2} - x \right]_1^2 \] \[ A = \left( \frac{2^2}{2} - 2 \right) - \left( \frac{1^2}{2} - 1 \right) \] \[ A = \left( \frac{4}{2} - 2 \right) - \left( \frac{1}{2} - 1 \right) \] \[ A = (2 - 2) - \left( -\frac{1}{2} \right) \] \[ A = 0 - \left( -\frac{1}{2} \right) = \frac{1}{2} \] Step 4: Final Answer:
The area is \( \frac{1}{2} \) square units.