Question

Find the area bounded by the curve \(y=|2-x|\), the \(x\)-axis, and the lines \(x=0\) and \(x=5\).

• A. \(12.5\) sq units
• B. \(4.5\) sq units
• C. \(6.5\) sq units
• D. \(8.5\) sq units

Hint:

For modulus functions: \[ |f(x)| = \begin{cases} f(x), & f(x)\ge0 -f(x), & f(x)<0 \end{cases} \] Always split the interval at points where the expression inside modulus becomes zero.

Answer 1

The Correct Option is C

Concept: For modulus functions, the expression inside the modulus changes sign at certain points. Therefore, we split the integral into intervals according to the sign of the expression. The area under a curve above the \(x\)-axis is given by: \[ \text{Area}=\int_a^b y\,dx \]

Step 1: Identify where the modulus changes sign. We are given: \[ y=|2-x| \] The quantity inside modulus becomes zero at: \[ 2-x=0 \] \[ x=2 \] Thus, we split the interval into: \[ 0\le x<2 \] and \[ 2\le x\le5 \]

Step 2: Write the function without modulus. For \(x<2\), \[ 2-x>0 \] Hence, \[ |2-x|=2-x \] For \(x>2\), \[ 2-x<0 \] Thus, \[ |2-x|=-(2-x)=x-2 \] Therefore, \[ y= \begin{cases} 2-x, & 0\le x\le2 \\ x-2, & 2\le x\le5 \end{cases} \]

Step 3: Set up the required integral. Hence total area is: \[ \text{Area} = \int_0^2 (2-x)\,dx + \int_2^5 (x-2)\,dx \]

Step 4: Evaluate the first integral. \[ \int_0^2 (2-x)\,dx = \left[ 2x-\frac{x^2}{2} \right]_0^2 \] Substituting limits: \[ = \left(4-2\right)-0 \] \[ =2 \]

Step 5: Evaluate the second integral. \[ \int_2^5 (x-2)\,dx = \left[ \frac{x^2}{2}-2x \right]_2^5 \] At \(x=5\): \[ \frac{25}{2}-10 = \frac52 \] At \(x=2\): \[ 2-4=-2 \] Thus, \[ \frac52-(-2) = \frac52+\frac42 = \frac92 \]

Step 6: Find the total area. \[ \text{Area} = 2+\frac92 \] \[ = \frac42+\frac92 \] \[ = \frac{13}{2} \] \[ =6.5 \] Hence, \[ \boxed{6.5\ \text{sq units}} \]