Question

The area (in square units) bounded by the line \(y = x\), the X-axis and the lines \(x = -2\) and \(x = 4\) is...

• A. \(6\)
• B. \(\frac{15}{2}\)
• C. \(\frac{17}{2}\)
• D. \(10\)

Hint:

The region lies on both sides of the x-axis, so add the two triangles.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The line \(y = x\) is below the x-axis for \(x < 0\) and above it for \(x > 0\). Area must be counted as a positive quantity for each part.

Step 2: Key Formula or Approach:
Area \(= \int_{-2}^{0}|x|\,dx + \int_0^4 x\,dx\).

Step 3: Detailed Explanation:
For \(x\) from \(-2\) to \(0\): the triangle has base \(2\) and height \(2\), so area \(= \frac12 \times 2 \times 2 = 2\).
For \(x\) from \(0\) to \(4\): base \(4\), height \(4\), area \(= \frac12 \times 4 \times 4 = 8\).
\[ \text{Total} = 2 + 8 = 10 \]
If the integral \(\int_{-2}^4 x\,dx = 6\) is taken without the modulus, we get option A. That is wrong because the negative part cancels the area.

Final Answer:
The area is \(10\) square units, option (D). \[ \boxed{10} \]