Question

The area bounded by the lines $y = x$, $x = -1$, $x = 2$ and the $X$-axis is

• A. $\frac{1}{2}\ \text{sq. units}$
• B. $\frac{3}{2}\ \text{sq. units}$
• C. $\frac{5}{2}\ \text{sq. units}$
• D. $\frac{7}{4}\ \text{sq. units}$

Hint:

Since $y = x$ forms standard straight lines, you can solve this visually using basic triangle geometry instead of calculus! The graph creates two separate right-angled triangles: a left triangle with base $1$ and height $1$ ($\text{Area} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$), and a right triangle with base $2$ and height $2$ ($\text{Area} = \frac{1}{2} \times 2 \times 2 = 2$). Adding them gives $\frac{1}{2} + 2 = \frac{5}{2}$ instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the total area bounded by the linear function $y = x$, vertical lines $x = -1$ and $x = 2$, and the horizontal $X$-axis ($y = 0$).

Step 2: Key Formula or Approach:
The area bounded by a curve $y = f(x)$ from $x = a$ to $x = b$ is given by the definite integral of its absolute value: $$\text{Area} = \int_{a}^{b} |f(x)| \, dx$$ Since the line $y = x$ crosses beneath the $X$-axis in the negative region ($x < 0$), we must split the integral at its root boundary $x = 0$ to prevent negative geometric cancellation.

Step 3: Detailed Explanation:
Partition the total integration interval $[-1, 2]$ into negative and positive segments: $$\text{Area} = \int_{-1}^{0} (-x) \, dx + \int_{0}^{2} x \, dx$$ Evaluate each definite integral separately using the power integration rule: $$\int_{-1}^{0} (-x) \, dx = \left[ -\frac{x^2}{2} \right]_{-1}^{0} = 0 - \left( -\frac{(-1)^2}{2} \right) = \frac{1}{2}$$ $$\int_{0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - 0 = \frac{4}{2} = 2$$ Sum these regional absolute areas together to find the net total area: $$\text{Total Area} = \frac{1}{2} + 2 = \frac{1 + 4}{2} = \frac{5}{2}\ \text{sq. units}$$ This matches the value in option (C).

Step 4: Final Answer:
The area bounded by the lines is $\frac{5}{2}\ \text{sq. units}$, which corresponds to option (C).