Question

If $ [\cdot] $ denotes the greatest integer function, then evaluate: $$ \int_{-1}^1 \left( x \cdot [1 + \sin \pi x] + 1 \right) dx $$

• A. 1
• B. 2
• C. \( \frac{5}{2} \)
• D. \( \frac{3}{2} \)

Hint:

Break integrals involving the greatest integer function into intervals where the function is constant. Evaluate piecewise.

Answer 1

The Correct Option is C

We are integrating: \[ \int_{-1}^1 \left( x \cdot \left[1 + \sin(\pi x)\right] + 1 \right) dx \] Let’s analyze the function \( f(x) = x [1 + \sin(\pi x)] + 1 \)
Note that: - \( \sin(\pi x) \in [-1, 1] \)
- So \( 1 + \sin(\pi x) \in [0, 2] \)
Now, consider behavior of \( [1 + \sin(\pi x)] \):
Case 1: \( x \in [-1, 0] \)
- \( \sin(\pi x) \in [0, -1] \Rightarrow [1 + \sin(\pi x)] = 0 \) (since \( 1 + \sin(\pi x)<1 \))
- So \( f(x) = x \cdot 0 + 1 = 1 \)
Case 2: \( x \in (0, 1) \)
- \( \sin(\pi x) \in [0, 1] \Rightarrow 1 + \sin(\pi x) \in (1, 2) \Rightarrow [1 + \sin(\pi x)] = 1 \)
- So \( f(x) = x \cdot 1 + 1 = x + 1 \)
At \( x = 0 \): \( \sin(0) = 0 \Rightarrow [1 + 0] = 1 \Rightarrow f(0) = 1 \)
At \( x = 1 \): \( \sin(\pi) = 0 \Rightarrow [1 + 0] = 1 \Rightarrow f(1) = 1 \cdot 1 + 1 = 2 \)
Now integrate: \[ \int_{-1}^0 f(x) dx = \int_{-1}^0 1 dx = 1 \] \[ \int_0^1 f(x) dx = \int_0^1 (x + 1) dx = \left[\frac{x^2}{2} + x\right]_0^1 = \frac{1}{2} + 1 = \frac{3}{2} \] Total: \[ 1 + \frac{3}{2} = \boxed{\frac{5}{2}} \]