Question

For a real number \( x \), define \( \left\lceil x \right\rceil \) to be the smallest integer greater than or equal to \( x \). Then \[ \int_0^1 \int_0^1 \left( \left\lceil x \right\rceil + \left\lceil y \right\rceil + \left| z \right| \right) dx \, dy \, dz = \text{.............}. \]

Hint:

When handling piecewise functions like \( \left\lceil x \right\rceil \), break them down into simple parts and evaluate the integrals piece by piece.

Answer 1

Correct Answer: 2.9 - 3.1

Step 1: Breaking down the expression.
The integrand \( \left\lceil x \right\rceil + \left\lceil y \right\rceil + |z| \) is composed of integer values for \( x \) and \( y \). For the interval \( [0,1] \), \( \left\lceil x \right\rceil \) and \( \left\lceil y \right\rceil \) will both be 1. So the expression simplifies to: \[ 1 + 1 + |z| = 2 + |z|. \]
Step 2: Evaluating the integral.
Now we compute the integral over the range \( [0,1] \) for \( x \) and \( y \), and for \( z \), which also ranges from 0 to 1: \[ \int_0^1 \int_0^1 \left( 2 + |z| \right) dx \, dy \, dz. \] Since \( |z| = z \) for \( z \in [0,1] \), the integral becomes: \[ \int_0^1 \int_0^1 \left( 2 + z \right) dx \, dy \, dz = \int_0^1 \left( 2 + z \right) \, dz = 3. \]