Question

The value of $\int_{0}^{3}x^{2}[x]dx$ is equal to

• A. 18
• B. 15
• C. 12
• D. 10
• E. 8

Hint:

Logic Tip: Always break integrals containing step functions or absolute values at their critical "jump" points. Once broken down, the step function simply becomes a constant multiplier for that specific interval.

Answer 1

The Correct Option is B

Concept:
The greatest integer function (or floor function), denoted by $[x]$, outputs the largest integer less than or equal to $x$. Since its value jumps at every integer, a definite integral containing $[x]$ must be split into continuous intervals between these integers.
Step 1: Split the integral at integer boundaries.
The limits of integration are from $0$ to $3$. We split the integral at $x=1$ and $x=2$: $$\int_{0}^{3} x^2[x] dx = \int_{0}^{1} x^2[x] dx + \int_{1}^{2} x^2[x] dx + \int_{2}^{3} x^2[x] dx$$
Step 2: Evaluate $[x]$ within each interval.
For $x \in [0, 1)$, $[x] = 0$. For $x \in [1, 2)$, $[x] = 1$. For $x \in [2, 3)$, $[x] = 2$. (Note: The single upper boundary point $x=3$ does not affect the area under the curve). Substitute these constants into the respective integrals: $$= \int_{0}^{1} x^2(0) dx + \int_{1}^{2} x^2(1) dx + \int_{2}^{3} x^2(2) dx$$ $$= 0 + \int_{1}^{2} x^2 dx + 2\int_{2}^{3} x^2 dx$$
Step 3: Integrate the polynomial terms.
The antiderivative of $x^2$ is $\frac{x^3}{3}$. Apply this to the remaining integrals: $$= \left[ \frac{x^3}{3} \right]_{1}^{2} + 2 \left[ \frac{x^3}{3} \right]_{2}^{3}$$
Step 4: Apply the limits of integration.
$$= \left( \frac{2^3}{3} - \frac{1^3}{3} \right) + 2 \left( \frac{3^3}{3} - \frac{2^3}{3} \right)$$ $$= \left( \frac{8}{3} - \frac{1}{3} \right) + 2 \left( \frac{27}{3} - \frac{8}{3} \right)$$ $$= \left( \frac{7}{3} \right) + 2 \left( \frac{19}{3} \right)$$ $$= \frac{7}{3} + \frac{38}{3} = \frac{45}{3} = 15$$