Question

Let $[x]$ be the greatest integer function. Then $\int_{-4}^{4}(x-[x])dx$ is equal to

• A. 8
• B. 6
• C. 4
• D. 16
• E. 12

Hint:

Math Tip: The graph of the fractional part function $\{x\}$ forms a series of identical right-angled triangles with a base of 1 and a height of 1. The area of one such triangle is $\frac{1}{2} \times 1 \times 1 = \frac{1}{2}$. For an interval length of 8, there are 8 triangles, so the total area is $8 \times \frac{1}{2} = 4$.

Answer 1

The Correct Option is C

Concept:
Calculus - Properties of the Fractional Part Function.
The expression $x - [x]$ represents the fractional part of $x$, denoted as $\{x\}$.
The fractional part function $\{x\}$ is periodic with a fundamental period of $T = 1$.
Step 1: Rewrite the integral using the fractional part function.
Substitute $x - [x]$ with $\{x\}$: $$ \int_{-4}^{4} (x - [x]) dx = \int_{-4}^{4} \{x\} dx $$
Step 2: Apply the property of periodic functions for definite integrals.
For a periodic function $f(x)$ with period $T$, the integral over $nT$ intervals is: $$ \int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx $$ Here, the total interval length is $4 - (-4) = 8$. Since the period $T = 1$, the interval contains exactly $8$ full periods. $$ \int_{-4}^{4} \{x\} dx = 8 \int_{0}^{1} \{x\} dx $$
Step 3: Evaluate the integral over a single period.
In the interval $[0, 1)$, the greatest integer $[x]$ is exactly $0$. Therefore, in this specific interval, $\{x\} = x - 0 = x$. Substitute this into the single-period integral: $$ \int_{0}^{1} \{x\} dx = \int_{0}^{1} x \,dx $$
Step 4: Calculate the final value.
Integrate $x$ and apply the limits: $$ \int_{0}^{1} x \,dx = \left[ \frac{x^2}{2} \right]_{0}^{1} $$ $$ = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} $$ Multiply by the $8$ periods found in Step 2: $$ 8 \times \frac{1}{2} = 4 $$