Animated Solution for Mathematics - Definite Integration: Let $f(x)=x-[x]$, for every real number $x$, where $[x]$ is the integral part of $x$. Then ${\int }_{-1}^{1}f(x)dx$ is

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that seems simple on the surface but hides a beautiful, rhythmic structure.

We are looking at the function $f(x)=x-[x]$. If you have spent any time in the world of functions, you might recognize this as the fractional part function, often denoted as $\{x\}$.

Imagine a sawtooth wave that climbs linearly from $0$ to $1$, then suddenly drops back to $0$ the moment it hits an integer. It is a pattern that repeats infinitely. Our mission is to find the area under this curve from $x=-1$ to $x=1$.

Calculus is often about breaking complex problems into manageable pieces. We are asked to evaluate:

Thanks to the linearity property of integrals, we do not have to struggle with the combined function. We can split this into two distinct, simpler integrals:

Now, we have two separate missions. Let us tackle them one by one.

First, consider ${\int }_{-1}^{1}xdx$. Before you rush to find the antiderivative, pause and look at the geometry.

The function $g(x)=x$ is an odd function, meaning $g(-x)=-g(x)$. When we integrate an odd function over a symmetric interval like $[-1,1]$, the area below the x-axis from $-1$ to $0$ is exactly equal in magnitude but opposite in sign to the area above the x-axis from $0$ to $1$.

They cancel each other out perfectly. Thus:

This is a powerful shortcut that saves us time and mental energy.

Now, we face the second part: ${\int }_{-1}^1[x]dx$. This is where many students stumble.

The greatest integer function $[x]$ is not continuous; it jumps at every integer. To integrate it, we must respect these jumps. We split the interval $[-1,1]$ at the point of discontinuity, $x=0$.

This gives us two intervals: $[-1,0)$ and $[0,1)$. In the interval $[-1,0)$, the greatest integer value is always $-1$. In the interval $[0,1)$, the greatest integer value is always $0$.

Now, the integral becomes:

The first part is simply $-1$ multiplied by the length of the interval, which is $1$, giving us $-1$. The second part is the integral of $0$, which is $0$. Adding these together, the total value of the second integral is $-1$.

We are almost at the finish line! Recall our original split: $I={\int }_{-1}^{1}xdx-{\int }_{-1}^1[x]dx$.

We found the first part to be $0$ and the second part to be $-1$. Substituting these back, we get:

It is elegant, isn't it? The complexity vanishes when you apply the right logic.

To truly own this concept, let us look at it geometrically. The function $f(x)=x-[x]$ creates a series of right-angled triangles.

Between $-1$ and $0$, we have a triangle with base $1$ and height $1$. Between $0$ and $1$, we have another identical triangle.

The area of one triangle is:

Since we have two such triangles, the total area is $\frac{1}{2}+\frac{1}{2}=1$. Our analytical result matches our geometric intuition perfectly.

Keep practicing this habit of verifying your calculus with geometry—it is the hallmark of a true physicist! The final answer is 1.