If [t] denotes the greatest integer ≤ t, then the value of 3(e-1)/e ∫₁² x² e^([x] + [x³]) dx is : (1

Q: Wait, why is $[x] = 1$ for $x \in [1, 2)$? Shouldn't it be $2$ at the endpoint?

That is a fantastic observation! The greatest integer function $[x]$ is defined as the largest integer less than or equal to $x$. For any $x$ in the interval $[1, 2)$, the value is always $1$. Even as $x$ approaches $2$, it never reaches it, so the value remains $1$. At exactly $x=2$, the value would be $2$, but in a definite integral, a single point does not change the area under the curve. Thus, we can safely treat $[x]$ as $1$ throughout the entire interval.

Q: Why did we change the limits of integration to $1$ and $8$?

When we perform the substitution $t = x^3$, we are essentially changing our coordinate system. We are no longer integrating with respect to $x$; we are integrating with respect to $t$. Therefore, we must map the boundaries of our original region to the new coordinate system. When $x=1$, $t = 1^3 = 1$. When $x=2$, $t = 2^3 = 8$. If we kept the limits as $1$ and $2$, we would be calculating the wrong area entirely!

Q: Can't we just integrate $e^{[t]}$ as a single function?

I wish we could! But $e^{[t]}$ is not a continuous function; it is a piecewise constant function. It jumps at every integer value of $t$. Because the derivative of a step function is zero (or undefined at the jumps), we cannot use standard integration rules like the power rule or exponential integration on the whole interval at once. We must break the 'staircase' into individual steps where the function is constant, integrate each step, and then sum them up.

Question

If $[t]$ denotes the greatest integer $\le t$, then the value of $\frac{3(e-1)}{e} \int_{1}^{2} x^2 e^{[x] + [x^3]} dx$ is :

Options:

(A) $e^9 - e$

(B) $e^8 - e$

(C) $e^7 - e$

(D) $e^8 - 1$

Sigma Percentile · Accepted Answer

### The Beauty of the Staircase: Taming the Greatest Integer Function

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an integral; we are peeling back the layers of a mathematical onion.

We are looking at the expression:

$$\frac{3(e-1)}{e} \int_{1}^{2} x^2 e^{[x] + [x^3]} dx$$

At first glance, this looks intimidating. The greatest integer function, denoted by the square brackets, often strikes fear into the hearts of students because it breaks the smoothness of calculus. But fear not! We are going to dismantle this problem piece by piece.

### Phase 1: The First Simplification

Let us look at the interval of integration: $x \in [1, 2]$. In this domain, the greatest integer function $[x]$ is remarkably well-behaved.

By definition, $[x]$ is the largest integer less than or equal to $x$. For any value of $x$ starting from $1$ up to, but not including $2$, the greatest integer is simply $1$.

This is our first victory! We can replace $[x]$ with $1$ immediately. Our integral now looks like this:

$$I = \int_{1}^{2} x^2 e^{1 + [x^3]} dx$$

Using the laws of exponents, we can pull that constant $e^1$ out of the integral, leaving us with:

$$I = e \int_{1}^{2} x^2 e^{[x^3]} dx$$

The mountain is already looking smaller, isn't it?

### Phase 2: The Art of Substitution

Now, we face the term $[x^3]$. This is the heart of the problem. Integrating $[x^3]$ directly is impossible because it changes values constantly.

We need a change of perspective. Let us introduce a substitution: $t = x^3$. When we differentiate both sides, we get $dt = 3x^2 dx$, which implies $x^2 dx = \frac{1}{3} dt$.

Look at how perfectly the $x^2$ term in our integral matches this! This is not a coincidence; it is the elegance of a well-designed problem.

We must also update our limits. When $x=1$, $t=1^3=1$. When $x=2$, $t=2^3=8$. Our integral transforms into:

$$I = \frac{e}{3} \int_{1}^{8} e^{[t]} dt$$

We have successfully moved from a complex variable $x$ to a much simpler variable $t$.

### Phase 3: The Summation of Steps

We are now staring at $\int_{1}^{8} e^{[t]} dt$. As we discussed, $[t]$ is a staircase function. It stays constant between integers.

To integrate this, we must break the path from $1$ to $8$ into unit intervals: $[1, 2), [2, 3), [3, 4), [4, 5), [5, 6), [6, 7),$ and $[7, 8)$. In each interval $[n, n+1)$, the value of $[t]$ is simply $n$.

Thus, the integral becomes a sum of integrals:

$$\int_{1}^{2} e^1 dt + \int_{2}^{3} e^2 dt + \dots + \int_{7}^{8} e^7 dt$$

Since the length of each interval is $1$, the integral of a constant $e^n$ over an interval of length $1$ is just $e^n$. We are left with the sum:

$$e^1 + e^2 + e^3 + e^4 + e^5 + e^6 + e^7$$

This is a classic Geometric Progression (GP) where the first term $a = e$, the common ratio $r = e$, and the number of terms $n = 7$. Using the sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$, we get:

$$\frac{e(e^7 - 1)}{e - 1}$$

### Phase 4: The Grand Finale

We are at the finish line. Let us assemble our pieces. Our integral $I$ is:

$$I = \frac{e}{3} \cdot \frac{e(e^7 - 1)}{e - 1}$$

The original question asks us to evaluate $\frac{3(e-1)}{e} \cdot I$. Substituting our value for $I$, we get:

$$\frac{3(e-1)}{e} \cdot \frac{e^2(e^7 - 1)}{3(e - 1)}$$

Now, watch the magic happen. The $3$ cancels with $3$, the $(e-1)$ cancels with $(e-1)$, and one $e$ cancels with $e$.

We are left with $e(e^7 - 1)$, which expands to **$e^8 - e$**.

We have conquered the integral! Remember, in JEE Advanced, the complexity is often just a mask for a beautiful, symmetrical structure waiting to be revealed. Keep practicing, keep visualizing, and keep falling in love with the process.

Animated Solution for Mathematics - Definite Integration: If $[t]$ denotes the greatest integer $\leq t$ , then the value of $\frac{3 ( e - 1 )}{e} \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$ is :

Select Answer:

Visualized Solution

Analyze the Interval for $[x]$

Simplify the Integral Expression

Apply Substitution $x^{3} = t$

Change the Limits of Integration

Break Integral into Unit Intervals

Evaluate the Sum of Integrals

Sum the Geometric Progression (GP)

Final Calculation and Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of the Staircase

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If [t] denotes the greatest integer ≤t, then the value of e3(e−1)​∫12​x2e[x]+[x3]dx is :