If ∫₀¹ (x²¹ + x¹⁴ + x⁷)(2x²¹ + 3x¹⁴ + 6x⁷)^(1/7) dx = (1/l) (11)^(m/n) where l, m, n ∈ ℕ, m and n ar

Q: Wait, how did you know to pick the entire expression $2x^{21} + 3x^{14} + 6x^7$ as $u$? Why not just part of it?

That is a fantastic question! In JEE Advanced problems, we look for the 'derivative relationship.' If you differentiate $2x^{21} + 3x^{14} + 6x^7$, you get $42x^{20} + 42x^{13} + 42x^6$. Notice that this derivative is just a constant multiple of the other part of the integral. If you had chosen only a part of the expression, the derivative would not have matched the remaining term, and you would be stuck with a messy integral. Always aim for the substitution that 'cleans up' the entire integrand.

Q: Why do we have to change the limits of integration? Can't I just integrate in terms of $u$ and then substitute $x$ back in?

You absolutely *can* substitute back, but changing the limits is a much safer and more elegant habit for JEE. When you substitute back, you have to deal with the original expression again, which increases the chance of algebraic errors. By changing the limits (from $0$ to $1$ for $x$, to $0$ to $11$ for $u$), you finish the problem entirely in the $u$-domain. It is cleaner, faster, and less prone to mistakes.

Q: What if the constant factor (the $42$) wasn't there? Would the substitution still work?

If the constant factor wasn't there, the substitution would still work, but you would have to adjust the integral by multiplying and dividing by the necessary constant. In this problem, the $42$ appeared naturally. If it hadn't, you would simply write $dx = \frac{du}{42(x^{20} + x^{13} + x^6)}$ and proceed. The core logic remains: as long as the variable parts match, the constants can always be handled.

Question

If $\int_{0}^{1} (x^{21} + x^{14} + x^7)(2x^{21} + 3x^{14} + 6x^7)^{1/7} dx = \frac{1}{l} (11)^{m/n}$ where $l, m, n \in \mathbb{N}$, $m$ and $n$ are coprime then $l+m+n$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Art of Seeing the Hidden Pattern

Welcome, student. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of high-degree polynomials.

When you see an integral like $I = \int_{0}^{1} (x^{20} + x^{13} + x^6)(2x^{21} + 3x^{14} + 6x^7)^{\frac{1}{7}} dx$, your instinct might be to panic. You might think, 'How on earth do I expand a term raised to the power of $1/7$?'

But in the world of JEE Advanced, we never brute-force our way through. We look for the elegance hidden beneath the surface.

### Phase 1: The Detective Work

Before you write a single line of algebra, I want you to pause. Look at the two components of the integrand. We have a polynomial $(x^{20} + x^{13} + x^6)$ multiplying another polynomial $(2x^{21} + 3x^{14} + 6x^7)$ raised to the power of $1/7$.

Do you see the relationship? Look at the exponents. The exponents in the second bracket are $21, 14,$ and $7$. The exponents in the first bracket are $20, 13,$ and $6$. They are exactly one degree lower!

This is not a coincidence; it is a mathematical invitation. It is the problem telling you, 'I am the derivative of the other part.' Whenever you see this pattern—a function and its derivative sitting side-by-side—you know exactly what to do: **Substitution**.

### Phase 2: The Substitution Strategy

Let us define our new variable. We choose the complex part, the one trapped inside the fractional power, to be our $u$. Let:

$$u = 2x^{21} + 3x^{14} + 6x^7$$

Now, we must find the differential $du$. We differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = 2(21)x^{20} + 3(14)x^{13} + 6(7)x^6$$

Performing the multiplication, we get:

$$\frac{du}{dx} = 42x^{20} + 42x^{13} + 42x^6$$

Look at that! Every single term shares a common factor of $42$. Let us factor it out:

$$du = 42(x^{20} + x^{13} + x^6) dx$$

This is the moment of truth. Our integral contains $(x^{20} + x^{13} + x^6) dx$. We can now replace this entire chunk with $\frac{du}{42}$. The monster has been tamed.

### Phase 3: The Transformation of Limits

Many students forget this step, but we are better than that. We are changing our universe from $x$ to $u$. Therefore, our boundaries must change too.

When $x = 0$, our new variable $u$ is:

$$u = 2(0)^{21} + 3(0)^{14} + 6(0)^7 = 0$$

When $x = 1$, our new variable $u$ is:

$$u = 2(1)^{21} + 3(1)^{14} + 6(1)^7 = 2 + 3 + 6 = 11$$

Our limits of integration are now $0$ to $11$. The integral is becoming incredibly simple.

### Phase 4: The Final Integration

Let us assemble our new, simplified integral:

$$I = \int_{0}^{11} u^{\frac{1}{7}} \frac{du}{42}$$

We pull the constant $\frac{1}{42}$ outside, and we are left with a standard power rule integration:

$$I = \frac{1}{42} \int_{0}^{11} u^{\frac{1}{7}} du$$

Applying the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$, we get:

$$I = \frac{1}{42} \left[ \frac{u^{\frac{1}{7} + 1}}{\frac{1}{7} + 1} ight]_{0}^{11} = \frac{1}{42} \left[ \frac{u^{\frac{8}{7}}}{\frac{8}{7}} ight]_{0}^{11}$$

Dividing by $\frac{8}{7}$ is the same as multiplying by $\frac{7}{8}$:

$$I = \frac{1}{42} \cdot \frac{7}{8} \left[ u^{\frac{8}{7}} ight]_{0}^{11}$$

Since $42 = 6 	imes 7$, the $7$ cancels out, leaving us with $6 	imes 8 = 48$ in the denominator:

$$I = \frac{1}{48} (11)^{\frac{8}{7}}$$

### Conclusion: The Victory

The problem asks us to match this with the form $\frac{1}{l} (11)^{\frac{m}{n}}$. By direct comparison, we see that $l = 48$, $m = 8$, and $n = 7$. We check the condition: are $8$ and $7$ coprime? Yes, they are.

Finally, we calculate the sum:

$$l + m + n = 48 + 8 + 7 = 63$$

And there you have it. What seemed like an impossible mountain was just a series of small, logical steps. Never fear the complexity of an integral; fear only the failure to look for the pattern. Keep practicing, keep questioning, and keep falling in love with the process.

Animated Solution for Mathematics - Definite Integration: If $\int_{0}^{1} (x^{21} + x^{14} + x^{7}) (2 x^{21} + 3 x^{14} + 6 x^{7})^{1/7} d x = \frac{1}{l} (11)^{m / n}$ where $l, m, n \in N$ , $m$ and $n$ are coprime then $l + m + n$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Analyze the Integral

The Substitution Method

Differentiating $u$

Factoring the Derivative

Changing the Limits

The New Integral

Evaluating the Integral

Simplifying the Result

Applying the Limits

Comparing with the Given Form

Final Calculation

The Sigma Insight: Evaluation of Special Integral Forms

The Art of Seeing the Hidden Pattern

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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