Let f be a real valued continuous function defined on the positive real axis such that g(x) = ∫_0^x

Q: Wait, why do we differentiate $g(x^3)$? Can't we just differentiate $g(x)$?

Great question! We differentiate $g(x^3)$ because that is the information provided. If you differentiate $g(x^3)$ with respect to $x$, you get $g'(x^3) \cdot 3x^2$. This is a valid path, but substituting $u = x^3$ first is often cleaner and less prone to algebraic errors. Both paths lead to the same destination!

Q: Why does the lower limit of the integral being zero matter?

It's a gift! When we apply the Leibniz rule, the derivative of the integral $\int_{a(x)}^{b(x)} f(t) dt$ is $f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$. Since our lower limit is a constant (zero), its derivative is zero, which simplifies our life significantly. It's the little things that make the math elegant.

Q: Is it always safe to divide by $u$?

In this problem, yes! We are given that $f$ is defined on the positive real axis, so $u > 0$. Division by zero is the cardinal sin of mathematics, but here, we are on safe ground.

Question

Let f be a real valued continuous function defined on the positive real axis such that $g(x)=\int_{0}^{x}tf(t)dt$. If $g(x^{3})=x^{6}+x^{7}$ then value of $\sum_{r=1}^{15}f(r^{3})$ is:

Options:

(A) 270

(B) 340

(C) 320

(D) 310

Sigma Percentile · Accepted Answer

### The Beauty of the Functional Integral

Imagine standing on the edge of a mathematical cliff, looking down at a function $f$ that is hidden behind an integral. The problem provides the definition $g(x) = \int_{0}^{x} t f(t) dt$ and the functional relationship $g(x^3) = x^6 + x^7$.

It looks intimidating, but every complex problem in JEE Advanced is just a series of small, logical steps waiting to be taken. Let's embark on this journey together.

### Phase 1: The Leibniz Key

Whenever you see a variable in the upper limit of an integral, your brain should immediately trigger the **Newton-Leibniz rule**. It is the master key that unlocks the integrand.

We differentiate both sides of $g(x) = \int_{0}^{x} t f(t) dt$ with respect to $x$. The derivative of the integral $\int_{0}^{x} t f(t) dt$ is simply $x f(x)$.

Thus, we find the beautiful, simple relation:
$$g'(x) = x f(x)$$
This is our primary tool.

### Phase 2: The Substitution Dance

Now, we look at the second piece of the puzzle: $g(x^3) = x^6 + x^7$. We need $g(x)$ to connect it with our derivative, but we currently have $g(x^3)$.

The solution is a clever substitution. Let $u = x^3$, which implies $x = u^{1/3}$.

Substituting this into our equation, we get:
$$g(u) = (u^{1/3})^6 + (u^{1/3})^7$$
This simplifies to $g(u) = u^2 + u^{7/3}$. Now, we have a clear expression for $g(u)$.

### Phase 3: The Derivative and the Link

With $g(u) = u^2 + u^{7/3}$, we differentiate with respect to $u$ to find $g'(u)$. Using the power rule, the derivative of $u^2$ is $2u$, and the derivative of $u^{7/3}$ is $\frac{7}{3} u^{4/3}$.

So, we obtain:
$$g'(u) = 2u + \frac{7}{3} u^{4/3}$$

Recall our earlier relation: $g'(x) = x f(x)$. Replacing $x$ with $u$, we have $g'(u) = u f(u)$.

Equating our two expressions for $g'(u)$, we get:
$$u f(u) = 2u + \frac{7}{3} u^{4/3}$$
Dividing by $u$ (which is safe since $u > 0$), we isolate $f(u)$:
$$f(u) = 2 + \frac{7}{3} u^{1/3}$$

### Phase 4: The Final Summation

The question asks for $\sum_{r=1}^{15} f(r^3)$. Substituting $u = r^3$ into our expression for $f(u)$, we get:
$$f(r^3) = 2 + \frac{7}{3} (r^3)^{1/3} = 2 + \frac{7}{3} r$$
The fractional power has vanished, leaving us with a simple linear expression.

Now, we calculate the sum:
$$\sum_{r=1}^{15} \left( 2 + \frac{7}{3} r ight) = \sum_{r=1}^{15} 2 + \frac{7}{3} \sum_{r=1}^{15} r$$

The first part is $2 	imes 15 = 30$. The second part is:
$$\frac{7}{3} 	imes \frac{15 	imes 16}{2} = \frac{7}{3} 	imes 120 = 7 	imes 40 = 280$$

Adding them together, $30 + 280 = 310$. We have arrived at the final answer: **310**.

Animated Solution for Mathematics - Definite Integration: Let f be a real valued continuous function defined on the positive real axis such that $g (x) = \int_{0}^{x} t f (t) d t$ . If $g (x^{3}) = x^{6} + x^{7}$ then value of $\sum_{r = 1}^{15} f (r^{3})$ is:

Select Answer:

Visualized Solution

Identify the Integral Equation

Apply Newton-Leibniz Rule

Analyze the Given Functional Form

Substitution for Simplification

Differentiate $g (u)$

Link $g^{'} (u)$ and $f (u)$

Solve for $f (u)$

Find $f (r^{3})$

Set up the Summation

Calculate the Constant Sum

Calculate the Sum of Natural Numbers

Final Calculation

The Sigma Insight: Newton-Leibniz & Reduction Formulas

The Beauty of the Functional Integral

Phase 1

Phase 2

Phase 3

Phase 4

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