Let f : (0, ∞) → R and F(x) = ∫(0 to x) t f(t) dt. If F(x²) = x⁴ + x⁵, then Σ(r=1 to 12) f(r²) is eq

Q: Wait, why do we substitute $x^2 = u$ instead of just differentiating $F(x^2)$ with respect to $x$?

You absolutely could differentiate with respect to $x$ using the chain rule! If you differentiate $F(x^2) = x^4 + x^5$ with respect to $x$, you get $F'(x^2) \cdot 2x = 4x^3 + 5x^4$. Since $F'(t) = t f(t)$, then $F'(x^2) = x^2 f(x^2)$. Substituting this in gives $x^2 f(x^2) \cdot 2x = 4x^3 + 5x^4$, which simplifies to $2x^3 f(x^2) = 4x^3 + 5x^4$. Dividing by $2x^3$ yields $f(x^2) = 2 + \frac{5}{2}x$. Both methods are mathematically sound; substitution just keeps the algebra cleaner!

Q: Why is the derivative of $\int_{0}^{x} t f(t) dt$ equal to $x f(x)$?

This is the Fundamental Theorem of Calculus, specifically the Leibniz Rule. When you differentiate an integral with respect to its upper limit $x$, the integrand $g(t)$ simply becomes $g(x)$. Here, our integrand is $t f(t)$. So, $\frac{d}{dx} \int_{0}^{x} t f(t) dt = x f(x)$. It is a powerful shortcut that saves us from having to perform the integration itself!

Q: Why can we divide by $t$ when solving for $f(t)$?

Great catch! In general, we must be careful about dividing by variables that could be zero. However, the problem defines $f$ on the domain $(0, \infty)$. Since $t$ is strictly greater than zero, it is perfectly safe to divide by $t$ without worrying about division by zero.

Question

Let $f : (0, \infty) 	o \mathbb{R}$ and $F(x) = \int_{0}^{x} t f(t) dt$. If $F(x^2) = x^4 + x^5$, then $\sum_{r=1}^{12} f(r^2)$ is equal to :

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with a function $f : (0, \infty) 	o \mathbb{R}$ and an integral definition:
$$F(x) = \int_{0}^{x} t f(t) dt$$

We are also given the functional relation:
$$F(x^2) = x^4 + x^5$$

Our objective is to calculate the sum $\sum_{r=1}^{12} f(r^2)$.

### The Master Key of Leibniz

To extract the function $f(t)$ from the integral, we apply the **Fundamental Theorem of Calculus** (Leibniz Rule). Differentiating $F(x)$ with respect to $x$ yields:

$$\frac{d}{dx} F(x) = \frac{d}{dx} \int_{0}^{x} t f(t) dt$$

This simplifies elegantly to:
$$F'(x) = x f(x)$$

This relation serves as our master key, allowing us to determine $f(x)$ once $F'(x)$ is known.

### Simplifying the Landscape

We are given $F(x^2) = x^4 + x^5$. To align this with our derivative formula, we perform the substitution $u = x^2$.

This transforms our functional relation into:
$$F(u) = u^2 + u^{5/2}$$

The landscape is now clear, as we have an explicit expression for $F(u)$ in terms of a simple power function.

### Unveiling the Function

Differentiating $F(u)$ with respect to $u$ using the power rule, we find:
$$F'(u) = 2u + \frac{5}{2}u^{3/2}$$

Recalling our master key $F'(u) = u f(u)$, we equate the expressions:
$$u f(u) = 2u + \frac{5}{2}u^{3/2}$$

Dividing both sides by $u$ (valid since $u > 0$), we arrive at the hidden function:
$$f(u) = 2 + \frac{5}{2}\sqrt{u}$$

### The Final Calculation

We now evaluate the summation $\sum_{r=1}^{12} f(r^2)$. Substituting $u = r^2$ into our derived function, we obtain:
$$f(r^2) = 2 + \frac{5}{2}\sqrt{r^2} = 2 + \frac{5}{2}r$$

The summation becomes:
$$\sum_{r=1}^{12} \left( 2 + \frac{5}{2}r ight) = \sum_{r=1}^{12} 2 + \frac{5}{2} \sum_{r=1}^{12} r$$

The first part is $2 	imes 12 = 24$. For the second part, we use the sum of the first $n$ natural numbers formula, $\frac{n(n+1)}{2}$:
$$\frac{5}{2} 	imes \left( \frac{12 	imes 13}{2} ight) = \frac{5}{2} 	imes 78 = 5 	imes 39 = 195$$

Adding these components together, we reach the final result:
$$24 + 195 = 219$$

The final answer is **219**.

Animated Solution for Mathematics - Definite Integration: Let $f : (0, \infty) \to R$ and $F (x) = \int_{0}^{x} t f (t) d t$ . If $F (x^{2}) = x^{4} + x^{5}$ , then $\sum_{r = 1}^{12} f (r^{2})$ is equal to :

Enter Numerical Value:

Visualized Solution

Problem Statement Analysis

Applying Leibniz Rule

Variable Substitution

Differentiating $F (t)$

Solving for $f (t)$

Evaluating $f (r^{2})$

Summation Setup

Sum of Constants

Sum of Linear Terms

Final Calculation

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Master Key of Leibniz

Simplifying the Landscape

Unveiling the Function

The Final Calculation

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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; } Let f:(0,∞)→R and F(x)=∫0x​tf(t)dt. If F(x2)=x4+x5, then ∑r=112​f(r2) is equal to :