(A) a point of inflection. (B) a point of local maxima (C) a point of local minima. (D) not a critic

Q: Why is $g(1) = 0$?

Think of the integral as the area under the curve. If you integrate from a point to itself, the width of the interval is zero, so the area must be zero. It is a fundamental property of definite integrals!

Q: How do I differentiate $F(x) = \int_1^x t^2 g(t) dt$?

The Leibniz Rule is your best friend here! It tells us that the derivative of an integral with a variable upper limit is simply the integrand evaluated at that limit. So, $F'(x) = x^2 g(x)$.

Q: Why does $F''(1) > 0$ mean local minima?

Think of the second derivative as the 'curvature'. A positive second derivative means the function is concave up, like a bowl. If the slope is zero at that point, the function must be at its lowest point locally, hence a local minimum!

Question

Let a function $f: [0, 5] 	o R$, be continuous, $f(1) = 3$ and $F$ be defined as: $F(x) = \int_1^x t^2 g(t) dt$, where $g(t) = \int_1^t f(u) du$. Then for the function $F$, the point $x = 1$ is

Options:

(A) a point of inflection.

(B) a point of local maxima

(C) a point of local minima.

(D) not a critical point

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a mathematical landscape, looking at a function $F(x)$ defined by two nested integrals. We are given $f: [0, 5] 	o \mathbb{R}$ as a continuous function, with $f(1) = 3$.

Our goal is to understand the behavior of the function defined as:
$$F(x) = \int_1^x t^2 g(t) \, dt$$
where $g(t)$ is defined by the inner integral:
$$g(t) = \int_1^t f(u) \, du$$

### The Inner Sanctum

First, let us focus on the inner function $g(t) = \int_1^t f(u) \, du$. This function acts as an accumulator, gathering the area under $f(u)$ from $1$ to $t$.

When we evaluate $g(1)$, we substitute $t = 1$:
$$g(1) = \int_1^1 f(u) \, du = 0$$
Since the upper and lower limits are identical, the area is zero. This is our first crucial piece of the puzzle: $g(1) = 0$.

Next, we determine the derivative of $g(t)$ using the Fundamental Theorem of Calculus (Newton-Leibniz Rule):
$$g'(t) = \frac{d}{dt} \int_1^t f(u) \, du = f(t)$$
Given that $f(1) = 3$, it follows that $g'(1) = 3$.

### The Leibniz Magic

Now, let us turn our attention to the main function $F(x) = \int_1^x t^2 g(t) \, dt$. To find its critical points, we calculate its first derivative.

Applying the Leibniz Rule, we differentiate with respect to $x$:
$$F'(x) = x^2 g(x)$$

Now, let us check the slope at $x = 1$:
$$F'(1) = (1)^2 \cdot g(1)$$
Since we already discovered that $g(1) = 0$, the entire expression becomes $F'(1) = 0$. This confirms that $x = 1$ is indeed a critical point.

### The Verdict

To determine if this point is a maximum or a minimum, we calculate the second derivative, $F''(x)$. We apply the product rule to $F'(x) = x^2 g(x)$:
$$F''(x) = 2x \cdot g(x) + x^2 \cdot g'(x)$$

Now, let us evaluate this at $x = 1$:
$$F''(1) = 2(1) \cdot g(1) + (1)^2 \cdot g'(1)$$

Substituting our known values, $g(1) = 0$ and $g'(1) = 3$:
$$F''(1) = 2(0) + 1(3) = 3$$

Since $F''(1) = 3$, which is strictly greater than zero, the function is concave upwards at $x = 1$. In the language of calculus, a positive second derivative at a critical point signifies a **local minimum**.

We have successfully navigated the nested integrals and uncovered the nature of the function at $x = 1$. The function $F(x)$ has a **local minimum** at $x = 1$.

Animated Solution for Mathematics - Definite Integration: Let a function $f : [0, 5] \to R$ , be continuous, $f (1) = 3$ and $F$ be defined as: $F (x) = \int_{1}^{x} t^{2} g (t) d t$ , where $g (t) = \int_{1}^{t} f (u) d u$ . Then for the function $F$ , the point $x = 1$ is

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Visualized Solution

Analyzing the Inner Function $g (t)$

Evaluating $g (1)$

Applying Leibniz Rule to $g (t)$

Differentiating the Main Function $F (x)$

Checking for Critical Point at $x = 1$

Finding the Second Derivative $F^{''} (x)$

Evaluating $F^{''} (1)$

Conclusion: Second Derivative Test

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Inner Sanctum

The Leibniz Magic

The Verdict

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Visualized Solution

Analyzing the Inner Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(t)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(1)

Applying Leibniz Rule to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(t)

Differentiating the Main Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F(x)

Checking for Critical Point at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Finding the Second Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F′′(x)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F′′(1)