Problem 26 - Calculus Single Variable

Question

In each of Exercises \(21-28,\) calculate the derivative of \(F(x)\) with respect to \(x\). $$ F(x)=\int_{-2}^{x} \sin \left(t^{3}\right) d t $$

Step-by-Step Solution

Answer

The derivative is \( F'(x) = \sin(x^3) \).

Step 1: Understand the Problem

We are tasked with finding the derivative of a function defined by an integral. The function is given as \( F(x) = \int_{-2}^{x} \sin(t^3) \, dt \). To find \( F'(x) \), we will apply the Fundamental Theorem of Calculus.

Step 2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). In this case, the function under the integral sign is \( \sin(t^3) \), so by the theorem, \( F'(x) = \sin(x^3) \).

Key Concepts

Integral Calculus

Integral calculus is focused on the concept of integration. Integration is essentially the reverse process of differentiation. While differentiation is about finding the rate at which values change, integration finds the total accumulation of these changes. The process of integrating can be visualized as finding the area under a curve where the curve is given by some function. In the exercise we're considering, the function is expressed as \( F(x) = \int_{-2}^{x} \sin(t^3) \, dt \). This indicates we're summing up tiny strips of areas under the graph of \( \sin(t^3) \) from \(-2\) to \(x\).

• Definite integral: Provides the accumulated value over an interval.
• Indefinite integral: Provides a general formula for the accumulated value without specific limits.

Integral calculus is a powerful tool for solving problems related to areas and accumulated quantities, like the one we solved using the Fundamental Theorem of Calculus.

Derivative

A derivative represents the rate at which a function changes with respect to a variable. Simply put, it tells us how fast something is changing at any given point. In the case of the exercise, finding the derivative of the function \( F(x) \) involves understanding how \( F(x) \) changes as \(x\) changes. This boils down to applying the Fundamental Theorem of Calculus. We begin with \( F(x) = \int_{-2}^{x} \sin(t^3) \, dt \), which describes a function in terms of an integral. By determining the derivative \( F'(x) \), we are essentially looking for \( f(x) = \sin(x^3) \), the function under the integral sign evaluated at \( x \).

• Derivatives are foundational for calculus, describing how quantities change.
• They are used in a variety of fields to understand growth, decay, and other rates of change.

Using derivatives, we gain powerful insights about the behavior and characteristics of various functions.

Definite Integral

A definite integral is a type of integral that computes the net area between the curve of a function and the x-axis over a specific interval. It gives an exact numerical value, rather than a general formula. In the given problem, we have a definite integral which is represented as: \[ F(x) = \int_{-2}^{x} \sin(t^3) \, dt \] This integral calculates the total accumulated area under the function \( \sin(t^3) \) from \( t = -2 \) to \( t = x \). Since we're dealing with a definite integral with one of its bounds as a variable \( x \), the Fundamental Theorem of Calculus helps us easily differentiate this and find \( F'(x) = \sin(x^3) \).

• Definite integrals have a significant role in applications involving total quantities between specified limits.
• They are crucial in fields requiring precise calculations of accumulated values.

Understanding definite integrals enables us to handle and solve real-world problems involving accumulation and aggregate sums.