Problem 61 - Calculus Early Transcendentals

Question

Simplify the following expressions. $$\frac{d}{d x} \int_{3}^{x}\left(t^{2}+t+1\right) d t$$

Step-by-Step Solution

Answer

Answer: The simplified expression is \(x^2 + x + 1\).

Step 1: Apply the Fundamental Theorem of Calculus part 1

To find the derivative, we will use the Fundamental Theorem of Calculus part 1, which states that if F is an antiderivative of f on an interval, and f has a continuous integral on the interval, then the derivative of F with respect to x is given by: $$\frac{d}{d x} \int_{a}^{x} f(t) d t = f(x)$$ Here, our function is \(f(t) = t^2 + t + 1\), so we will apply the theorem to our given expression: $$\frac{d}{d x} \int_{3}^{x} \left(t^{2} + t + 1\right) d t = \left(x^2 + x + 1\right)$$

Step 2: Simplify the expression

Now that we have applied the Fundamental Theorem of Calculus part 1, the expression is already in its simplest form: $$\left(x^2 + x + 1\right)$$ So the simplified expression is: $$x^2 + x + 1$$

Key Concepts

Derivative

A derivative signifies the rate at which a function changes at any given point. Essentially, it provides the slope of the tangent line to the function at a particular spot on its curve. This fundamental concept forms the core of differential calculus.
It is expressed as \( \frac{d}{dx} \) of a function \( f(x) \). For a simple function such as \( f(x) = x^2 + x + 1 \), its derivative, or \( \frac{d}{dx}\), would be calculated by finding the rates of change for each power of \( x \).

• The derivative of \( x^2 \) is \( 2x \).
• The derivative of \( x \) is 1.
• For constants, like 1, the derivative is simply zero.

So, the derivative \( \frac{d}{dx}(x^2 + x + 1) \) would converge to \( 2x + 1 \).
Understanding derivatives is critical when analyzing how changes occur in functions, and it serves as a fundamental tool across disciplines from physics to engineering.

Integral

An integral can be considered as the reverse operation of a derivative and is central to integral calculus. It represents the accumulation of quantities and is often visualized as the area under a curve.
In calculus, an integral of a function \( f(t) \) over an interval \([a, b]\) is written as \( \int_{a}^{b} f(t) \, dt \). This can be interpreted as the total accumulation of the function's value as \( t \) moves from \( a \) to \( b \).
For instance, in the expression \( \int_{3}^{x} (t^2 + t + 1) \, dt \), the integral computes the area under the curve of \( f(t) = t^2 + t + 1 \) from \( t = 3 \) to \( t = x \). This process of finding integrals is also known as integration.

• Definite integrals provide actual numerical values representing the area.
• Indefinite integrals include a constant \( C \) and represent a family of functions.

Integrals are immensely valuable in calculating areas, volumes, and other quantities where a simple geometric approach doesn't suffice.

Antiderivative

An antiderivative is a function whose derivative equals the original function. In essence, it is the process of reversing differentiation.
Given a function \( f(x) \), an antiderivative is denoted as \( F(x) \) such that \( F'(x) = f(x) \). Understanding antiderivatives is crucial as it sets the foundation for solving integrals.

• If \( F(x) \) is an antiderivative of \( f(x) \), then \( \int f(x) \, dx = F(x) + C \) for any constant \( C \).
• This "constant of integration" is a vital part of the antiderivative since differentiation of any constant is zero.

In the problem, the function \( t^2 + t + 1 \) is integrated from 3 to \( x \), essentially reversing the process that would have originally derived this polynomial. The Fundamental Theorem of Calculus synthesizes both derivatives and integrals by ensuring that differentiation and integration are inverse operations.
Antiderivatives help in finding solutions to various real-world problems, like computing total growth from a rate function.