Problem 31 - Thomas Calculus

Question

Find \(d y / d x\) in Exercises \(31-36\) $$ y=\int_{0}^{x} \sqrt{1+t^{2}} d t $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \sqrt{1+x^{2}} \)

Step 1: Identify the Function

We have the function related to the problem: \[ y = \int_{0}^{x} \sqrt{1+t^{2}} \, dt \] This tells us that \( y \) is defined as a definite integral from \( 0 \) to \( x \).

Step 2: Recall the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(t) \) is an antiderivative of \( f(t) \), then: \[ \frac{d}{dx}\int_{a}^{x} f(t) \, dt = f(x) \] In our case, the integrand function is \( \sqrt{1+t^{2}} \).

Step 3: Apply the Fundamental Theorem of Calculus

Applying the fundamental theorem, we directly find the derivative of the integral from \( 0 \) to \( x \): \[ \frac{dy}{dx} = \sqrt{1+x^{2}} \] We replace \( t \) with \( x \) to find this expression since we are differentiating with respect to \( x \).

Key Concepts

Understanding Definite Integrals

A definite integral is a fundamental concept in calculus that represents the accumulation of quantities, such as area under a curve, over a specific interval. In mathematical terms, it is denoted as \( \int_{a}^{b} f(t) \, dt \), where \( f(t) \) is the function being integrated over the interval from \( a \) to \( b \).
This can be visualized as calculating the total area between the function \( f(t) \) and the x-axis, from \( t = a \) to \( t = b \).
Definite integrals are useful in a variety of real-world applications, such as physics, engineering, and economics, where they help to compute quantities that accumulate over time or distance.

• They have both upper and lower limits, which define the interval of integration.
• Unlike indefinite integrals, definite integrals provide a specific numerical value.

The definite integral simplifies to a difference of the values of the antiderivative at these points, giving us an important tool in problem-solving.

Grasping the Concept of an Antiderivative

An antiderivative is a function whose derivative yields the original function. When you differentiate an antiderivative, you get back to the original function. This is a key idea behind the Fundamental Theorem of Calculus.
The notation \( F(t) \) is often used to represent an antiderivative of a function \( f(t) \). In other words, if \( F'(t) = f(t) \), then \( F(t) \) is an antiderivative of \( f(t) \).
Antiderivatives are crucial when solving problems related to accumulations, such as finding areas or solving differential equations. They essentially "undo" differentiation.

• An antiderivative may not be unique; it can differ by a constant.
• Finding antiderivatives requires understanding different functions and formulas.

Antiderivatives help in computing definite integrals, as they allow us to evaluate the accumulation from one point to another.

Exploring Differentiation

Differentiation is a process in calculus used to find the derivative of a function. The derivative represents the rate at which a function changes with respect to a variable, often symbolized as \( \frac{dy}{dx} \) for a function \( y = f(x) \).
Derivatives can provide valuable insights into the behavior of a function, including identifying maximum and minimum points, understanding how variables change relative to each other, and modeling rates of change in applications.

• The process involves finding the slope of the tangent line to the curve at any given point.
• Differentiation is a cornerstone of calculus and has extensive uses in fields like physics, engineering, and economics.

The Fundamental Theorem of Calculus ties definitely into differentiation by linking it to antiderivatives. It essentially shows how differentiation and integration are inverse processes.
In the given exercise, differentiation is used to find the change of the integral function \( y = \int_{0}^{x} \sqrt{1+t^{2}} \ dt \), effectively applying the Fundamental Theorem to find \( \frac{dy}{dx} = \sqrt{1+x^{2}} \).