Problem 45 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=\int_{0}^{x} \sqrt{1+t^{2}} d t$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Relation

We are given that \( y = \int_{0}^{x} \sqrt{1 + t^2} \, dt \). Notice that this expression represents a definite integral with an upper limit \( x \) and a constant lower limit \( 0 \). We will be differentiating this with respect to \( x \).

Step 2: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(t) \) is an antiderivative of \( f(t) \), then \( \frac{d}{dx} \left( \int_{a}^{x} f(t) \, dt \right) = f(x) \). Here, \( f(t) = \sqrt{1+t^2} \).

Step 3: Differentiate the Integral

Based on the Fundamental Theorem of Calculus, differentiate to get \( \frac{dy}{dx} = \sqrt{1+x^2} \), because \( x \) is the variable upper limit of the integral.

Key Concepts

Definite Integral

A definite integral is a way to find the accumulation of a certain quantity, like area under a curve. It involves two limits, a lower limit and an upper limit, defining a specific interval on the x-axis. For example, \[ y = \int_{0}^{x} \sqrt{1+t^{2}} \, dt \]has lower limit 0 and upper limit \(x\).

• The integral gives a numerical value based on the function and its limits.
• The process computes the net area between the x-axis and the function, from the lower limit to the upper limit.
• In our case, the integral of \(\sqrt{1+t^{2}}\) from 0 to \(x\) finds the area under this curve as \(t\) goes from 0 to \(x\).

When you differentiate a definite integral with the upper limit as a variable, you apply the fundamental theorem standing at mathematical crossroads between integration and differentiation.

Differentiation

Differentiation is the process of finding the derivative, which is a measure of how a function changes as its input changes. If you differentiate a function, it tells you how much that function's output will change for a small change in its input.

• The derivative shows the rate of change or the slope of the function at any point.
• It is foundational in calculus and helps solve real-world problems involving rates of change.

In the context of the provided problem, we differentiated the integral using the fundamental theorem of calculus. This allowed us to find \(\frac{dy}{dx}\), which in this scenario measures how much the area under the curve \(\sqrt{1+t^{2}}\) changes as \(x\) changes.In summary, differentiation plays a crucial role in understanding how quantities change, and it is instrumental in solving the exercise by finding \(\sqrt{1+x^{2}}\) as the result.

Antiderivative

An antiderivative is a function whose derivative is the original function given. Consider it as reverse engineering differentiation. It helps in finding indefinite integrals, the general form of integration without limits.

• Finding an antiderivative is essential for calculating both definite and indefinite integrals.
• If \( F(t) \) is an antiderivative of \( f(t) \), then \( \frac{d}{dt}F(t) = f(t) \).

Here, we use the concept implicitly by applying the fundamental theorem of calculus. This theorem shows that the derivative of the integral of a function returns the original function, like finding \( \sqrt{1+x^{2}} \) by differentiating.In essence, the antiderivative helps transition from derivatives back to original functions. It is a vital part of understanding and executing integration and differentiation in calculus.