Problem 14 - Calculus for Biology and Medicine

Question

Find \(\frac{d y}{d x} .\) $$ y=\int_{\pi / 4}^{x} \cos ^{2}(u-3) d u $$

Step-by-Step Solution

Answer

The derivative is \(\cos^2(x-3)\).

Step 1: Identify the type of function

We need to differentiate a function defined by a definite integral with a variable upper bound, which is a classic application of the Fundamental Theorem of Calculus.

Step 2: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if a function is defined as an integral with respect to a variable upper bound, the derivative of the function is the integrand evaluated at that upper bound. Specifically, if \(F(x) = \int_{a}^{x} f(u) \, du\), then \(\frac{d F}{d x} = f(x)\). In this problem, the integrand is \(\cos^2(u-3)\).

Step 3: Differentiate the function

According to the theorem, we take the integrand \(\cos^2(u-3)\) and replace \(u\) with \(x\). Therefore, the derivative \(\frac{d y}{d x} = \cos^2(x-3)\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the rate at which a quantity changes. This is most often called the derivative. In practical terms, differentiation helps us understand how a function behaves at any point.Key points about differentiation include:

• The process involves finding the derivative, indicated by symbols like \( rac{d}{dx} \) or \( f'(x) \).
• The derivative tells us the slope of the tangent line to the function at any given point.
• For a function \(y = f(x)\), the derivative \( rac{dy}{dx} \) represents the immediate rate of change of \(y\) with respect to \(x\).

In this exercise, we are differentiating a function that is initially defined in terms of a definite integral. Differentiating such functions often utilizes the power of the Fundamental Theorem of Calculus, which simplifies the process.

Definite Integrals

A definite integral is used to calculate the area under the curve of a function between two points. Unlike indefinite integrals, definite integrals have specific limits of integration. They result in a numerical value rather than a general expression.Characteristics of definite integrals:

• In a definite integral, \( \int_a^b f(u) \, du \), \(a\) is the lower limit, and \(b\) is the upper limit.
• It computes the net area under the function \(f(u)\) between \(a\) and \(b\).
• Definite integrals are used in various applications, including calculating areas, volumes, and other quantities.

In the given exercise, the function \( y = \int_{\pi/4}^{x} \cos^2(u-3) \, du \) represents a definite integral with a variable upper bound, which is crucial for differentiating through the Fundamental Theorem of Calculus.

Variable Upper Bound

The concept of a variable upper bound in an integral is essential for applying the Fundamental Theorem of Calculus. This occurs when the upper limit of the integral is a variable, rather than a fixed number.The implications of having a variable upper bound:

• A variable upper bound means the upper limit of integration changes depending on another variable (commonly \(x\)). This makes it a function of that variable.
• For \( \int_a^{x} f(u) \, du \), differentiating with respect to \(x\) using the Fundamental Theorem of Calculus results in the integrand evaluated at \(x\). The derivative is \( f(x) \).

In our exercise, \( y = \int_{\pi/4}^{x} \cos^2(u-3) \, du \), changing the upper limit from a constant to a variable makes the function fit perfectly into the context of the Fundamental Theorem. It makes it straightforward to find the derivative \( \frac{dy}{dx} = \cos^2(x-3) \).