Problem 27 - Calculus for Biology and Medicine

Question

Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{2 x}^{3}(1+\sin t) d t $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -2(1 + \sin(2x)) \)

Step 1: Understanding Leibniz's Rule

Leibniz's rule helps us differentiate an integral whose limits are functions of the variable with respect to which we differentiate. It states that if \( y = \int_{a(x)}^{b(x)} f(t) \, dt \), then \( \frac{dy}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f(t)}{\partial x} \, dt \). Since \( t \) is independent of \( x \), \( \frac{\partial f(t)}{\partial x} = 0 \).

Step 2: Identify the Limits of Integration

In the given problem, the integral is \( y = \int_{2x}^{3} (1 + \sin t) \, dt \). Here, the upper limit \( b(x) = 3 \) and the lower limit \( a(x) = 2x \).

Step 3: Apply Leibniz's Rule

Using Leibniz's rule, the derivative \( \frac{dy}{dx} \) is given by \(-f(a(x)) \cdot a'(x) \) because \( f(b(x)) \cdot b'(x) = 0 \) (since \( b(x) = 3 \) is a constant and thus \( b'(x) = 0 \)).

Step 4: Evaluate the Function at the Lower Limit

Evaluate the function at the lower limit: \( f(a(x)) = f(2x) = 1 + \sin(2x) \).

Step 5: Compute the Derivative of the Lower Limit

Calculate \( a'(x) \): since \( a(x) = 2x \), \( a'(x) = 2 \).

Step 6: Calculate the Derivative

Using Leibniz's rule, we find \( \frac{dy}{dx} = - (1 + \sin(2x)) \cdot 2 \). Therefore, \( \frac{dy}{dx} = -2(1 + \sin(2x)) \).

Key Concepts

Integration

Integration is one of the core concepts in calculus, which essentially allows us to find the total accumulation of quantities. In the context of continuous functions, it is the reverse process of differentiation. Integrals can help us find areas under curves, evaluate numerous physical properties, and solve various practical problems.
To solve the integral \( y = \int_{2x}^{3} (1 + \sin t) \, dt \), we are dealing with a definite integral. This means we are looking to evaluate the "net" area under the curve from the lower limit \(2x\) to the upper limit \(3\).

• The limits of integration can vary if they are defined as functions of another variable, like \(x\) in our problem.
• Often, solving such integrals directly requires specific techniques like substitution or parts, but we focus here on the application of Leibniz's rule for derivatives.

Understanding the fundamental theorem of calculus is key here, as it bridges the gap between differentiation and integration, illustrating how they are inverse operations.

Differentiation

Differentiation is the process of finding the derivative, which represents the rate of change of a function with respect to one of its variables. It's a fundamental tool in calculus for understanding how functions behave.
For the problem at hand, we're using Leibniz's rule to find the derivative with respect to \(x\) of an integral with variable limits.

• Generally, differentiating a function provides us with a new function that gives the slope of the tangent line to the curve at any given point.
• In our specific problem, the differentiation involves calculating the effect of changing the limits of integration.

Leibniz's rule helps by taking into account these variable bounds, providing a way to correctly determine the derivative without directly solving the integral first. Such techniques are powerful for tackling complex integrals in applied mathematics.

Calculus Problem Solving

Solving calculus problems like the one given requires an organized approach to employ different techniques and rules. In this exercise, we specifically use Leibniz's rule, a tool for differentiation, to tackle an integral with varying limits.

• First, understand the problem and identify the functions involved, especially the limits of integration that depend on the variable \(x\).
• Apply the relevant rules or theorems. In this case, Leibniz's rule allows us to differentiate under the integral sign appropriately.
• Calculate necessary derivatives of any variable limits, and evaluate the functions at these limits.

When applying Leibniz’s rule, always confirm the zero impact of differentiation for terms that do not depend on \(x\), which simplifies calculations. This structured approach helps in managing potentially intimidating calculus problems by breaking them down into simpler steps.