Problem 16 - Calculus for Biology and Medicine

Question

Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{0}^{2 x-1}\left(t^{3}-2\right) d t $$

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dx} = 2(2x-1)^3\).

Step 1: Understand the Problem

We are asked to differentiate the integral with respect to \(x\). The given integral is a definite integral with a variable upper limit of integration, \(2x-1\).

Step 2: Recall Leibniz's Rule

Leibniz's rule for differentiating under the integral sign states that if \(y(x) = \int_{a(x)}^{b(x)} f(t, x) \, dt\), then \(\frac{dy}{dx} = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x} \, dt\). In our case, \(a(x) = 0\), and \(b(x) = 2x - 1\). The function inside the integral is \(t^3 - 2\), which does not explicitly depend on \(x\).

Step 3: Apply Leibniz's Rule

Since the function \(f(t, x) = t^3 - 2\) does not depend on \(x\), the partial derivative \(\frac{\partial f}{\partial x} = 0\). Thus, we only consider the terms involving the limits. At the upper limit, \(f(b(x), x) = ((2x-1)^3 - 2)\) and \(b'(x) = 2\). At the lower limit, \(a(x) = 0\), so it does not contribute as \(f(a(x), x) = 0\) and \(a'(x) = 0\).

Step 4: Substitute Values into Leibniz's Rule

Since the lower limit function and its derivative contribute nothing, the derivative becomes:\[\frac{dy}{dx} = (2x-1)^3 \cdot 2 - 0 = 2(2x-1)^3.\]

Step 5: Simplify the Expression

The expression \(2(2x-1)^3\) is our final result for \(\frac{dy}{dx}\). To ensure clarity, you could expand \((2x-1)^3\), but it is not necessary for the differentiation process.

Key Concepts

Differentiation

Differentiation is a cornerstone concept in calculus, focusing on finding the rate at which a quantity changes. It's like asking how fast a car is moving at a particular moment. In calculus terms, this is expressed as the derivative of a function.

When we differentiate a function, we are looking for how it varies with respect to another variable, like time or, in our example, the variable \(x\). The derivative is often denoted by \(\frac{dy}{dx}\), representing the change in \(y\) with respect to \(x\).

• The process of differentiation follows a set of rules and techniques, like the power rule, product rule, and chain rule.
• In cases with more complex functions, we use advanced techniques like implicit differentiation or, as in our problem, differentiation under the integral sign with Leibniz's rule.

Mastering differentiation requires practice in applying these methods to a range of functions. It is an essential tool in analysis, physics, engineering, and beyond, helping us to model and predict behaviors and changes.

Integral Calculus

Integral calculus is another fundamental area of calculus. It involves finding the total accumulation of quantities, akin to finding the total distance traveled by a car from its speed over time. The integral of a function gives us this accumulated value.

There are two primary types of integrals:

• Indefinite integral: Represents a family of functions and includes a constant of integration. It's the antiderivative of a function.
• Definite integral: Computes the actual quantity over a specific interval. It's the type of integral we encountered in the problem.

In our problem, the definite integral \(\int_{0}^{2x-1}(t^3 - 2) dt\) involves a variable upper limit. The goal is to derive an expression for \(y\) in terms of \(x\) by considering these limits, which leads us to employ Leibniz's rule for differentiation under the integral sign. Integral calculus, particularly definite integrals, is pivotal in many real-world applications, including calculating areas, volumes, and other cumulative quantities.

Variable Limits of Integration

In calculus, the integration limits determine the boundaries over which the function is evaluated. These limits can either be constants or variables.

When integration limits are variables, as in the case of \(\int_{0}^{2x-1}(t^3 - 2) dt\), the integral's value becomes dependent on these variables. This dependency requires special techniques when differentiating, particularly Leibniz’s rule.

Leibniz's rule helps us differentiate integrals with respect to a parameter, especially when these integrals have variable limits.

• In our problem, the integral has a fixed lower limit \(a(x) = 0\) and a variable upper limit \(b(x) = 2x-1\). The function inside, \(f(t) = t^3 - 2\), remains independent of \(x\).
• Leibniz's rule simplifies the differentiation by considering the rate of change with respect to these limits.

Utilizing this rule effectively requires understanding the relationship between the integral’s limits and the variable of differentiation. By applying it correctly, we can handle more complex scenarios efficiently in both mathematics and applied settings.