Problem 33 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = 1 + 7x^2\)

Step 1: Identify the Limits and the Function

The given function is an integral with variable limits. We have the lower limit as \(x\) and the upper limit as \(2x\). The integrand is \(1 + t^2\).

Step 2: Write Down Leibniz's Rule

Leibniz's rule for differentiating under the integral sign with variable limits \(a(x)\) and \(b(x)\) is given by:\[\frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t) \, dt \right) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)\]

Step 3: Determine the Derivatives of the Upper and Lower Limits

The upper limit is \(b(x) = 2x\), so \(b'(x) = 2\). The lower limit is \(a(x) = x\), so \(a'(x) = 1\).

Step 4: Evaluate the Integrand at the Limits

Substitute the upper limit \(b(x) = 2x\) into the integrand \(f(t) = 1 + t^2\):\[ f(2x) = 1 + (2x)^2 = 1 + 4x^2 \]Substitute the lower limit \(a(x) = x\) into the integrand:\[ f(x) = 1 + x^2 \]

Step 5: Apply Leibniz's Rule

Using Leibniz's Rule, calculate:\[\frac{dy}{dx} = f(2x) \cdot b'(x) - f(x) \cdot a'(x)\]This becomes:\[\frac{dy}{dx} = (1 + 4x^2) \cdot 2 - (1 + x^2) \cdot 1\]

Step 6: Simplify the Expression

Compute the expression:\[\frac{dy}{dx} = 2(1 + 4x^2) - (1 + x^2) = 2 + 8x^2 - 1 - x^2 = 1 + 7x^2\]

Step 7: Final Answer

Combine and simplify the terms to get the final derivative:\[\frac{dy}{dx} = 1 + 7x^2\]

Key Concepts

Variable Limits Integral

The concept of a variable limits integral plays a crucial role when the bounds of integration depend on a variable. In this case, the integral is not just a static area calculation but instead changes as the limits move. For instance, if you have an integral from \( x \) to \( 2x \), as given in the exercise, the boundaries of the integral depend directly on the variable \( x \). These variable limits impact the overall value of the integral as \( x \) changes.
This is different from a standard integral with constant limits, where the solution is a fixed value. Instead, when dealing with variable limits, the integral is more dynamic, providing a function of the variable within the limits. This dynamic nature is why methods like Leibniz's rule are essential—they allow us to assess how the value of the integral changes concerning the variable, enabling us to calculate the derivative of such integrals.

Differentiate Under the Integral Sign

To differentiate under the integral sign with variable limits, we use Leibniz's Rule.
Leibniz's Rule provides a systematic approach to calculate the derivative of an integral whose limits are functions of a variable. It states:

• First, evaluate the integrand at the upper limit \(b(x)\) and multiply it by the derivative of the upper limit \(b'(x)\).
• Then, do the same for the lower limit \(a(x)\), but we subtract this result.

So, mathematically, this becomes:\[\frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t) \, dt \right) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)\]This rule elegantly simplifies the complexity of working with integrals that vary with \(x\).
When applied to our specific problem, it allows the differentiation of the initial integral, directly handling the changes in both the upper and lower limits with respect to \(x\).

Integral Calculus

Integral calculus is a fundamental branch of calculus focusing on finding the antiderivative, known as integration. It complements differential calculus, which is about calculating rates of change. Integrals allow us to calculate accumulated quantities, such as areas under curves or total accumulated values over a certain range.
Two primary types of integrals exist: definite integrals, which compute take into account specific limits, and indefinite integrals, which represent a family of functions and a constant of integration. In this exercise, we dealt with a definite integral involving variable limits, adding an extra layer of complexity.

• It involved not just integrating but differentiating the result concerning its limit dependencies.
• This makes integral calculus not only important for solving problems of accumulation but also crucial for understanding how those accumulations change, as outlined through Leibniz's Rule.

Thus, understanding and applying the principles of integral calculus allows for the exploration of relationships and interactions in real-world situations, such as computing the changing area expressed by variable limit integrals.